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PHP - Select Value In Combobox From Db...!!!
Hello Friends,
i have work with combobox by inserting value from DB dynamically... i just stuck with the situation as m populating combobox as statically , successfully inserting value in DB... but the point is m not sure for how to select perticular value stored in DB for combobox for displaying the data... here is my attempt... Code: [Select] Name: <input type="text" name="txtName" value="<?php print $uName; ?>"/> <br /> User Type: <select name="selType"> <option value="Admin">Admin</option> <option value="User">User</option> </select> i m getting the value of textbox from select query in the loop bt wht abt combobox value? if for 1 record the name will be krazyk n usertype will be admin then how to select the value from DB? please suggest me the way to accomplish it... awaiting for your better response... Thank you...!!! Similar TutorialsHi Everyone. I'm using the following code to populate a combobox with salutations. The List is populated and the initial value is 'select a salutation'. When the submit button is clicked, the selection is saved in the database. my issue is that when i open the page based on the id number, the list does not remember my choice. it just lists all the salutations in the combobox. How can i change this code to remember the value selected? thanks. I am saving the value of Salutation_1 from the combobox into records.Salutation_1. <?php $result=mysql_query("select Salutation from salutations"); $options=""; while ($row=mysql_fetch_array($result)) { $categoryname=$row["Salutation"]; $options.="<OPTION VALUE=\"$categoryname\">".$categoryname.'</option>'; } ?> <select name="Salutation_1"> <option >< select salutation > <?php echo $options ?></option> I have a form gathering data from the database. There is one field 'country' which is a combo box. The combo box is populated from the countries table which joins the customers table to provide the country specific to the customer. I'm trying to have the combo box display the country associated with the customer. I think I'm close in my code, but unsure. This is a playground to so I can implement on our other site.
<?php
require_once('database.php');
$sql2 = "SELECT * From countries
";
$countries = $db->query($sql2);
if(isset($_GET['customerID'])) {
$customerID = filter_input(INPUT_GET, 'customerID', FILTER_SANITIZE_NUMBER_INT);
$sql = "SELECT * FROM customers WHERE customerID =$customerID ";
//$sql2 = "SELECT * From countries
//INNER JOIN customers ON countries.countryCode=customers.countryCode
//WHERE customers.customerID = $customerID";
$stmt = $db->query($sql);
}
if(isset($_GET['customerID'])){
$customerID = filter_input(INPUT_GET, 'customerID', FILTER_SANITIZE_NUMBER_INT);
$countryQuery = "
{$sql2} INNER JOIN customers ON countries.countryCode = customers.countryCode WHERE customers.customerID = $customerID";
$countriesQuery = $db->prepare($countryQuery);
$countriesQuery->execute(['customerID' => $_GET['customerID']]);
$selectedCountry = $countriesQuery->fetch(PDO::FETCH_ASSOC);
var_dump($selectedCountry);
}
?>
<!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">
<!-- the head section -->
<head>
<title>My Guitar Shop</title>
<link rel="stylesheet" type="text/css" href="main.css" />
</head>
<!-- the body section -->
<body>
<div id="page">
<div id="header">
<h1>SportsPro Technical Support</h1>
<p>Sports management software for the sports enthusiast.</p></h1>
</div>
<div id="main">
<h1>View/Update Customer</h1>
<form action="update.php" method="get" >
<?php
?>
<div id="content">
<!-- display a table of products -->
<h2>Customers</h2>
<form name="customerInfo">
<?php foreach ($stmt as $cust) { ?>
<div>
<label>First Name</label>
<input type="text" name="name" class ="form-control" value ="<?php echo $cust['firstName']; ?>">
</div><br>
<div>
<label>Last Name</label>
<input type="text" name="name" class ="form-control" value ="<?php echo $cust['lastName']; ?>">
</div><br>
<div>
<label>Address</label>
<input type="text" name="address" class ="form-control" value ="<?php echo $cust['address']; ?>">
</div><br>
<div>
<label>City</label>
<input type="text" name="city" class ="form-control" value ="<?php echo $cust['city']; ?>">
</div><br>
<div>
<label>State</label>
<input type="text" name="state" class ="form-control" value ="<?php echo $cust['state']; ?>">
</div><br>
<form action="update.php" method="get">
<select name="country">
<option value=""></option>
<?php
foreach ($countries->fetchAll() as $country): ?>
<option value="<?php echo $country['customerID']; ?>
<?php echo isset($customerID) == $selectedCountry['customerID'] ? ' selected':''?>
"><?php echo $country['countryName']; ?></option>
<?php endforeach;?>
</select>
</form>
</div>
<br>
<div>
<label>Country Code</label>
<input type="text" name="countryCode" class ="form-control" value ="<?php echo $cust['countryCode']; ?>">
</div><br>
<div>
<label>Zip Code</label>
<input type="text" name="postalCode" class ="form-control" value ="<?php echo $cust['postalCode']; ?>">
</div><br>
<div>
<label>Email </label>
<input type="text" name="email" class ="form-control" value ="<?php echo $cust['email']; ?>">
</div><br>
<div>
<label>Phone Number </label>
<input type="text" name="phone" class ="form-control" value ="<?php echo $cust['phone']; ?>">
</div><br>
<div>
<label>Password </label>
<input type="text" name="password" class ="form-control" value ="<?php echo $cust['password']; ?>">
</div><br>
<div>
<?php }
?>
</div>
</div>
<form action="UpdateCustomer.php" method="get">
<input type="submit" name="data" value="Update_Data"></input>
</form>
<div id="footer">
<p>
© <?php echo date("Y"); ?> SportsPro, Inc.
</p>
</div>
</div><!-- end page -->
</body>
</html>
I trying to display data from a database for editing I have 2 table: 1. tbl_proizvodi_karakteristike (Translate: tbl_product_caracteristics) - In this table i have list of product caracteristic relatet to product 'id' in table 'tbl_product_list' (this 'tbl_product_list' is not important now) - field: id, opis, vrednsot, proizvod_id 2. tbl_karakteristike (Translate: tbl_caracteristic) - In this table i have list of all caracteristic - field: id, opis in field 'opis' in table 'tbl_proizvodi_karakteristike' i insert 'id' from 'tbl_karkateristike' and that works ok with combo box. But, when i try to edit value in 'tbl_proizvodi_karakteristike', i do not know how to show current value in combo box (i have list of value from 'tbl_karakteristike' in combox, starting from first) For examle, if i know that value in selected caracteristics is 'opis = 25', it means that i in combo box on load edit form mast have first value 25 and rest of posible value. For now i have value '1 and rest of value' Hiar is current code in edit form Code: [Select] <form action='product_carasteristic.php?product_id=<?php echo $p_id; ?>&product_name=<?php echo $p_name; ?>' method='POST'> <?php $p_id = $_GET['id']; $karakteristike = mysql_query("SELECT * FROM `tbl_proizvodi_karakteristike` WHERE `id`='$p_id'") or die(mysql_error()); while($row = mysql_fetch_array($karakteristike)){ $p_id = $row['id']; $p_opis = $row['opis']; $p_vrednost = $row['vrednost']; $p_proizvod_id = $row['proizvod_id']; ?> <table> <tr> <td > Karakteristika </td> <td> <select name="opis" style="font-family: verdana; font-size:15px; width:342px;"> <?php $tbl_karakteristike = mysql_query("SELECT * FROM `tbl_karakteristike`") or die(mysql_error()); while($row = mysql_fetch_array($tbl_karakteristike)){ $k_id = $row['id']; $k_opis = $row['opis']; ?> <option value='<?php echo $k_id; ?>'><?php echo $k_opis; ?></option> <?php } ?> </select> </td> </tr> <tr> <td> Vrednost karakteristike </td> <td> <input type="text" name="vrednost" value='<?php echo $p_vrednost; ?>'> </td> </tr> <tr> <td> <input type="submit" name="submit" value="Kreiraj"> </td> <td> Neophodno je popuniti sva polja </td> </tr> </table> <?php } ?> </form> thanks forward I use firebug to detect how javascript work. It same code. the java script :
<script type="text/javascript">
$(function() {
$("#cmbNegara").change(function(){
$("img#imgLoad").show();
var kodejeniskayu = $(this).val();
$.ajax({
type: "POST",
dataType: "html",
url: "getProvinsi.php",
data: "kodejeniskayu="+kodejeniskayu,
success: function(msg){
if(msg == ''){
$("select#cmbProvinsi").html('<option value="">--Pilih Ukuran--</option>');
$("select#cmbKota").html('<option value="">--Pilih Ukuran--</option>');
}else{
$("select#cmbProvinsi").html(msg);
}
$("img#imgLoad").hide();
getAjaxAlamat();
}
});
});
$("#cmbProvinsi").change(getAjaxAlamat);
function getAjaxAlamat(){
$("img#imgLoadMerk").show();
var kodeukuran = $("#cmbProvinsi").val();
$.ajax({
type: "POST",
dataType: "html",
url: "getKota.php",
data: "kodeukuran="+kodeukuran,
success: function(msg){
if(msg == ''){
$("select#cmbKota").html('<option value="">--Stok Kosong--</option>');
}else{
$("select#cmbKota").html(msg);
}
$("img#imgLoadMerk").hide();
}
});
};
});
</script>
getprovinsi.php
<?php require_once('Connections/karyo.php'); ?>
<?php
ini_set('display_errors',0);
//ambil parameter
$kodejeniskayu = $_POST['kodejeniskayu'];
if($kodejeniskayu == ''){
exit;
}else{
$sql ="SELECT kodeukuran,ukuran FROM tukuran WHERE kodejeniskayu='$kodejeniskayu'";
$getNamaProvinsi = mysql_query($sql,$karyo) or die ('Query Gagal');
while($data = mysql_fetch_array($getNamaProvinsi)){
echo '<option value="'.$data['kodeukuran'].'">'.$data['ukuran'].'</option>';
}
exit;
}
?>
getkota.php
<?php require_once('Connections/karyo.php'); ?>
<?php
ini_set('display_errors',0);
//ambil parameter
$kodeukuran = $_POST['kodeukuran'];
if($kodeukuran == ''){
exit;
}else{
$sql ="SELECT idstok,harga FROM tstok WHERE kodeukuran = '$kodeukuran'";
$getNamaKota = mysql_query($sql,$karyo) or die ('Query Gagal');
while($data = mysql_fetch_array($getNamaKota)){
echo '<option value="'.$data['idstok'].'">'.$data['harga'].'</option>';
}
exit;
}
?>
the problem at online is, the combobox at getprovinsi not showing up.
hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks I have 2 queries that I want to join together to make one row
Dear All, I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box. All is working fine except that the locality select box is not being populated. I know that the problem is in the sql statement WHERE country_id='$co' because i am having an error that $co is an undefined variable. All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement? Thank you for your help in advance. MySQL Tables indicated below: CREATE TABLE countries( country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT, country_name VARCHAR(30) NOT NULL, PRIMARY KEY(country_id), UNIQUE KEY(country_name), INDEX(country_id), INDEX(country_name)) ENGINE=MyISAM; CREATE TABLE localities( locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT, country_id INT(3) UNSIGNED NOT NULL, locality_name VARCHAR(50), PRIMARY KEY (locality_id), INDEX (country_id), INDEX (locality_name)) ENGINE=MyISAM; Extract PHP script included below: // connect to database require_once(MYSQL); if(isset($_POST['submitted'])) { // trim the incoming data /* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */ $trimmed=array_map('trim',$_POST); // clean the data $co=mysqli_real_escape_string($dbc,$trimmed['country']); $lc=mysqli_real_escape_string($dbc,$trimmed['locality']); } ?> <form action="form.php" method="post"> <p>Country <select name="country"> <option>Select Country</option> <?php $q="SELECT country_id, country_name FROM countries"; $r=mysqli_query($dbc,$q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($r)) { $country_id=$row[0]; $country_name=$row[1]; echo '<option value="' . $country_id . '"'; if(isset($trimmed['country']) && ($trimmed['country']==$country_id)) echo 'selected="selected"'; echo '>' . $country_name . '</option>\n'; } ?> </select> </p> <p>Locality <select name="locality"> <option>Select Locality</option> <?php $ql="SELECT locality_id, country_id, locality_name FROM localities WHERE country_id='$co' ORDER BY locality_name"; $rl=mysqli_query($dbc,$ql) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($rl)) { $locality_id=$row[0]; $country_id=$row[1]; $locality_name=$row[2]; echo '<option value="' . $locality_id . '"'; if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id)) echo 'selected="selected"'; echo '>' . $locality_name . '</option>\n'; } // close database connection mysqli_close($dbc); ?> </select> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> </form> Code: [Select] id player_id nat nt_caps 13740 28664 97 24 13741 28664 68 0 13742 28664 79 0 16252 42904 15 40 16253 42904 68 0 16254 42904 241 0 That's how my table looks. I want to select the player_id's that have either nt_caps = "0" for every nat OR player_id's that have nt_caps != "0" only for nat = "68". The SQL query I try to use is: SELECT player_id FROM x WHERE nat = '68' AND (nat != '68' AND nt_caps = '0') But then I get player_id '42904' and '28664' because they both have 1 entry that matches the query but I don't want them because they have nt_caps for another nat than nat "68". I hope you understand what I try to achieve. I'm trying to run a readout of a db which runs fine as an individual script. When I embed it in PHP inside an html div container, it bails when it encounters the first ">" and simply outputs the PHP characters from there through the "?>". The rest of the html runs fine before and after. For example, this line echo "<select>"; would output "; and any other php script up to the ?> end, after which it renders html fine. If I run the script as a separate php file, it runs as expected. Any help would be appreciated. Thanks. hi. just started a website talkietaco.com and at the momment my code selects the first row and displays it. this is all well and good but when i add a new row to the table it gos to the bottom. I want to be able to select the last row and echo it out. Any ideas? would i need to add an id row or something? Heres the code and thank you in advance for any help people can offer. Code: [Select] <?php $con = mysql_connect("localhost","",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mainbase", $con); $result = mysql_query("SELECT * FROM matteroffact"); echo "<table border='0'> <tr> <th></th> </tr>"; $row = mysql_fetch_array($result) or die(mysql_error()); echo "<tr>"; echo "<td><strong>" . $row['question']. "</strong> ". $row['answer']; "</td>"; echo "<tr>"; echo "</table>"; ?> ok here's my problem $Sql1 returns two values: 8 and 10 and these numbers get put into a <select>. So far so good. I assign a onchange to it. When i select 8 it makes the changes but when I select 10 nothing happends. I preciate some help. Code: [Select] <?php require("status.php"); require("id.php"); $Link = mysql_connect($Host, $User, $Password); mysql_select_db('sportsportal', $Link); $Sql1 = "select distinct week, (select max(week) from coupons where user='$User') max from coupons where user='$User'"; $Result1 = mysql_query($Sql1, $Link); print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top>"; print "<select name=current_week onchange=window.location='coupon.php?curwk='+this.value>"; while($Row1 = mysql_fetch_array($Result1)){ //if($Row1[week] == $Row1[max]){ //print "<option value='$Row1[week]' selected>Vecka $Row1[week]</option>"; //} else { //print "<option value='$Row1[week]'>Vecka $Row1[week]</option>"; //} print "<option value='$Row1[week]'>Vecka $Row1[week]</option>"; } print "</select>"; print "<p></td>"; print "</tr>"; if(isset($_REQUEST['curwk'])){ $Curwk = $_REQUEST['curwk']; } else { $Curwk = 0; } $Sql = "select home, away, home_score, away_score, winner from coupons where user='$User' and week='$Curwk'"; $Result = mysql_query($Sql, $Link) or die(mysql_error()); while($Row = mysql_fetch_array($Result)){ if(@$Row[home] == @$Row[winner]){ print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top><b>$Row[home]</b> - $Row[away] $Row[home_score]-$Row[away_score]</td>"; print "</tr>"; } else { print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top>$Row[home] - <b>$Row[away]</b> $Row[home_score]-$Row[away_score]</td>"; print "</tr>"; } } mysql_close($Link); ?> Hi guys, I'm having a issue with a little bit of coding involving SELECT the problem is I want it to select from the database and say limit it to 5 so basicly it prints 5 results from the database but its only printing 1 result Here is the code i'm using Code: [Select] $q=mysql_query("SELECT * FROM papercontent LIMIT 5",$c); $content=mysql_result($q,0,0); print " <center><table width = '40%' border = '1'><tr><th> <center><u>Latest Announcements</u></center> $content </tr></td></table> "; Thanks hi guys ive just finished this task after hours of head scratching since svg is only really supported good in firefox and opera ive chosen firefox as my browser to view this url www.deansignori.com/phpsvgpie/index.php i do need more help with this task but a different problem (creating select box to call different stylesheet and to change from 2d - 3d i have the code set out so that i can explode any segment or change size of slices or change from 2d-3d but i have to do this manually in the code to render different piecharts im wanting to use 1 but change it using a select box and echo my variable into it basically im unsure of the syntax for this problem psuedo code for style maybe something like if select box value isset onchange stylecolour echo stylecolour if select box value isset onchange stylegrey echo stylegrey and for 3d-2d if select box value isset onchnage format3 echo format3 if select box value isset onchange format2 echo format2 this would be on my index page can anyone advise me please regards Dean Code: [Select] $query ="SELECT id,today,uuendus,username FROM u_data WHERE today < '$datex' AND uuendus='YES' ORDER BY id ASC";i want select older then $datex all other code is correct i just want to know is Code: [Select] WHERE today < '$datex' correct to use? I'm trying to write a select to match a few certain words... $sql = "SELECT * FROM podcasts WHERE `type` = 'podcast' AND recap LIKE 'Men%' AND recap LIKE '%Hockey%' ORDER BY date DESC"; That's what I have so far...which isn't working... I need to match the string "Mens" OR "Men's" with the word "Hockey" BUT...I can't match "Women's" or "Womens" Any idea what I'm doing wrong? The above code only returns 6 results...when it should return nearly 100. Thanks! Hi guys, I'm trying to do make this code so that IF a user owns a property (in this code a bulletfactory) then the BF CP Shows up... Here's the code so far..... Code: [Select] <?php session_start(); include_once"includes/db_connect.php"; include_once"includes/functions.php"; logincheck(); $username=$_SESSION['username']; $query=mysql_query("SELECT * FROM users WHERE username='$username'"); $fetch=mysql_fetch_object($query); $query_bf=mysql_query("SELECT * FROM bf WHERE location='England, Japan, Colombia, USA, Russia, Italy, Turkey'"); $fetch_bf=mysql_fetch_object($query_bf); if (strtolower($fetch_bf->owner) == (strtolower($fetch->username))){ require_once"bulletCP.php"; exit(); } ?> Hi, I'm not quite sure how to do this, so i thought i'd ask you guys from some assistance. Basically i am inserting Author's into a table successfully using an array. The reason for this is that i have multiple authors being added and there is no limit as to how many. An example of what i mean can be seen he http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/test.html You can click on "Add author" to add however many necessary.. Anyway, i have got the insert working, however when i edit i want to be able to see all the authors that have been added but obviously i don't know how many there are.. Typically i would like to see something like this: http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/edit.jpg So i would click on an edit link and it would pre populate the text boxes. I don't have a problem with doing this, but how can i show the correct amount of input textboxes based on how many authors exist for that specific record? Can someone offer some advice please? The input elements are like so: Quote <input type="text" name="author[]" id="author1"/> <input type="text" name="author[]" id="author2"/> <input type="text" name="author[]" id="author3"/> ... ... ... <input type="text" name="author[]" id="author10"/> Some records may have 1 author some may have 10, so how can i do this? Thanks again.. Hi Guys
I have a table which in it's shortened form has the following columns:
id | postID | title | content | version
The column for postID has a number that can be shared by multiple rows - differentiated by version number.
I want to run a query to select all records that are like a given keyword (i.e. %LIKE%) but where results share the same postID I only want to return the highest version number for that record.
The difficulty is some records may have multiple version numbers that match the like statement and some may have only one. So this variance with the LIKE search is causing me some confusion.
I've tried this in a few ways using a sub-query but for the life of me I cannot work out how to do it.
Any help would be appreciated,
Drongo
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