How to populate html combo box with mysql data

Question

Please help, I'm having trouble loading mysql data on the combo box. The data that I'm loading is 1 column from a table. Here is my current code, and it crashed firefox for some reason:

<td colspan=”2″>Religion</TD>
<td>
<select name="REL" onClick="submitCboSemester();">
<?php
$query_disp="SELECT * FROM Religion ORDER BY RID";
$result_disp = mysql_query($query_disp, $conn);
while($query_data = mysql_fetch_array($result_disp))
{
?>
<option value="<? echo $query_data["RID"]; ?>"<?php if ($query_data["RID"]==$_POST['REL']) {?>selected<? } ?>><? echo $query_data["RELIGION"]; ?></option>
<? } ?>
</select>
</td>

The column is RELIGION and it ID is RID How do I populate the combo box with all the data in the column RELIGION

In other words, it works in browsers other than Firefox? Sometimes you don't need to be too specific ;) — BalusC
– BalusC, Commented May 5, 2010 at 2:30

Jacob Relkin · Accepted Answer · 2010-05-05 03:24:22Z

4

Chances are, you don't have short_open_tag enabled in php.ini, this is probably what was causing your code to fail.

Also, you should output selected="selected", not just selected

Try this:

<td colspan=”2″>Religion</TD>
<td>
<select name="REL" onClick="submitCboSemester();">
<?php
$query_disp="SELECT * FROM Religion ORDER BY RID";
$result_disp = mysql_query($query_disp, $conn);
while($query_data = mysql_fetch_array($result_disp))
{
?>
<option value="<?php echo $query_data["RID"]; ?>"<?php if ($query_data["RID"]==$_POST['REL']) {?> selected="selected"<? } ?>><?php echo $query_data["RELIGION"]; ?></option>
<? } ?>
</select>
</td>

edited May 5, 2010 at 3:24

answered May 5, 2010 at 2:17

Jacob Relkin

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Comments

Usha · Accepted Answer · 2011-07-19 13:47:17Z

<html>
<title>
demo </title>
<body>
<?
// Load field datas into List box
$cn=mysql_connect("localhost",root) or die("Note: " . mysql_error());
echo "Conn ok<br>";
$res=mysql_select_db("test",$cn) or die("Note: " . mysql_error());
echo " Database opened<br>";
$q="insert into usha2 (`sno`, `city name`) values (NULL, 'Nagercoil');";
//$q="INSERT into `test`.`usha2` (`sno`, `city name`) VALUES (NULL, 'Delhi');";
$res=mysql_query($q) or die("Note: " . mysql_error());
echo " record inserted<br>";
$res=mysql_query("select * from usha2;") or die("Note: " . mysql_error());
echo " qry executed<br>";
?>
<select name="pno" size=1>
<?
while($ri = mysql_fetch_array($res))
{
echo "<option value=" .$ri['city name'] . ">" . $ri['city name'] . "</option>";
}
echo "</select> "
?>
</body>

mexique1 · Accepted Answer · 2010-05-05 19:13:56Z

1

Extract the data before displaying it !

And also, use the PHP Alternative Syntax when mixing PHP & HTML :

<?php
$query_disp="SELECT * FROM Religion ORDER BY RID";
$result_disp = mysql_query($query_disp, $conn);
$options = array();
while ($query_data = mysql_fetch_array($result_disp)) {
    $options[$query_data["RID"]] = $query_data["RELIGION"];
}
?>

<select name="REL" onClick="submitCboSemester();">
<?php foreach ($options as $key => $value) : ?>
    <?php $selected = ($key == $_POST['REL']) ? 'selected="selected"' : ''; ?>
    <option value="<?php echo $key ?>" <?php echo $selected ?>>
    <?php echo $value ?>
    </option>
<?php endforeach; ?>
</select>

answered May 5, 2010 at 19:13

mexique1

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Comments

Joshua Pinter · Accepted Answer · 2010-05-05 03:16:39Z

0

Looks like there is no space between the end of your value and the selected.

answered May 5, 2010 at 3:16

Joshua Pinter

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Comments

castor520 · Accepted Answer · 2014-03-27 19:40:51Z

0

When you pull the data from your database you should use the following format as the format you are currently using is being phase out as MySqli advances

    $religion = $db->query("SELECT Religion FROM TableName ORDER BY RID");

$db is a connection to the database created as such

    $db = new Mysqli($server, $username, $password, $database);

Just set the variable to the correct value.

answered Mar 27, 2014 at 19:40

castor520

1

1 Comment

boisvert Over a year ago

The point is correct but is not an answer to the question: why does the code crash firefox?

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