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PHP - Update Data With Dropdown Menu?
I'm not sure what to search for or if this can be done...
Let's say I have a mysql table named "genre". Now I have "Male" and "Female" in a dropdown menu like this in a form: <select name="genre"> <option>Male</option> <option>Female</option> </select> How would you display, for example, the option Female in a update.php file if that's the genre stored in the mysql database when you fetch the results? Similar TutorialsHi, I'm trying to get data from one field in a table (database). But I get undesirable result: Here is my code -> <?php $result2 = mysql_query("SELECT DISTINCT theme FROM mytable ") or die(mysql_error()); while($row2 = mysql_fetch_array( $result2 )) { ?> <form method="post" action='<?php echo $_SERVER["PHP_SELF"]; ?>'> <select name='themes'"> <?php $arr= array($row2['theme']); foreach($row2 as $value) { echo "<option value='$value'><b>". $value."</b> </option><br> "; } } ?> The attached image file show the result that I don't wont. (It's not a dropdown). Is there anyone who may help me, I spent a lot of time to find out but I can't. Thanks a lot for your help Hi, I have a MySQL database called "2011_database" that has a table called "2011_list." In that table I have fields, amongst others, called "name" and "district." I need to find way to get the data from the table and put them into a drop down list on other PHP page. But they need to be listed as "name - district" on one line. I am PHP beginner and if I understand it correctly there need to be two references to get all the data in all records, a third reference to merge them together with " - " in between; and what eludes me the most, putting them in a drop down menu. Any help is greatly appreciated Thanks Hi, after banging my head against the wall for a while thinking this would be a simple task, I'm discovering that this is more complicated than I thought. Basically what I have is a link table linking together source_id and subject_id. For each subject there are multiple sources associated with each. I had created a basic listing of sources by subject... no problem. I now need a way of having a form to create an ordered list in a user-specified way. In other words, I can currently order by id or alphabetically (subject name lives on a different table), but I need the option of choosing the order in which they display. I added another row to this table called order_by. No problem again, and I can manage all of this in the database, however I want to create a basic form where I can view sources by subject and then enter a number that I can use for sorting. I started off looping through each of the entries and the database (with a where), and creating a foreach like so (with the subject_id being grabbed via GET from the URL on a previous script) Code: [Select] while($row = mysqli_fetch_array($rs)) { //update row order if (isset($_POST['submit'])) { //get variables, and assign order $subject_id = $_GET['subject_id']; $order_by = $_POST['order_by']; $source_id = $row['source_id']; //echo 'Order by entered as ' . $order_by . '<br />'; foreach ($_POST['order_by'] as $order_by) { $qorder = "UPDATE source_subject set order_by = '$order_by' WHERE source_id = '$source_id' AND subject_id = '$subject_id'"; mysqli_query($dbc, $qorder) or die ('could not insert order'); // echo $subject_id . ', ' . $order_by . ', ' . $source_id; // echo '<br />'; } } else { $subject_id = $_GET['subject_id']; $order_by = $row['order_by']; $source_id = $row['source_id']; } And have the line in the form like so: Code: [Select] echo '<input type="text" id="order_by" name="order_by[]" size="1" value="'. $order_by .'"/> (yes I know I didn't escape the input field... it's all stored in an htaccess protected directory; I will clean it up later once I get it to work) This, of course, results in every source_id getting the same "order_by" no matter what I put into each field. I'm thinking that I need to do some sort of foreach where I go through foreach source_id and have it update the "order_by" field for each one, but I must admit I'm not sure how to go about this (the flaws of being self-taught I suppose; I don't have anyone to go to on this). I'm hoping someone here can help? Thanks a ton in advance Yesterday, I created a topic about how I could update records and I managed to achieve that successfully. Now I have another dilemma. When I have a specific record I want to update, I want to change a category ID of an product (e.g. change it from 1 to 2) but how do I go about doing this? Here is my code thus far: Code: [Select] <?php require_once ('./includes/config.inc.php'); require_once (MYSQL); $id=$_GET['prodID']; $results = mysqli_query($dbc, "SELECT * FROM product WHERE prodID=".$_GET['prodID'].""); $row = mysqli_fetch_assoc($results); ?> <form action="" method='POST'> Product ID: <input type="text" value="<?php echo $row['prodID'];?>" name="prodID" /> <br /> Product: <input type="text" value="<?php echo $row['product'] ;?>" name="product" /> <br /> Product Description: <input type="text" value="<?php echo $row['prod_descr'] ;?>" name="prod_descr" /> <br /> Category: <select name="category"> <option value="<?php echo $row['catID'];?>"></option> </select> Price: <input type="text" value="<?php echo $row['price'] ;?>" name="price" /> <br /> In Stock: <input type="text" value="<?php echo $row['stock'] ;?>" name="stock" /> <br /> <br /><input type="submit" value="save" name="save"> </form> <?php if(isset($_POST['save'])) { $id = $_POST['prodID']; $product = $_POST['product']; $descr = $_POST['prod_descr']; $price = $_POST['price']; $stock = $_POST['stock']; // Update data $update = mysqli_query($dbc, "UPDATE product SET product='$product', prod_descr='$descr', price='$price', stock='$stock' WHERE prodID=".$_GET['prodID'].""); header( 'Location: update_save.php' ) ; } ?> Hello all,
For the UPDATE portion of my CRUD WebApp what I would like to do is to bring in (and display) the values (of a selected row) from my transaction table.
This is working just fine for all fields which are of the "input type". The problem I'm having is with two fields which are of the "select type" i.e. dropdown listboxes.
For those two fields, I would like to bring in all the valid choices from the respective lookup/master tables, but then have the default/selected value be shown based on what's in the transaction table. The way I have it right now, those two fields are showing (and updating the record with) the very first entry's in the two lookup tables/select query.
The attached picture might make things a little bit clearer. You'll notice in the top screenshot that the first row (which is the one I'm selecting to update) has a "Store Name" = "Super Store" and an "Item Description" = "Old Mill Bagels". Now, when I click the "update" botton and I'm taken to the update screen, the values for those two fields default to the very first entries in the SELECT resultset i.e. "Food Basics" and "BD Cheese Strings". Cricled in green (to the top-left of that screenshot) is the result of an echo that I performed, based on the values that are in the transaction record.
I cannot (for the life of me) figure out how to get those values to be used as default/selected values for the two dropdown's...so that if a user does not touch those two dropdown fields, the values in the transaction table will not be changed.
Your help will be greatly appreciated.
Here's a portion of the FORM code:
<form class="form-horizontal" action="update.php?idnumber=<?php echo $idnumber?>" method="post">
<?php
// Connect to Store_Name (sn) table to get values for dropdown
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT DISTINCT store_name FROM store_master ORDER BY store_name ASC";
$q_sn = $pdo->prepare($sql);
$q_sn->execute();
$count_sn = $q_sn->rowCount();
$result_sn = $q_sn->fetchAll();
Database::disconnect();
// Connect to Item_Description (id) table to get values for dropdown
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT DISTINCT product_name FROM product_master ORDER BY product_name ASC";
$q_id = $pdo->prepare($sql);
$q_id->execute();
$count_id = $q_id->rowCount();
$result_id = $q_id->fetchAll();
Database::disconnect();
foreach($fields AS $field => $attr){
$current_store_name = $values['store_name'];
$current_item_description = $values['item_description'];
//Print the form element for the field.
if ($field == 'store_name')
{
echo $current_store_name;
echo '<br />';
echo $current_item_description;
echo '<div class="control-group">';
echo "<label class='control-label'>{$attr['label']}: </label>";
echo '<div class="controls">';
//Echo the actual input. If the form is being displayed with errors, we'll have a value to fill in from the user's previous submission.
echo '<select class="store-name" name="store_name">';
// echo '<option value="">Please select...</option>';
foreach($result_sn as $row)
{
echo "<option value='" . $row['store_name'] . "'>{$row['store_name']}</option>";
}
// $row['store_name'] = $current_store_name;
echo "</select>";
echo '</div>';
echo '</div>';
}
elseif ($field == 'item_description')
{
echo '<div class="control-group">';
echo "<label class='control-label'>{$attr['label']}: </label>";
echo '<div class="controls">';
echo '<select class="item-desc" name="item_description">';
// echo '<option value="">Please select...</option>';
foreach($result_id as $row)
{
echo "<option alue='" . $row['product_name'] . "'>{$row['product_name']}</option>";
}
echo "</select>";
echo '</div>';
echo '</div>';
}
else
{
echo '<div class="control-group">';
echo "<label class='control-label'>{$attr['label']}: </label>";
echo '<div class="controls">';
//Echo the actual input. If the form is being displayed with errors, we'll have a value to fill in from the user's previous submission.
echo '<input type="text" name="'.$field.'"' . (isset($values[$field]) ? ' value="'.$values[$field].'"' : '') . ' /></label>';
echo '</div>';
echo '</div>';
}
And here's some other declarations/code which I think might be required.
$fields = array(
'store_name' => array('label' => 'Store Name',
'error' => 'Please enter a store name'),
'item_description' => array('label' => 'Item Description',
'error' => 'Please enter an item description'),
'qty_pkg' => array('label' => 'Qty / Pkg',
'error' => 'Please indicate whether it\'s Qty or Pkg'),
'pkg_of' => array('label' => 'Pkg. Of',
'error' => 'Please enter the quantity'),
'price' => array('label' => 'Price',
'error' => 'Please enter the price'),
'flyer_page' => array('label' => 'Flyer Page #',
'error' => 'Please enter the flyer page #'),
'limited_time_sale' => array('label' => 'Limited Time Sale',
'error' => 'Please enter the days for limited-time-sale'),
'nos_to_purchase' => array('label' => 'No(s) to Purchase',
'error' => 'Please enter the No. of items to purchase')
);
...
...
....
{
// If [submit] isn't clicked - and therfore POST array is empty - perform a SELECT query to bring in
// existing values from the table and display, to allow for changes to be made
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT * FROM shoplist where idnumber = ?";
$q = $pdo->prepare($sql);
$q->execute(array($idnumber));
$values = $q->fetch(PDO::FETCH_ASSOC);
if(!$values)
{
$error = 'Invalid ID provided';
}
Database::disconnect();
}
If there's anything I've missed please ask and I'll provide.
Thanks
So i have a <drop down> menu, and a <nav> menu on my left side of the page. I have a problem when i click at the first column of my drop down menu and it explores submenu, the submenu mixes with the nav menu that is under the drop down menu on the left. i could solve it with margin-top of the nav menu but i don't like the empty space beetwen them. I tried with putting overflow:visible; in CSS of my dropdown menu but it is still the same. drop down menu is seen but there is stil seen nav menu under and they are mixed. So basicly i want my dropdown menu to be priority, so when i clicks on my first column (only first column is problem because there under is nav menu the other are open nice) it will explore submenu and the part of nav menu that covers the submenu will be hidden.
Here is the screen shot:
Here is the code of head <dropdown> menu if someone finds the problem.
#menu, #menu ul {
margin: 0;
padding: 0;
list-style: none;
}
#menu {
width: 900px;
margin-top:20px;
margin-left:auto;
margin-right:auto;
border: 1px solid #222;
background-color: #111;
background-image: linear-gradient(#444, #111);
border-radius: 6px;
box-shadow: 0 1px 1px #777;
}
#menu:before,
#menu:after {
content: "";
display: table;
}
#menu:after {
clear: both;
}
#menu {
zoom:1;
}
#menu li {
float: left;
border-right: 1px solid #222;
box-shadow: 1px 0 0 #444;
position: relative;
}
#menu a {
float: left;
padding: 12px 30px;
color: #999;
text-transform: uppercase;
font: bold 12px Arial, Helvetica;
text-decoration: none;
text-shadow: 0 1px 0 #000;
}
#menu li:hover > a {
color: #fafafa;
}
*html #menu li a:hover { /* IE6 only */
color: #fafafa;
}
#menu ul {
margin: 20px 0 0 0;
_margin: 0; /*IE6 only*/
opacity: 0;
visibility: hidden;
position: absolute;
top: 38px;
left: 0;
z-index: 1;
background: #444;
background: linear-gradient(#444, #111);
box-shadow: 0 -1px 0 rgba(255,255,255,.3);
border-radius: 3px;
transition: all .2s ease-in-out;
}
#menu li:hover > ul {
opacity: 1;
visibility: visible;
margin: 0;
}
#menu ul ul {
top: 0;
left: 150px;
margin: 0 0 0 20px;
_margin: 0; /*IE6 only*/
box-shadow: -1px 0 0 rgba(255,255,255,.3);
}
#menu ul li {
float: none;
display: block;
border: 0;
_line-height: 0; /*IE6 only*/
box-shadow: 0 1px 0 #111, 0 2px 0 #666;
}
#menu ul li:last-child {
box-shadow: none;
}
#menu ul a {
padding: 10px;
width: 130px;
_height: 10px; /*IE6 only*/
display: block;
white-space: nowrap;
float: none;
text-transform: none;
}
#menu ul a:hover {
background-color: #0186ba;
background-image: linear-gradient(#04acec, #0186ba);
}
#menu ul li:first-child > a {
border-radius: 3px 3px 0 0;
}
#menu ul li:first-child > a:after {
content: '';
position: absolute;
left: 40px;
top: -6px;
border-left: 6px solid transparent;
border-right: 6px solid transparent;
border-bottom: 6px solid #444;
}
#menu ul ul li:first-child a:after {
left: -6px;
top: 50%;
margin-top: -6px;
border-left: 0;
border-bottom: 6px solid transparent;
border-top: 6px solid transparent;
border-right: 6px solid #3b3b3b;
}
#menu ul li:first-child a:hover:after {
border-bottom-color: #04acec;
}
#menu ul ul li:first-child a:hover:after {
border-right-color: #0299d3;
border-bottom-color: transparent;
}
#menu ul li:last-child > a {
border-radius: 0 0 3px 3px;
}
Here is the code of <nav> menu if someone finds the problem.
#cssmenu {
width:15%;
padding: 0;
margin-top: 50px;
margin-left:auto;
margin right:auto;
float:left;
border: 0;
line-height: 1;
}
#cssmenu ul,
#cssmenu ul li,
#cssmenu ul ul {
list-style: none;
margin: 0;
padding: 0;
}
#cssmenu ul {
position: relative;
z-index: 597;
float: left;
}
#cssmenu ul li {
float: left;
min-height: 1px;
line-height: 1em;
vertical-align: middle;
position: relative;
}
#cssmenu ul li.hover,
#cssmenu ul li:hover {
position: relative;
z-index: 599;
cursor: default;
}
#cssmenu ul ul {
visibility: hidden;
position: absolute;
top: 100%;
left: 0px;
z-index: 598;
width: 100%;
}
#cssmenu ul ul li {
float: none;
}
#cssmenu ul ul ul {
top: -2px;
right: 0;
}
#cssmenu ul li:hover > ul {
visibility: visible;
}
#cssmenu ul ul {
top: 1px;
left: 99%;
}
#cssmenu ul li {
float: none;
}
#cssmenu ul ul {
margin-top: 1px;
}
#cssmenu ul ul li {
font-weight: normal;
}
/* Custom CSS Styles */
#cssmenu {
width: 200px;
background: #333333;
font-family: 'Oxygen Mono', Tahoma, Arial, sans-serif;
zoom: 1;
font-size: 12px;
}
#cssmenu:before {
content: '';
display: block;
}
#cssmenu:after {
content: '';
display: table;
clear: both;
}
#cssmenu a {
display: block;
padding: 15px 20px;
color: #ffffff;
text-decoration: none;
text-transform: uppercase;
}
#cssmenu > ul {
width: 200px;
}
#cssmenu ul ul {
width: 200px;
}
#cssmenu > ul > li > a {
border-right: 4px solid #1b9bff;
color: #ffffff;
}
#cssmenu > ul > li > a:hover {
color: #ffffff;
}
#cssmenu > ul > li.active a {
background: #1b9bff;
}
#cssmenu > ul > li a:hover,
#cssmenu > ul > li:hover a {
background: #1b9bff;
}
#cssmenu li {
position: relative;
}
#cssmenu ul li.has-sub > a:after {
content: '+';
position: absolute;
top: 50%;
right: 15px;
margin-top: -6px;
}
#cssmenu ul ul li.first {
-webkit-border-radius: 0 3px 0 0;
-moz-border-radius: 0 3px 0 0;
border-radius: 0 3px 0 0;
}
#cssmenu ul ul li.last {
-webkit-border-radius: 0 0 3px 0;
-moz-border-radius: 0 0 3px 0;
border-radius: 0 0 3px 0;
border-bottom: 0;
}
#cssmenu ul ul {
-webkit-border-radius: 0 3px 3px 0;
-moz-border-radius: 0 3px 3px 0;
border-radius: 0 3px 3px 0;
}
#cssmenu ul ul {
border: 1px solid #0082e7;
}
#cssmenu ul ul a {
font-size: 12px;
color: #ffffff;
}
#cssmenu ul ul a:hover {
color: #ffffff;
}
#cssmenu ul ul li {
border-bottom: 1px solid #0082e7;
}
#cssmenu ul ul li:hover > a {
background: #4eb1ff;
color: #ffffff;
}
#cssmenu.align-right > ul > li > a {
border-left: 4px solid #1b9bff;
border-right: none;
}
#cssmenu.align-right {
float: right;
}
#cssmenu.align-right li {
text-align: right;
}
#cssmenu.align-right ul li.has-sub > a:before {
content: '+';
position: absolute;
top: 50%;
left: 15px;
margin-top: -6px;
}
#cssmenu.align-right ul li.has-sub > a:after {
content: none;
}
#cssmenu.align-right ul ul {
visibility: hidden;
position: absolute;
top: 0;
left: -100%;
z-index: 598;
width: 100%;
}
#cssmenu.align-right ul ul li.first {
-webkit-border-radius: 3px 0 0 0;
-moz-border-radius: 3px 0 0 0;
border-radius: 3px 0 0 0;
}
#cssmenu.align-right ul ul li.last {
-webkit-border-radius: 0 0 0 3px;
-moz-border-radius: 0 0 0 3px;
border-radius: 0 0 0 3px;
}
#cssmenu.align-right ul ul {
-webkit-border-radius: 3px 0 0 3px;
-moz-border-radius: 3px 0 0 3px;
border-radius: 3px 0 0 3px;
}
Edited by Dorkmind, 26 November 2014 - 05:00 PM. Hi i have a online/offline script that works great no issues at all. In order to change the html, php file to display when offline I need to point it to that file via input field. What i would like is to change the input filed and have a dropdown that list all the folders and subfolders within the the directory where all the templates are located. I have found a php snippet on the forum that does exactly what i want. The issue is now is to remove the input filed in the script and insert the snippet found here and thats where the issue comes. I'm hoping some guru here can easily help me out . please see below for the code online/offline code ( input filed i would like to replace with dropdown selector <input type="text" name="txttemplate" value="%s" id="txttemplate" />) Code: [Select] <?php session_start(); $page_offline = TRUE; $page_offline_title = " "; $page_offline_message = " Offline"; $admin_ip = "1.1.1.1"; $admin_password = "*****************"; $page_offline_template = "template/path/index.php"; $page_offline_foradmin = TRUE; ini_set("display_errors", "0"); error_reporting(E_COMPILE_ERROR|E_ERROR|E_CORE_ERROR); $page = strstr($_SERVER['PHP_SELF'], "off.php") ? "admin" : "website"; // Try login $login_error = FALSE; if(isset($_POST['txtpassword'])){ if(md5($_POST['txtpassword']) == $admin_password){ $_SESSION['offline_admin'] = TRUE; }else{ $login_error = TRUE; } } // Save config settings $offline = $page_offline ? 'checked="checked"' : ""; $title = $page_offline_title; $message = $page_offline_message; $saved = FALSE; $template = $page_offline_template; $adminoffline = $page_offline_foradmin ? 'checked="checked"' : ""; if(isset($_POST['btnsave']) && $_SESSION['offline_admin']){ $val_offline = isset($_POST['chkoffline']) && ($_POST['chkoffline'] == "true") ? TRUE : FALSE; $val_title = htmlspecialchars($_POST['txttitle']); $val_message = htmlspecialchars($_POST['txtmessage']); $val_template = htmlspecialchars($_POST['txttemplate']); $val_password = htmlspecialchars($_POST['txtpassword']); $val_adminoffline = isset($_POST['chkconstruct']) && ($_POST['chkconstruct'] == "true") ? TRUE : FALSE; $offline = $val_offline ? 'checked="checked"' : ""; $title = $val_title; $message = $val_message; $template = $val_template; $adminoffline = $val_adminoffline ? 'checked="checked"' : ""; if(empty($val_password)){ page_offline_save($val_offline, $val_title, $val_message, $val_template, $adminoffline); }else{ page_offline_save($val_offline, $val_title, $val_message, $val_template, $adminoffline, $val_password); } $saved = TRUE; } // Templates $admin_style_tpl = <<<ADMINSTYLETPL body{ font-family:Arial, Helvetica, sans-serif; font-size:11px; text-align: center; background-color:#F9F9F9; color:#555555; } #wrapper{ width:400px; padding:20px; margin: 0 auto; border:1px solid #DADADA; background-color: #FFF; text-align: left; } h2{ margin: 0px 0px 10px 0px; padding: 0px; font-size: 24px; color:#464646; font-family:Georgia,"Times New Roman",Times,serif; font-style: italic; font-weight: normal; } label{ display: block; width: 150px; font-weight: bold; padding: 14px 0px 3px 0px; } input{ -moz-border-radius-bottomleft:3px; -moz-border-radius-bottomright:3px; -moz-border-radius-topleft:3px; -moz-border-radius-topright:3px; width: 300px; background:#F5F5F5 none repeat scroll 0 0; border:1px solid #CCCCCC; color:#666666; padding: 4px 8px; } #chkoffline, #chkconstruct{ width: 20px; } .error{ border: solid 1px #CC0000; } ADMINSTYLETPL; $admin_login_tpl = <<<ADMINLOGINTEMPLATE <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Login</title> <style> %s </style> </head> <body> <div id="wrapper"> <h2>Login</h2> <form id="form1" name="form1" method="post" action=""> <label for="txtpassword">Password</label> <input type="password" name="txtpassword" %s id="txtpassword" /><br /><br /> <input name="btnlogin" type="submit" value="Login" /> </form> </div> </body> </html> ADMINLOGINTEMPLATE; $admin_config_tpl = <<<ADMINCONFIGTEMPLATE <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title></title> <style> %s </style> </head> <body> <div id="wrapper"> <h2></h2> <form id="frmadmin" name="frmadmin" method="post" action=""> <label for="chkoffline">Offline</label> <input name="chkoffline" id="chkoffline" %s type="checkbox" value="true" /> <label for="chkconstruct">Show offline to admin</label> <input name="chkconstruct" id="chkconstruct" %s type="checkbox" value="true" /> <label for="txttitle">Title</label> <input type="text" name="txttitle" value="%s" id="txttitle" /> <label for="txtmessage">Message</label> <input type="text" name="txtmessage" value="%s" id="txtmessage" /> <label for="txttemplate">Template</label> <input type="text" name="txttemplate" value="%s" id="txttemplate" /> <label for="txtpassword">Admin password</label> <input type="password" name="txtpassword" id="txtpassword" /> <br /> Leave this field empty if you don't want to change the admin password. <br /><br /> <input name="btnsave" type="submit" value="Save" /> </form> </div> </body> </html> ADMINCONFIGTEMPLATE; $admin_saved_tpl = <<<ADMINSAVEDTEMPLATE <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Offline Admin</title> <style> %s </style> </head> <body> <div id="wrapper"> <h2>Configuration saved</h2> The configuration is saved succesfully. You can continue to <a href="off.php">config</a> or go to your <a href="../">Homepage</a> or back to <a href="index.php">admin panel</a>. </div> </body> </html> ADMINSAVEDTEMPLATE; if($page == "admin"){ // If user not is loggedin then show login screen if(!isset($_SESSION['offline_admin']) || !$_SESSION['offline_admin']){ echo sprintf($admin_login_tpl, $admin_style_tpl, ($login_error) ? 'class="error"' : ''); }else{ if(!$saved){ echo sprintf($admin_config_tpl, $admin_style_tpl, $offline, $adminoffline, $title, $message, $template); }else{ echo sprintf($admin_saved_tpl, $admin_style_tpl); } } exit(); }else{ if($page_offline && (($_SERVER['REMOTE_ADDR'] != $admin_ip) || $page_offline_foradmin)){ ob_start(); include($page_offline_template); $template = ob_get_contents(); ob_end_clean(); // Replace variables $template = str_replace(array("{TITLE}", "{MESSAGE}"), array($page_offline_title, $page_offline_message), $template); // Show template echo $template; exit(); } } // Write settings to file function page_offline_save($offline, $title, $message, $template, $adminoffline, $password=""){ $lines = explode("\n", file_get_contents("offline.php")); // Set configuration $offline_state = ($offline) ? "TRUE" : "FALSE"; $lines[2] = '$page_offline = ' . $offline_state . ';'; $lines[3] = '$page_offline_title = "' . $title . '";'; $lines[4] = '$page_offline_message = "' . $message . '";'; $lines[5] = '$admin_ip = "' . $_SERVER['REMOTE_ADDR'] . '";'; $lines[7] = '$page_offline_template = "' . $template . '";'; $adminoffline_state = ($adminoffline) ? "TRUE" : "FALSE"; $lines[8] = '$page_offline_foradmin = ' . $adminoffline_state . ';'; if(!empty($password)){ $lines[6] = '$admin_password = "' . md5($password) . '";'; } // Save to file file_put_contents("off.php", implode("\n", $lines)); } ?> snippet found here on the forum Code: [Select] <?php $parent_directory = 'path/to/directory'; $file_types = 'html,htm,php,etc'; //===================================================// // FUNCTION: directoryToArray // // // // Parameters: // // - $root: The directory to process // // - $to_return: f=files, d=directories, b=both // // - $file_types: the extensions of file types to // // to return if files selected // //===================================================// function directoryToArray($root, $to_return='b', $file_types=false) { $array_items = array(); if ($file_types) { $file_types=explode(',',$file_types); } if ($handle = opendir($root)) { while (false !== ($file = readdir($handle))) { if ($file != "." && $file != "..") { $add_item = false; $type = (is_dir($root. "/" . $file))?'d':'f'; $name = preg_replace("/\/\//si", "/", $file); if ($type=='d' && ($to_return=='b' || $to_return=='d') ) { $add_item = true; } if ($type=='f' && ($to_return=='b' || $to_return=='f') ) { $ext = end(explode('.',$name)); if ( !$file_types || in_array($ext, $file_types) ) { $add_item = true; } } if ($add_item) { $array_items[] = array ( 'name'=>$name, 'type'=>$type, 'root'=>$root); } } } // End While closedir($handle); } // End If return $array_items; } if (isset($_POST[pickfile])) { // User has selected a file take whatever action you want based // upon the values for folder and file } else { echo ' <html> <head> <script type="text/javascript"> function changeFolder(folder) { document.pickFile.submit(); } </script> </head> <body>'; echo "<form name=\"pickFile\" method=\"POST\">\n"; $directoryList = directoryToArray($parent_directory,'d'); echo "<select name=\"folder\" onchange=\"changeFolder(this.value);\">\n"; foreach ($directoryList as $folder) { $selected = ($_POST[folder]==$folder[name])? 'selected' : ''; echo "<option value=\"$folder[name]\" $selected>$folder[name]</option>\n"; } echo '</select><br><br>'; $working_folder = ($_POST[folder]) ? $_POST[folder] : $directoryList[0][name]; $fileList = directoryToArray($parent_directory.'/'.$working_folder,'f',$file_types); echo "<select name=\"file\">\n"; foreach ($fileList as $file) { echo "<option value=\"$file[name]\">$file[name]</option>\n"; } echo '</select><br><br>'; echo "<button type=\"submit\" name=\"pickfile\">Submit</button>\n"; echo "</form>\n"; echo "</body>\n"; echo "</html>\n"; } ?> Thank you in advance for any help... Hi all, I was wondering if someone knows a clean way to make a dropdown menu using just php and html with the following functionality. If someone selects a value from the dropdown options it should be visible on refresh. So in case a have a dropdown menu with: oranges, apples, grapes and someone selects grapes and submits it, grapes should be selected. I have made someonthing but it looks redundant. I think there should be something with a foreach loop or something but i can't figure it out yet. Thanks in advance, cheers!! I currently have an HTML form where the options for a certain drop-down menu are hard-coded. Instead, I want to use PHP to... 1. Look up the values in a column (cities) in a MySQL table (locations) 2. Make those values the only options in the dropdown menu. Any ideas how I would do this? This is what I have so far. <label for="city">What is your destination city?</label> <select class="form-control" id="city" name="city"> <?php //connect to the database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); //grab the city names from the MySQL table $query = "SELECT cities FROM locations"; $res = mysqli_query($dbc, $query); while ($data = mysqli_fetch_assoc($res)) { echo '<option value="'.$data['cities'].'">'.$data['cities'].'</option>'; } //close the db connection mysqli_close($dbc); ?> </select> I have a page in my admin control panel which allows for adding and editing an existing entry on the same page. At the moment, it displays the entries on the page in a table like format. After I edit an entry and hit Submit, I display a message on that page that says "Link updated successfully!", but the table on the page doesn't load the new entry until I refresh again - obviously since the data isn't entered into the database until I hit the submit button to process the form. But I'm wondering if it's possible to update the table on the page and display that new updated table all after hitting the submit button?? I am caught between a rock and a hard place here... My update page (to update an existing entity in the database) is a normal selection page, which then directs you to the page where you can update or change the data (1st srcipt snippet below), up to here everything works well, but on submit the page must be directed to a page that verifies the data (2nd script snippet below) and if correct, update the database or redirect to the 2nd page to correct the faulty data entry - easy right (well I have been struggling to get this working for 2 days straight. Pleeaaaaase Help. After the selection page, this is the input page-script:- Code: [Select] $labels = array("company_descr" => "Description:"); include("../includes/xxxxxxxxx.inc"); // database login details $cxn = mysqli_connect($host,$user,$password,$dbname) or die ("Couldn't connect to server."); $query = "SELECT * FROM `company` WHERE `company_id`=\"{$_POST['comp']}\""; $result = mysqli_query($cxn,$query) or die ("Cannot execute query."); $row = mysqli_fetch_assoc($result); ?> <html> <head> <title>Update Location</title> <style type='text/css'> <!-- form { margin: 1.5em 0 0 0; padding: 0; } .field { padding-top: .5em; } label { font-weight: bold; float: left; margin-right: 1em; text-align: right; } #submit { margin-left: 20%; padding-top: 1em; } --> </style> <body> <?php echo "<div style='text-align: centre'> <h3>Update Company - \"{$_POST['comp']}\"</h3>\n"; echo "<p style='font-color: #FF0000; font-size: 12; font-align: center'> Please change the field(s) that you want to update below.</p> <hr /></div>\n"; echo "<form action='../update/updateCompany.php' method='POST'>"; foreach($labels as $field => $label) { echo "<div class='field'> <label for='$field'>$label</label> <input type='text' name='$field' id='$field' value='$row[$field]' size='35' maxlength='35' /></div>\n"; } echo "<div id='submit'><input type='submit' value='Update' />\n"; echo "</div>\n</form>\n</body>\n</html>"; ?> This is the script that is supposed to verify the data and then send you back to the input page or update the database if the data is correct Code: [Select] foreach($_POST as $field => $value) { if(empty($value)) { $blanks[] = $field; } else { $good_data[$field] = strip_tags(trim($value)); } } if(isset($blanks)) { $message_2 = "The following fields are blank. Please enter the required information: "; foreach($blanks as $value) { $message_2 .="$value, "; } extract($good_data); include("../updateInput/updateInputCompanyTest.php"); exit(); } /* validate data */ foreach($_POST as $field => $value) { if(empty($value)) { if(preg_match("/descr/i",$field)) { if(!preg_match("/^[A-Za-z0-9.,' -]{1,50}$/",$value)) { $errors[] = "$value is not a valid address or city."; } } } // end if not empty } foreach($_POST as $field => $value) { $$field = strip_tags(trim($value)); } if(@is_array($errors)) { $message_2 = ""; foreach($errors as $value) { $message_2 .= $value." Please try again<br />"; } include("../updateInput/updateInputCompanyTest.php"); exit(); } // end if errors are found else // add new member to database { include("../includes/xxxxxxxx.inc"); // database login details $cxn = mysqli_connect($host,$user,$password,$dbname) or die ("Couldn't connect to server."); $query = "INSERT INTO `company` (`company_descr`) VALUES ('$company_descr') WHERE `company_id`=\"{$_POST['comp']}\""; $result = mysqli_query($cxn,$query) or die ("Couldn't execute query."); header("Location: ../backtomainpage.php"); } include("../updateInputCompanyTest.php"); ?> I have 2 websites currentweb.com copyweb.com (say) Using php can I update 2 databases and upload files in both these 2 websites ( for both these web sites I have user id and password exmaple I am uploading a file (Image), in currentweb.com can I upload it to copyweb.com at the same ( I have database and ftp password for this ) If Yes Please help with code how do i this? i have no idea....i need to incorporate it on an html form pleas help? i will get the data from this line and must be able to select one from the dropdown and get the id/name from the one selected [COLOR="Red"]$supplier_names = LpmAdnetworkPeer::getByName($catcher_id);[/COLOR] thank you I think after a lot of trouble I've finally managed to get my hierarchical php driven hierarchical dropdown menu working. I now have 1 problem with it which I simply can't figure out. Rather than explain this I think a picture is worth a thousand words so here's the script in action:- http://www.coolvisiontest.com/garyssite/rhrvouchers/menu4.php If you notice in that example when you hover over categories then level1a then level2a the menu's and their children appear 1 at a time as they should. However if you hover over level1a then level2b then all at once all of the child categories of 2b and all of their child catagories and their child categories etc appear at once rather than one at a time. You should have to hover over level3e to see level4a and you should have to hover over level4a to see level5a etc. The code I have is a combination of 2 sets of code I came across to be honest so I'm having trouble figuring it out. At the moment I only have 3 problems with it:- 1. I only want subcategories to be shown when their parent category is hovered over as explained above 2. I only want the bottom level of categories to act as links e.g. levels 2c 3a 3b 3c 3d and 5a would act as links in this example as they are the last of all the children, the rest of the categories should be plain text. I'm not sure if this is achievable though as I think links are required for the hover functionality. 3. I want some way of indicating to the user that a category has children perhaps by including a little right arrow or plus sign and of course this should only be included for the categories that have children and not the others. Can anyone help me out with these three things please? Here's my code so far:- <?php $link = mysqli_connect('localhost','',''); mysqli_select_db($link,''); $query = mysqli_query($link,'SELECT * FROM nested_categories'); while ( $row = mysqli_fetch_assoc($query) ){ $menu_array[$row['id']] = array('id' => $row['id'],'name' => $row['name'],'parent' => $row['parent_id']); } function generate_menu($parent){ $has_childs = false; global $menu_array; foreach($menu_array as $key => $value){ if ($value['parent'] == $parent){ if ($has_childs === false){ $has_childs = true; echo '<ul>'; } echo '<li><a href="menu4.php?id='.$value['id'].'">'.$value['name'].'</a>'; generate_menu($key); echo '</li>'; } } if ($has_childs === true) echo '</ul>'; } ?> <div id="menu"> <ul> <li><h2>categories</h2> <?php generate_menu(0); ?> </li> </ul> </div> And the Css:- #menu { width: 100%; background: #eee; float: left; } #menu ul { list-style: none; margin: 0; padding: 0; width: 12em; float: left; } #menu a, #menu h2 { font: bold 11px/16px arial, helvetica, sans-serif; display: block; border-width: 1px; border-style: solid; border-color: #ccc #888 #555 #bbb; margin: 0; padding: 2px 3px; } #menu h2 { color: #fff; background: #000; text-transform: uppercase; } #menu a { color: #000; background: #efefef; text-decoration: none; } #menu a:hover { color: #a00; background: #fff; } #menu li {position: relative;} #menu ul ul { position: absolute; z-index: 500; } #menu ul ul ul { top: 0; left: 100%; } div#menu ul ul, div#menu ul li:hover ul ul, div#menu ul ul li:hover ul ul {display: none;} div#menu ul li:hover ul, div#menu ul ul li:hover ul, div#menu ul ul ul li:hover ul {display: block;} Hello,
I'm new to php, I don't really have much experience, just have used it in combination with dreamweaver. I'm tackling a new problem now.
Basically, it's about a dropdown menu with pre-defined values. There are multiple dropdowns, having different value choices. People are supposed to be able to select more than one option and then search for the results. these should then be generated according to the selected values, or if no value is selected, ALL is shown.
When the user clicks on search, the values create a msql query that will then be executed and the results shown on the page.
I've been searching high and low for code ideas, but see different stuff everywhere I look, and none of the code seems to be what i want. Some people are using Javascript, some Ajax, or PHP. ideally, I want to use just php and sql
Can someone point me in the right direction on how to tackle this problem, or maybe someone has some code ideas?
Thanks in advance.
Mio
hello everybody..
i need help..
i want to make a dropdown menu with php coding that will show the result from my database.. can anyone help me to generate the coding?
this is an example form:
<form action="" method="get"><select name=""> <option value="select">select</option> <option value="1">1 </option> <option value="2">2</option> <option value="3">3</option> </select> <label> <input type="submit" name="button" id="button" value="Submit" /> </label> </form> if user select 1, the php code will show a list of name by their identification card, if user select 2 the php code will show a list of name by country. I want make the following, (I have already a database with three tables (Countries, Timeline and Category)). 1: list of countries (drop down menu 1), Timeline of the countries history (drop down 2) and Category (drop down 3). 2: The selected values of the drop down menus must show take the information from the database. Can any one help me with the coding? Hi, I don't know if this is the right place, but I used a code for a dropdown menu witch works fine on the website (geloven.eu/ABK2021) who uses PHP 7.2.24, but the dropdown does NOT work on another site bijbelkamp.eu/ABK2021 who uses PHP 7.3.28. The index page just includes the menu.php (here menu.txt) and the css file bootstrap21.css (here bootstrap21.txt). Can this be some coding issue or does this have nothing to do with the code? bootstrap21.txt menu.txt Hi All, I have an insert/update that i am using in several places across my site, it works fine everywhere apart from on one page. I am clearly missing something but cant for the life of me work it out. My php:
if ($_SERVER['REQUEST_METHOD']=='POST') {
$jobId = $_SESSION['current_job_id'];
$qty = $_POST['drinkItemQty'];
// prepare insert query
$stmt = $conn->prepare("INSERT INTO ssm_drink_order (drink_qty, job_id, drink_id)
VALUES (?,?,?)
ON DUPLICATE KEY UPDATE
drink_qty = VALUES(drink_qty)"
);
foreach ($_POST['drinkItemId'] as $k => $diid) {
if ($qty[$k] > 0) {
$stmt->bind_param("iii", $qty[$k], $jobId, $diid);
$stmt->execute();
}
if ($qty[$k] < 1) {
$stmt1 =$conn->prepare("DELETE FROM ssm_drink_order WHERE job_id = ?");
$stmt1->bind_param('i', $jobId);
$stmt1->execute();
}
}
}
the page html
<tbody>
<tr>
<th style="width:70%;" class="text-center">Drink Item</th>
<th class="text-center">Quantity</th>
</tr>
<tr>
<td>
House Gin
<input name="drinkItemId[]" type="hidden" value="2">
</td>
<td class="text-center">
<input name="drinkItemQty[]" type="number" value="999" class="text-center">
</td>
</tr>
<tr>
<td>
House Brandy
<input name="drinkItemId[]" type="hidden" value="4">
</td>
<td class="text-center">
<input name="drinkItemQty[]" type="number" value="" class="text-center">
</td>
</tr>
<tr>
<td>
House Vodka
<input name="drinkItemId[]" type="hidden" value="1">
</td>
<td class="text-center">
<input name="drinkItemQty[]" type="number" value="" class="text-center">
</td>
</tr>
<tr>
<td>
House Whiskey
<input name="drinkItemId[]" type="hidden" value="3">
</td>
<td class="text-center">
<input name="drinkItemQty[]" type="number" value="" class="text-center">
</td>
</tr>
</tbody>
Any help is as ever greatly appreciated Hi guys, I have a trouble of updating the data in the database by input the data at the end of the url. It have not been added in the table. Here it is updated: <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'myusername'); define('DB_PASSWORD', 'mypassword'); define('DB_DATABASE', 'mydatabasename'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $login = clean($_GET['user']); $password = clean($_GET['pass']); $firstname = clean($_GET['firstname']); if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'PASSWORD ID missing'; $errflag = true; } if($value == '') { $errmsg_arr[] = 'THE first name IS missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE login='$login' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result) > 0) { $row = mysql_fetch_row($result); echo $row[0]; } else { echo 'Username, Password or first name is not found.'; close; $sql="INSERT INTO members (firstname, lastname) VALUES ('$_POST[firstname]','$_POST[lastname]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "first name have been updated"; } ?> It have only login in the database but did not doing any update when I have input the data at the end of the url to get the database update. Do you know why I can't update the data in the database by input the data in the url like this? http://www.mysite.com/updatemysql.php?user=test&pass=test&firstname=shitorwhateveriwanttoinputandgetitupdate Do you have any idea and maybe if I have miss something? |
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