RE: [PHP] Populating HTML List boxes From DB
| From: | Peter Houchin | Date: | Tue, 17 Apr 2001 02:05:51 +0000 |
| Subject: | RE: [PHP] Populating HTML List boxes From DB | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-48863@lists.php.net to get a copy of this message | ||
Thanks Every one it now works like a charm (o:
-----Original Message-----
From: Jason Murray [mailto:Jason.Murray@melbourneit.com.au]
Sent: Tuesday, April 17, 2001 11:55 AM
To: 'Peter Houchin'; Jason Murray; Php-General@Lists. Php. Net
Subject: RE: [PHP] Populating HTML List boxes From DB
> <select name="device">
> <?php
> $query ="SELECT DISTINCT device FROM 3backup ORDER BY device";
> $rs =mysql_query($query);
$rs is now the result set for your SQL.
> $device = mysql_result($rs, 'device');
This is wrong. Check the manual page for how you use mysql_result.
http://www.php.net/mysql-result
> while($row = mysql_fetch_object($rs)){
> echo '<option value="';
> echo $device;
> echo '">';
> echo $device;
> echo '</option>\n';
> }
This is odd. Having pulled out the device (or tried to), you're now creating
an object and then not using it.
Try this:
<select name="device">
<?php
$query = "SELECT DISTINCT device FROM 3backup ORDER BY device";
$rs = mysql_query($query);
while($row = mysql_fetch_row($rs))
{
echo '<option
value="'.$row[device].'">'.$row[device].'</option>\n';
}
?>
</select>
Now, in order for this to produce more than one option in the
dropdown box, you have to be sure that your SELECT DISTINCT does
in fact return more than one row.
Jason
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