Have you checked the query in an editor to see if they are working?
Just substitute known values in the place of the variables you are
using.
Jef
> -----Original Message-----
> From: C. Reeve [mailto:creeve.2005@stripped]
> Sent: Wednesday, July 16, 2003 10:16 AM
> To: mysql@stripped
> Subject: MySQL query problems
>
>
> Hi,
>
> Not sure if the problem here is PHP or MySQL, but here we go. I am
> trying to do two queries on a database - one after the other, but the
> second one never seems to get executed. The two queries are identical
> except for two variables. I have checked my form and they are correct
> and are being sent to the PHP script the way they should be.
>
> The variables are $playerto, $playerfrom, $nameto, $namefrom. The 'TO'
> query is the one that doesn't work.
>
> Here is my code. Any help is appreciated.
>
> if ($namefrom != $nameto) {
> if ($playerfrom != $playerto) {
>
> include("2004server.inc");
> if($error) {
> include("trades-input.php");
> exit;
> }
>
> $query1 = "select manager.idn, manager.total,
> roster.idp, position, points from roster join reference
> join manager where manager.idn=reference.idn and
> reference.idp=roster.idp and manager.idn like '$namefrom' and
> roster.idp like '$playerfrom'";
>
> $result1 = mysql_query($query1) or $mysqlerror =
> mysql_error();
> if ($mysqlerror) {
> $error = "$d_base_error$email_error";
> include("trades-input.php");
> exit;
> }
>
> $line = mysql_fetch_row($result1);
>
> mysql_free_result($result1);
>
> $query2 = "select manager.idn, manager.total,
> roster.idp, position, points from roster join reference
> join manager where manager.idn=reference.idn and
> reference.idp=roster.idp and manager.idn like '$nameto' and
> roster.idp like '$playerto'";
>
> $result2 = mysql_query($query2) or $mysqlerror =
> mysql_error();
> if ($mysqlerror) {
> $error = "$d_base_error$email_error";
> include("trades-inputs.php");
> exit;
> }
>
> $row = mysql_fetch_array($result2);
>
> mysql_free_result($result2);
> }
> }
>
>
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