On Tue, 2002-02-05 at 16:23, Michael O'Neal wrote:
> Hi.
>
> I'm working on an "edit" page where the pull down menu is populated from a
> database. I can't figure out how to print "selected" when that particular
> record is the one associated with the current ID. Can anyone help?
>
> Here's my current code:
>
> <!-- Dynamic Career Menu-->
>
> <?php
>
> include("common.inc");
>
>
> $connection = @mysql_connect("$db_host","$db_user","$db_pass") or
This isn't your problem, but you do not really need to quote the above
variables.
> die("Couldn't Connect.");
> $db = @mysql_select_db($db_name, $connection) or die("Couldn't select
> database.");
>
>
> $sql ="SELECT job_id,category,description,job_id AS my_job_id FROM
> $cat_table_name";
> $result = @mysql_query($sql, $connection) or die("Error #". mysql_errno() .
> ": " . mysql_error());
>
> while ($row = mysql_fetch_array($result)) {
>
> $job_id=$row['job_id'];
> $category=$row['category'];
> $description=$row['description'];
> $my_job_id=$row['my_job_id'];
>
>
> echo "<option value=\"$job_id\" ";
>
> if ($job_id == '$my_job_id') {
^^^^^^^^^^^^
This is your problem: remove the quotes. PHP does not interpolate
single-quoted strings:
http://www.php.net/manual/en/language.types.string.php#language.types.string.syntax.single
[snip]
Hope this helps,
Torben
> --
>
> Michael O'Neal
> Web Producer
--
Torben Wilson <[EMAIL PROTECTED]>
http://www.thebuttlesschaps.com
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