Re: Challenge: MySql result to drop down box

I used a different table and so different column names and your last
version modifed only that way works for me just fine
otoh i'm using Platform: mysql 3.x, php4


"Jim S" <try@xxxxxxxx> wrote in message news:2Zstg.163$vk.64@xxxxxxxxxxx
I have made the two changes you suggested but am still not receiving a
result set. Let me repost my modified code.

<?php
$res=mysql_query("SELECT DISTINCT job_name FROM
oats_jobs_users_laborCode where user='$username' order by job_name");
echo "<select name=jobname> <option default='default'>Choose
One</option>";
for ($i=0; $row=mysql_fetch_row($res); $i++)
{
echo "<option value='$row[0]'>$row[0]</option>";
}
echo "</select>";
?>

And thank you for the mysql_fecth_rows command--it was helpful in other
places as well :-)

Rik wrote:
And now I see this:

echo "<option value='<? echo $row[0];?>'><? echo
$row[0];?></option>";

That will not echo what you want it to echo.
Think about it, <?php ?> tags aren't nestable, why should they?

echo "<option value='$row[0]'>$row[0]</option>";

Grtz,


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