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);
}
}

Change $email_error to $mysqlerror and hopefully some error message will
appear.

Beauford.2005 wrote:
> if ($mysqlerror) {
> $error = "$d_base_error$email_error";
> include("trades-input.php");
> exit;
> }

I tried that, but there is no error to stop the script at this point.
The script continues past this point and to the end of the script -
there are just no values in the variables for the second query, which
obviously gives me the wrong results.

-----Original Message-----
From: Marek Kilimajer [mailto:kilimajer@webglobe.sk]
Sent: July 16, 2003 1:18 PM
To: Beauford.2005
Cc: PHP
Subject: Re: [php] MySQL query problems


Change $email_error to $mysqlerror and hopefully some error message will

appear.

Beauford.2005 wrote:
> if ($mysqlerror) {
> $error = "$d_base_error$email_error";
> include("trades-input.php");
> exit;
> }



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But what's your prob ?
When you say the second one seems to never be executed ...
Does the line:
$row = mysql_fetch_array($result2);
launches a Php Error ?

And pay attention, because you're using mysql_fetch_array and
mysql_fetch_row, be sure that
you are not treating the result in the same way ;-)

"Beauford.2005" <beauford.2005@rogers.com> a écrit dans le message de news:
000001c34bb5$5772a200$6401a8c0@p1...
> 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);
> }
> }
>

Not sure what the problem was, but I dug up a similar script I used for
something else and modifed it for this project, and voila. I then put
the two side by side and I still can't see where the problem is.

Thanks for the input.

-----Original Message-----
From: Nomadeous [mailto:nomadeous@free.fr]
Sent: July 16, 2003 5:31 PM
To: php-general@lists.php.net
Subject: [php] Re: MySQL query problems


But what's your prob ?
When you say the second one seems to never be executed ...
Does the line:
$row = mysql_fetch_array($result2);
launches a Php Error ?

And pay attention, because you're using mysql_fetch_array and
mysql_fetch_row, be sure that you are not treating the result in the
same way ;-)

"Beauford.2005" <beauford.2005@rogers.com> a écrit dans le message de
news: 000001c34bb5$5772a200$6401a8c0@p1...
> 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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