Jeremy:
Have you tried echoing your $sql variable to ensure it has in it what
you expect? In other words, does $PHP_AUTH_USER have the value you need
it to have?
hth
Happy programming!
Chris Mulcahy
> -----Original Message-----
> From: Jeremy Morano [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, September 04, 2001 2:57 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP] option block
>
>
> Hello everyone, I'm having a problem with my option block...
> I'm trying to
> pass the value of uid to the next page. I always get the last
> uid value in
> table user. How can I change this so I get to pass the value
> of the user's
> choice?
>
>
>
>
> $sql = "SELECT user.uid, user.first_name, user.last_name
> FROM user, company
> Where company.uid = user.companyUid and company.company
> = '$PHP_AUTH_USER'
> ORDER BY last_name
> ";
>
>
>
> $result = @mysql_query($sql,$connection) or die("Couldn't
> execute query.");
>
>
> while ($row = mysql_fetch_array($result)) {
> $uid = $row['uid'];
> $first_name = $row['first_name'];
> $last_name = $row['last_name'];
>
> $option_block .= "<option value=\"$uid\">$last_name,
> $first_name</option>";
> }
>
> $display_block = "
>
> <FORM METHOD=\"post\" ACTION=\"show_moduser.php\">
>
> <P><strong>User:</strong>
> <select name=\"uid\">
> $option_block
> </select>
>
> <INPUT TYPE=\"SUBMIT\" NAME=\"submit\" VALUE=\"Select this User\"></P>
> </form>
>
> ";
>
>
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