Reply to Re: Challenge: MySql result to drop down box

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Posted by Johnny on 10/11/12 11:53

Hmmm... u got some explainin to do.
what is wrong $username as a variable name? it's not a reserved word AFAIK
in either PHP or MySql
perhaps you are using it in another part of your code that you did not
disclose....thus making it a challenge for us to help you ;-)
As I said back on the 13th your revised code works for me.


"Jim S" <try@to.guess> wrote in message news:vY5vg.161$XR1.102@fe02.lga...
> Ok--I found the problem. After beating my head on it for a few days I
> realized the $username variable should have actually been $user_name O.o
>
> Thank you for all who helped, I am sure the solutions I did implement
> helped me to avoid other problems.
>
> Jim S wrote:
> > I think I am just missing something silly. I have tested my query and
> > it seems to be functioning fine. I have another similar query that runs
> > just before this one so the issue isn't a connection. If you have any
> > other questions I'd be happy to answer them.
> >
> > Platform: mysql 3.x, php5
> >
> > <?php
> > $res=mysql_query("SELECT job_name FROM oats_jobs_users_laborCode
where
> > user='$username' order by job_name");
> > echo "<select name=jobname> <option default='default'>Choose
> > One</option>";
> > for ($i=0; $row=mysql_fetch_assoc($res); $i++)
> > {
> > echo "<option value='<? echo $row[0];?>'><? echo
> > $row[0];?></option>";
> > }
> > echo "</select>";
> > ?>

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