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PHP - Mysql Php Dropdown
Hi,
I'm trying to create a dropdown box that is filled by a mysql column. I found this example, which fills the dropdown box with a list of columns in the database, but I want to fill the drop down box with the content in a column. Any guidance on how to do this would be appreciated. Thank you! Example: $connection = mysql_connect("localhost","username","password"); $fields = mysql_list_fields("database", "table", $connection); $columns = mysql_num_fields($fields); echo "<form action=page_to_post_to.php method=POST><select name=Field>"; for ($i = 0; $i < $columns; $i++) { echo "<option value=$i>"; echo mysql_field_name($fields, $i); } echo "</select></form>"; Similar TutorialsHey guys. I am new to PHP and I have figured out how to populate a dropdown box with data from my database but now I want to create another dropdown box with data from my database but I want it to be data that depends on the data from the first drop down box. I am making code for a supply order for a website I am making for tafe and I filled in the first drop down box with the supplier names and now I want a second drop down box of all the products that that supplier is supplying us. I cannot for the life of me work out how to do this and I have no idea what to search to get this. Any help will be greatly appreciated. I will show you the code for my first dropdown box. Code: [Select] <?php $db_host = "localhost"; $db_username = "root"; $db_password = ""; $db_name = "ocs"; @mysql_connect("$db_host","$db_username","$db_password") or die("Could not connect to MySQL"); @mysql_select_db("$db_name") or die("Cannot find database"); $query = "SELECT ProductID FROM products"; $result = mysql_query($query) or die(mysql_error()); $dropdown = "<select name='products'>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['ProductID']}'>{$row['ProductID']}</option>"; } $dropdown .= "\r\n</select>"; echo $dropdown; ?> Thanks a lot -Teammuffin Hi, I don know how to explain this in words too well, so i have create 2 images to show what I need. What I need to do is use whats in the first image (table.jpg)(the database) to create whats in the second image (dropdown.jpg)(the drop-down menu) The results of the form will be entered into the database. category_id will be the value thats inserted. Many kind regards Eoin [attachment deleted by admin] Hi guys,
Does anyone know why my code isn't working?:
$servername="127.0.0.1";
$username="root";
$password="";
$dbname="testdb";
$connection = new mysqli($servername,$username,$password);
$result = "SELECT 'Country' from 'w'";
echo '<select>';
foreach($result as $res) {
echo '<option value="'.$res['Country'].'">' . $res['Country'] . '</option>';
}
echo '</select>';
}
There's not a credential issue, the table is called 'w' and the collumn I'm trying to get the values from is 'Country'
Any help is appreciated.
Cheers,
Broken
i have been trying to get this code to get a list of usernames from a database and i have now got that to work but when i try and save it it saves all the usernames from the drop down list and not just the one i have selected how can i get it to just use the one i have selected Code: [Select] <?php include "connect.php"; //connection string include("include/session.php"); print "<link rel='stylesheet' href='style.css' type='text/css'>"; print "<table class='maintables'>"; print "<tr class='headline'><td>Post a message</td></tr>"; print "<tr class='maintables'><td>"; // Write out our query. $query = "SELECT username FROM users"; // Execute it, or return the error message if there's a problem. $result = mysql_query($query) or die(mysql_error()); $dropdown = "<select name='username'>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['username']}'>{$row['username']}</option>"; } $dropdown .= "\r\n</select>"; if(isset($_POST['submit'])) { $name=$session->username; $yourpost=$_POST['yourpost']; $subject=$_POST['subject']; $to=$dropdown; if(strlen($name)<1) { print "You did not type in a name."; //no name entered } else if(strlen($yourpost)<1) { print "You did not type in a post."; //no post entered } else if(strlen($subject)<1) { print "You did not enter a subject."; //no subject entered } else { $thedate=date("U"); //get unix timestamp $displaytime=date("F j, Y, g:i a"); //we now strip HTML injections $subject=strip_tags($subject); $name=strip_tags($name); $yourpost=strip_tags($yourpost); $to=strip_tags($to); $insertpost="INSERT INTO forumtutorial_posts(author,title,post,showtime,realtime,lastposter,name) values('$name','$subject','$yourpost','$displaytime','$thedate','$name','$to')"; mysql_query($insertpost) or die("Could not insert post"); //insert post print "Message posted, go back to <A href='forum.php'>Forum</a>."; } } else { print "<form action='newtopic.php' method='post'>"; print "Your name:<br>"; print "$session->username<br>"; print "User to send to:<br>"; print "$dropdown"; print "Subject:<br>"; print "<input type='text' name='subject' size='20'><br>"; print "Your message:<br>"; print "<textarea name='yourpost' rows='5' cols='40'></textarea><br>"; print "<input type='submit' name='submit' value='submit'></form>"; } print "</td></tr></table>"; ?> MOD EDIT: Changed PHP manual link [m] . . . [/m] tags to [code] . . . [/code] tags. Hi. I am using this script to populate a dropdown list box from sql, it works but does anyone know how to sort the list in alphabetical order? $sql="SELECT * FROM Fish WHERE ***** = '".$_GET['stocktype']."'"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $ID=$row["ID"]; $Stock=$row["Commonn"]; $Options.="<OPTION VALUE=\"$ID\">".$Stock; } <SELECT NAME='stock1'> <OPTION VALUE='$Options'>$Options</option> </SELECT> Hi Guys, I am a typical newbie with a twist, i program in VB, recently moved to PHP. I need help, preferable an example, but first let me explain what i have. I have a .tpl page that has 3 drop down boxes, one of these is populated from within the .tpl page itself. The other 2 i waant to pupulate from a SELECT on a database from a .php file. The .TPL file i have so far is:- Code: [Select] <style type="text/css"> <!-- .style7 { color: #006600; font-weight: bold; font-size: 18px; } .style14 {color: #003300; font-weight: bold; } .style16 { color: #000066; font-style: italic; } .style25 { color: #FF0000; font-weight: bold; font-size: 12px; } .style28 { font-size: 12px; color: #330000; } .style29 { font-size: 12px } .style31 {color: #003300; font-weight: bold; font-size: 16px; } .style33 { color: #000099; font-weight: bold; font-style: italic; font-size: 18px; } .style34 { color: #FF0000; font-weight: bold; font-size: 24px; } .style35 {font-size: 12px; color: #003300; } .style36 { color: #FF0000; font-weight: bold; font-size: 14px; font-style: italic; } .style38 {color: #003300; font-weight: bold; font-size: 16px; font-style: italic; } --> </style> <script src="Scripts/AC_RunActiveContent.js" type="text/javascript"></script> <h1 align="center" class="style34">Scotbirds Alertz - Rare and Scarce</h1> <h3 align="center" class="style36">Soon Only VIP Members will be able to Access this Page</h3> <h3 align="center" class="style14">If you have seen something unusual / rare then please call it in on -- <span class="style33">Hotline: 0333 5772473</span></h3> <p align="center" class="style14"> <table width="90%" border="0" align="center" cellpadding="5"> <tr> <?php print $_SERVER['PHP_SELF']; $the_date_filter = $_GET["DATE_FILTER"];?> <td width="1%"><td width="5%"><td width="2%"><td width="2%"><td width="2%"><form action="alertz_VIP.php" method="post"> <td width="2%"><td width="2%"><td width="5%"><span class="style38">Date</span><td width="3%"></td> <td width="12%"><span class="style14"> <select name=DATE_FILTER size="1" id=DATE_FILTER onchange="this.form.submit()"> <option value="All Dates">All Dates</option> <option value="Today">Today</option> <option value="Last 48hrs">Last 48hrs</option> <option value="Last week">Last Week</option> <option value="last month">Last Month</option> </select> </span></td> <td width="1%"> </td> <td width="7%"><span class="style16"><span class="style14"><span class="style31">Region</span></span></span></td> <td width="22%"><span class="style16"><span class="style14"> <select name=REGION_FILTER size="1" id=REGION_FILTER onchange="this.form.submit()"> <option value="AllRg">All Regions</option> <option value={REGION_FILTER}></option> </select> </span></span></td> <td width="1%"> </td> <td width="8%"><span class="style16"><span class="style14"><span class="style31">Species</span></span></span></td> <td width="20%"><span class="style16"><span class="style14"> <select name=SPECIES_FILTER size="1" id=SPECIES_FILTER onchange="this.form.submit()"> <option value="AllSpec">All Species</option> <option value={SPECIES_FILTER}></option> </select> </span></span></td> <td width="9%"><input type="Submit" name="submit2" value="Search" method="get" action="alertz_VIP.php" /></td> </table> <p> </p> <table width="870" border="1" align="center" cellpadding="0" cellspacing="0" bordercolor="#99CC99"> <tr> <th width="178" class="style31" scope="col">Region</th> <th width="184" class="style31" scope="col">Species</th> <th width="96" class="style31" scope="col">Date</th> <th width="96" class="style31" scope="col">Time</th> <th width = "304" class="style31" scope="col">Comments</th> </tr> <!-- BEGIN alerts --> <tr> <th class="style7 style29" scope="col">{alerts.REGION}</th> <th class="style25" scope="col">{alerts.SPECIES}</th> <th class="style35" scope="col">{alerts.DATE} </th> <th scope="col"><span class="style35">{alerts.TIME} </span></th> <th scope="col"><span class="style28">{alerts.COMMENTS} </span></th> </tr> <!-- END alerts --> </table> <p align="center" class="style14"> </p> Now from here i expect the user to select a date from the date box, ( Ideally i would like the other 2 options to only become visible once a selection on the date has been made, i would also like to retain the selected in the date box once the form has been submited. Now the code i have for the PHP so far is quite long winded as my PHP skills are not so good, although i am learning fast. Code: [Select] <?php /*************************************************************************** * Alertz.php * ------------------- * begin : 30/10/04/10 * copyright : (C) 2010 Andy Guppy * email : webmaster@scotbird.co.uk * * ***************************************************************************/ define('IN_ICYPHOENIX', true); if (!defined('IP_ROOT_PATH')) define('IP_ROOT_PATH', './'); if (!defined('PHP_EXT')) define('PHP_EXT', substr(strrchr(__FILE__, '.'), 1)); // Include files include(IP_ROOT_PATH . 'common.' . PHP_EXT); include_once(IP_ROOT_PATH . 'includes/functions_groups.' . PHP_EXT); include(IP_ROOT_PATH . '/alerts/alertsconfig.'.PHP_EXT); // Page Authorise $cms_page_id = 'scotalertz'; $cms_page_nav = (!empty($cms_config_layouts[$cms_page_id]['page_nav']) ? true : false); $cms_global_blocks = (!empty($cms_config_layouts[$cms_page_id]['global_blocks']) ? true : false); $cms_auth_level = (isset($cms_config_layouts[$cms_page_id]['view']) ? $cms_config_layouts[$cms_page_id]['view'] : AUTH_ALL); check_page_auth($cms_page_id, $cms_auth_level); // Obtain Select Criteria $temp_region_filter = $_REQUEST["REGION_FILTER"]; $temp_species_filter = $_REQUEST["SPECIES_FILTER"]; $temp_date_filter = $_REQUEST["DATE_FILTER"]; // Filter characters if required $region_filter =str_replace(" & ", "&", $temp_region_filter); $species_filter = $temp_species_filter; $date_filter = $temp_date_filter; // standard session management $userdata = session_pagestart($user_ip); // Check to see if user is logged in if ((!$userdata['session_logged_in']) ) // No he isnt { redirect(append_sid(LOGIN_MG . '?redirect=alerts.' . PHP_EXT)); } else // Yes they are { init_userprefs($userdata); // set page title $page_title = "ScotBird alerts - VIP's ONLY !! "; // standard page header include(IP_ROOT_PATH . 'includes/page_header.'.PHP_EXT); // Connect to the database $db = new sql_db($alerts_mysql_host,$alerts_mysql_username,$alerts_mysql_password,$alerts_mysql_db,false); if(!$db) { die("Database Connection Failed:- Please Contact Site Admin" . mysql_error()); } $template->set_filenames(array('body' => 'alertz_VIP.tpl')); if (!isset($_REQUEST['DATE_FILTER'])) { // if date is NOT set then perform this echo 'Date has not been set'; $sql2 = "SELECT * FROM alerts ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts GROUP BY region"; echo $sql2; } else { // if date is set then perform this $year =date("Y"); $month = date("m"); $day = date("d"); $theenddate = $year . '-' .$month . '-' . $day ; echo 'the choice selected was :- '.$_REQUEST['DATE_FILTER']; switch ($date_filter) { case "All Dates": $sql2 = "SELECT * FROM alerts ORDER BY Date DESC, time DESC "; break; case "Today": $sql2 = "SELECT * FROM alerts WHERE Date = '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date = '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date ='".$theenddate ."' GROUP BY region"; break; case "Last 48hrs": $year =date("Y"); $month = date("m"); $day = date("d")-1; $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; case "Last week": $year =date("Y"); $month = date("m"); $day = date("d")-7; $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; case "last month": $year =date("Y"); $month = date("m")-1; $day = date("d"); $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; case "last 3 months": $year =date("Y"); $month = date("m")-3; $day = date("d"); $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; default: // Default is the last 48hrs $year =date("Y"); $month = date("m"); $day = date("d")-1; $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; } } $result2 = $db->sql_query($sql2); if (!($result2 = $db->sql_query($sql2))) { die("Database Query Failed" . mysql_error()); } $i = 1; while ( $row2 = $db->sql_fetchrow($result2) ) { $template->assign_block_vars('alerts', array( 'POS' => $i , 'REGION' => str_replace("&", " & ", $row2['region']), 'SPECIES' => str_replace("%", "'",$row2['species']), 'DATE' => $row2['Date'], 'TIME' => $row2['time'], 'COMMENTS' => str_replace("'", "%",$row2['comments']), ) ); $i++; } $template->assign_vars(array( 'USERNAME' => htmlspecialchars($userdata[username]), 'REGION_FILTER' => str_replace("&", " & ", $region_filter_options), 'SPECIES_FILTER' => $species_filter_options, ) ); $result = $db->sql_query($sql); if (!($result = $db->sql_query($sql))) { die("Database Query Failed" . mysql_error()); } // This is where you would add a new VARS Array if you intend to use your own custom VARS. while ($row = $db->sql_fetchrow($result)) { $region_filter_options .= '<option value="' . $row['region'] . '">' . $row['region'] . '</option>'; } $result = $db->sql_query($sql3); if (!($result = $db->sql_query($sql3))) { die("Database Query Failed" . mysql_error()); } // This is where you would add a new VARS Array if you intend to use your own custom VARS. while ($row = $db->sql_fetchrow($result)) { $species_filter_options .= '<option value="' . $row['species'] . '">' . $row['species'] . '</option>'; } $template->assign_vars(array( 'USERNAME' => htmlspecialchars($userdata[username]), 'REGION_FILTER' => str_replace("&", " & ", $region_filter_options), 'SPECIES_FILTER' => $species_filter_options, ) ); // Build the page $template->pparse('body'); // standard page footer include(IP_ROOT_PATH . 'includes/page_tail.'.PHP_EXT); } ?> Now my questions are these:- 1) In the .tpl file hows can i submit from any of the dropdown boxes and retain the selection after submission 2) How can i have it so that the second and third drop down boxes are only visible after the previous one, ie the first box ( date ) has to have a selection before the Region ( 2nd one ) is visible and so 3) How can i reduce the amount of code to cover all options in building a select statement or is Switch - case the best way. I would greatly appreciate help with this and even more so for some example of what i am asking for so i can learn from them. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=351349.0 Hello,
I have been trying to figure this for a while now and reading the tutorials are not helping, I think I'm a little over my head on this one and was hoping someone could help me out with this issue.
I am making a User Edit page and would like to have the access level part of the form show the users access current access level thats set in the database when the page loads, and if it needs to be changed you can press the dropdown box and select a new access level. I can't figure out how to show the current access level as default and populate the drop down box with the other access levels in my table.
My Tables look like this
Users table (users):
---------------------------------------------------------------------------------------------------
| id | username | password | flag | realfirst | reallast | dept |
---------------------------------------------------------------------------------------------------
1 loderd 9 test guy Service
Auth Table (auth):
--------------------------------------------
| id | auth_level | descrip |
--------------------------------------------
1 1 Service Tech
2 2 Office Staff
3 9 Super Admin
My SQL Query looks like this
$users = $db->fetch_all_array("SELECT users.id, users.username, users.password, users.realfirst, users.reallast, users.dept, users.flag, auth.auth_level, auth.descrip, auth.id
FROM users LEFT JOIN auth
ON users.flag = auth.auth_level
WHERE users.id = '".$_GET['id']."'");
I can't seem to figure out how I can do this for the Access Level dropdown box.
<tr>
<td width="19%" align="right" valign="top">Access Level :</td>
<td width="1%" align="left"> </td>
<td width="80%" align="left" valign="top">
<?php
echo "<select name='flag' id='flag'>";
foreach ($users as $row){
if($row[auth_level]==$row[auth_level]){
echo "<option value=$row[auth_level] selected>$row[auth_level] - $row[descrip]</option>";
}else{
echo "<option value=$row[auth_level]>$row[auth_level] - $row[descrip]</option>";
}
}
echo "</select>"; ?>
</td>
</tr>
Any help would be greatly appreciatedHi all, I'm trying to display a form based on a dropd own selection. The drop down is propagated from a database query. I have the drop down list displaying, but when I click submit, I can't get the result to display. Code: [Select] $sql="SELECT id, company_name FROM webprojects ORDER BY company_name ASC"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $company_name=$row["company_name"]; $options.="<OPTION VALUE=\"$id\">".$company_name; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>PHP Dynamic Drop Down Menu</title> </head> <body> <form id="form1" name="form1" method="post" action="selected.php"> <SELECT NAME=id> <OPTION VALUE=0>Choose <?=$options.="<OPTION VALUE=\"$id\">".$company_name.'</option>';?> </SELECT> <label> <input type="submit" name="submit" id="submit" value="Submit" /> </label> </form> </body> </html> Any help is greatly appreciated Hello there! I've been banging my head on this for a while and I just can't seem to get it to work properly. I have a dropdown menu which selects information from table1 using a select statement (this table is called 'lid'). It selects the firstname, lastname and member id from this table and shows it in the dropdown menu. I'm glad I got that part working but the hard thing is inserting the data that the user selects into another table. So when you select the id member from this dropdown menu it only inserts a blank row into table2 (which is called 'teamlid'). Can you guys help me? How can I insert the id member into my table2? What am I doing wrong here? Thanks a million! This is my first post so if I'm doing anything wrong, let me know and I'll fix it asap! My code:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Boast & Drive</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
<style type="text/css">
.wrapper{
width: 650px;
margin: 0 auto;
}
.page-header h2{
margin-top: 0;
}
table tr td:last-child a{
margin-right: 15px;
}
</style>
</head>
<body>
<div class="container-fluid">
<div class="row">
<div class="col-md-12">
<div class="page-header clearfix">
<h2 class="pull-left">Teamleden</h2>
<div class="btn-toolbar">
<a href="read.php" class="btn btn-primary btn-lg pull-right">Terug</a>
</div>
</div>
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
error_reporting(E_ALL);
//Verbinding maken met de database
require_once "login.php";
$sql = "SELECT tl.teamnaam,
tl.tl_ID,
tl.lidnummer,
l.voornaam,
l.achternaam
FROM teamlid tl
JOIN lid l ON tl.lidnummer = l.lidnummer
ORDER BY tl.teamnaam;";
if($result = mysqli_query($conn, $sql)) {
if(mysqli_num_rows($result) > 0) {
echo "<table class='table table-bordered table-striped'>";
echo "<thead>";
echo "<tr>";
echo "<th>Teamnaam</th>";
echo "<th>Tl_ID</th>";
echo "<th>Lidnummer</th>";
echo "<th>Voornaam</th>";
echo "<th>Achternaam</th>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['teamnaam'] . "</td>";
echo "<td>" . $row['tl_ID'] . "</td>";
echo "<td>" . $row['lidnummer'] . "</td>";
echo "<td>" . $row['voornaam'] . "</td>";
echo "<td>" . $row['achternaam'] . "</td>";
echo "<td>";
echo "<a href='update.php?id=". $row['lidnummer'] ."' title='Gegevens wijzigen' data- toggle='tooltip'><span class='glyphicon glyphicon-pencil'></span></a>";
echo "<a href='delete.php?id=". $row['lidnummer'] ."' title='Lid verwijderen' data- toggle='tooltip'><span class='glyphicon glyphicon-trash'></span></a>";
echo "</td>";
echo "</tr>";
}
echo "</tbody>";
echo "</table>";
mysqli_free_result($result);
} else{
echo "<p class='lead'><em>Er zijn geen gegevens om weer te geven.</em></p>";
}
} else{
echo "De volgende fout is gevonden: " . mysqli_error($conn);
}
?>
<form name="dropdown" method="post">
<div class="page-header clearfix">
<h2 class="pull-left">Teamlid toevoegen</h2>
</div>
<p>Selecteer hieronder met behulp van het dropdown menu een lid welke je aan bovenstaand team wilt toevoegen</p>
<div class="container-fluid">
<div class="row">
<?php
// Variabelen aanmaken en tonen met lege waardes
$teamnaam = $lidnummer = '';
// Code voor dropdown. Selecteert voornaam, achternaam en lidnummer van tabel lid)
$sql = "SELECT voornaam, achternaam, lidnummer FROM lid ORDER BY achternaam";
$result = mysqli_query($conn, $sql);
echo "<select id='teamLid' name='teamLid'>";
echo "<option>--Selecteer Lid--</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['lid'] . "'>" . $row['voornaam'] . " " . $row['achternaam'] . " " . $row['lidnummer'] . "</option>";
}
echo "</select>";
if
(isset($_POST["id"]) && !empty($_POST["id"])) {
$id = $_POST["teamLid"];
$stmt = $conn->prepare("INSERT INTO teamlid (teamnaam, lidnummer) VALUES (?,?)");
$stmt->bind_param('si', $param_teamnaam, $param_lidnummer);
$param_teamnaam = $teamnaam;
$param_lidnummer = $lidnummer;
$stmt->execute();
}
// Verbinding sluiten
mysqli_close($conn);
?>
<div>
<input type="hidden" name="id" value="<?php echo $id; ?>" />
<input type="submit" name="submit" class="btn btn-primary" value="Toevoegen">
</div>
</div>
</div>
</form>
</div>
</div>
</body>
</html>
Hello everyone, So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help. A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right. Some of this probably doesnt even make sense but I'll show you the kind of things I've tried... <select name="gender" size="1" id="gender"> <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option> <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option> </select> OR <select name="gender" id="gender"> <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option> <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option> <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option> </select> OR <select name="gender" size="1" id="gender"> <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option> <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option> <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option> </select> OR <select name="gender" id="gender"> <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option> <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option> </select> Honestly man, I've got no idea. The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen. And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol. Any help would be tremendous and greatly appreciated. Thanks very much, Learner P.S Man! hi. I am stuck on the dropdown box value calculation. anyone can help me on this problem ? I appreciate it. I have two dropdown boxes on my webpage, and the values that in the dropdown boxes are retrived from the mysql database. dropdown box "A" displays two differnt values, which are "Chassise_Name" and "Chassise_price", and The dropdown box "B" displays the quantity of the item. Dropdown box "A" Dropdown box "B" SuperServer 7036A-T(Black) $90 3 "Superserver 7036A-T(Black)" is retrieved from the Chassis_Name ; "$90" is retrieved from the chassis_Price; "3" is retrieved from the Quantity in other table. Question: how can I take only the price on the Dropdown box "A" multiply the number in the Dropdown box "B"? Here is my code: Code: [Select] echo "<table border='1'>"; echo "<tr><td>"; $result = mysql_query("SELECT Chassis_Name, Chassis_Price FROM Chassis where Chassis_Form_Factor='ATX'") or die(mysql_error()); echo '<select name="Chassis" onChange="change()">'; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo '<option Name="Chassis">',$row['Chassis_Name'],' ',$row['Chassis_Price'],' ','</option>'; } echo '</select>'; echo "</td>"; echo "<td>"; $result= mysql_query("SELECT Number FROM Quantity") or die(mysql_error()); echo '<select name="Quantity" Value="Quantity" onChange="change()" >'; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result)) { // Print out the contents of each row into a table echo '<option Name="Quantity">',$row['Number'],'</option>'; } echo '</select>'; echo "</td></tr>"; echo "</table>"; ?> Hi, I'm a php newbie, with some mysql experience. I have a mysql database as follows: Database=watch, Table=events - fields id, reportno, sdate, comments What I need is: 1. A dropdown list to display reportno from mysql database. 2. Depending on which reportno I choose, I'd like to open a popup(or separate) page to display the stored information. Tks in advance for any help I have found postings close, but not close enough to find my error. I am looking up data from a MySql table and putting it in a dropdown box on a form. I can select the item, but apparently not really. I am not able to echo it, or post it to a record. I'm sure I am missing something simple, but... Code attached if anyone can show me the errors of my ways. Thank you. Hello to all. I'm new here and i am very glad to be a part of this community. I am new in PHP/mysql but i try to learn as much as i can in php & mysql. And here is the problem: Into a php or html page i want to display 2 dropdown list to act as a filter for displaying a table. I managed to display a table based on option select from first dropdown but i don'y know how to make or activate the second dropdown list to act as a filter, and when i press "result" to display on the same page the results. Thanks a lot. hello. i have an issue where the data stored with an image is not saving to a mysql table. the image data is ok, just not the selections from the dropdown lists. here is the code <?php include ('connect.php'); // Insert any new image into database if(isset($_POST['xsubmit']) && $_FILES['imagefile']['name'] != "") { $fileName = $_FILES['imagefile']['name']; $fileSize = $_FILES['imagefile']['size']; $fileType = $_FILES['imagefile']['type']; $content = addslashes (file_get_contents($_FILES['imagefile']['tmp_name'])); $jeweltype = $_POST['jeweltype']; $jewelsize = $_POST['jewelsize_in']; $jewelcolour = $_POST['jewelcolour_in']; $jewelmaterial = $_POST['jewelmaterial_in']; $jewelgender = $_POST['jewelgender_in']; if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } // Checking file size if ($fileSize < 150000) { mysql_query ("INSERT into jewel_images (name,size,type,content,jeweltype,jewelsize,jewelcolour,jewelmaterial,jewelgender) " . "values ('$fileName','$fileSize','$fileType','$content','$jeweltype','$jewelsize','$jewelcolour','$jewelmaterial','$jewelgender')"); } else { $err = "The Image file to too large!"; } } // Find out about latest image $gotten = mysql_query("select * from jewel_images order by row_id desc"); $row = mysql_fetch_assoc($gotten); $bytes = $row['content']; // If this is the image request, send out the image if ($_REQUEST['pic'] == 1) { header("Content-type: $row[type];"); print $bytes; } ?> <html> <head> <title>Upload an image to a database</title> </head> <body> <font color="#FF3333"><?php echo $err ?></font> <table> <form name="Upload" enctype="multipart/form-data" method="post"> <tr> <td>Upload <input type="file" name="imagefile"><br /> Jewelery Type: <select> <?php $sql="SELECT jeweltype FROM jeweltypes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jeweltype" ><?php echo $data['jeweltype'] ?></option> <?php } ?> </select> <br /> Jewelery Size: <select> <?php $sql="SELECT jewelsize FROM jewelsizes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelsize" ><?php echo $data['jewelsize'] ?></option> <?php } ?> </select> <br /> Jewelery Colour: <select> <?php $sql="SELECT jewelcolour FROM jewelcolours"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelcolour_in" ><?php echo $data['jewelcolour'] ?></option> <?php } ?> </select> <br /> Jewelery Material: <select> <?php $sql="SELECT jewelmaterial FROM jewelmaterials"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelmaterial_in" ><?php echo $data['jewelmaterial'] ?></option> <?php } ?> </select> <br /> Jewelery Gender: <select> <?php $sql="SELECT jewelgender FROM jewelgenders"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelgender_in" ><?php echo $data['jewelgender'] ?></option> <?php } ?> </select> <br /> <input type="submit" name="xsubmit" value="Upload"> </td> </tr> <tr> <td>Latest Image</td> </tr> <tr> <td><img src="?pic=1"></td> </tr> </form> </table> </body> </html> ============================================== here is the query =============================================== <html> <head><title>Your Page Title</title></head> <body> <?php $database="josh_jewel"; mysql_connect ("localhost", "xxxxxxxxx", "yyyyyyyyyyyy"); @mysql_select_db($database) or die( "Unable to select database"); $result = mysql_query( "SELECT jewelcolour FROM jewel_images" ) or die("SELECT Error: ".mysql_error()); $num_rows = mysql_num_rows($result); print "There are $num_rows records.<P>"; print "<table width=400 border=1>\n"; while ($get_info = mysql_fetch_row($result)){ print "<tr>\n"; foreach ($get_info as $field) print "\t<td><font face=arial size=1/>$field</font></td>\n"; print "</tr>\n"; } print "</table>\n"; ?> </body> </html> I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID* is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID. Index.php
<!doctype html>
<?PHP
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass", "Database" =>
"database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
?>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<section id="formaT2" class="formaT2 formContent">
<div class="row">
<div class="col-md-2 col-3 row-color remove-mob"></div>
<div class="col-md-5 col-9 bg-img" style="padding-left: 0;
padding-right: 0;">
<h1>Form</h1>
<div class="rest-text">
<div class="contactFrm">
<p class="statusMsg <?php echo
!empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
<form action="connection.php" method="post">
<div>machinery</div>
<select id="machinery">
<option value="0">--Please Select Machinery--</option>
<?php
// Fetch Department
$sql = "SELECT Machinery FROM T013";
$machanery_data = sqlsrv_query($conn2,$sql);
while($row = sqlsrv_fetch_array($machanery_data) ){
$id = $row['Id'];
$machinery = $row['Machinery'];
// Option
echo "<option value='".$id."' >".$machinery."</option>";
}
?>
</select>
<div class="clear"></div>
<div>Sub Machinery</div>
<select id="sub_machinery">
<option value="0">- Select -</option>
</select>
<input type="submit" name="submit"
id="submit" class="strelka-send" value="Insert">
<div class="clear"> </div>
</form>
</div>
</div>
</div>
</div>
</section>
</script>
<script type="text/javascript">
$(document).ready(function(){
$("#machinery").change(function(){
var machinery_id = $(this).val();
$.ajax({
url:'ajaxfile.php',
type: 'post',
data: {machinery:machinery_id},
dataType: 'json',
success:function(response){
var len = response.length;
$("#sub_machinery").empty();
for( var i = 0; i<len; i++){
var machinery_id = response[i]['machinery_id'];
var machinery = response[i]['machinery'];
$("#sub_machinery").append("<option
value='"+machinery_id+"'>"+machinery+"</option>");
}
}
});
});
});
</script>
</body>
</html>
Ajaxfile.php
<?php
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass",
"Database" => "database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
$machineryID = $_POST['machinery']; // department id
$sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
$result = sqlsrv_query($conn2,$sql);
$machinery_arr = array();
while( $row = sqlsrv_fetch_array($result) ){
$machinery_id = $row['ID'];
$machinery = $row['MachineID'];
$machinery_arr[] = array("ID" => $machinery_id, "MachineID" =>
$machinery);
}
// encoding array to json format
echo json_encode($machinery_arr);
?>
Edited May 6, 2019 by davidd
Hi freaks, I'm new to php first of all. I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database. The code I'm using to populate dropdown: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if(!$con) { die ('Can not connect to : '.mysql_error()); } mysql_select_db("ims",$con); $result=mysql_query("select cat_id,cat_name from category"); echo "<select name=cat>"; while($nt=mysql_fetch_array($result)) { echo "<option value=$nt[cat_id]> $nt[cat_name] </option>"; } echo "</select>"; mysql_close($con); ?> Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]"); Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow. Help on this will be highly appreiable . (Tell me if I'm not clear with my question) |
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