Quick problem

Encenta.com · 09-12-2005, 10:37 AM

Hey, Can anybody tell me what's wrong with the following? I'd be very grateful

PHP Code:

  $query = mysql_query("SELECT * FROM FORUM_users WHERE username='$sessionusername'");

????: NamePros.com http://www.namepros.com/programming/123073-quick-problem.html
$sql = mysql_query($query) or die(mysql_error());

    $obj = mysql_fetch_object($sql);

    $postcount = $obj->user_posts; 

 
echo "$postcount";

Returns

PHP Code:

  You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #5' at line 1

dgaussin · 09-12-2005, 10:55 AM

I think it's because $sessionusername contains 'Resource id #5' and # should be escaped. Anyway I'm not sure it's normal you have $sessionusernam set to that... Try to display the value before execute the query...

Encenta.com · 09-12-2005, 11:08 AM

Thanks for the reply. I have retyped te code in the following way where $username is the username for the session:

PHP Code:

  $query = mysql_query("SELECT * FROM FORUM_users WHERE username='$username'");

????: NamePros.com http://www.namepros.com/showthread.php?t=123073
$result=mysql_query($query);

????: NamePros.com http://www.namepros.com/showthread.php?t=123073
$num=mysql_num_rows($result);

 
$i=0;

while ($i < $num) {

 
$posts=mysql_result($result,$i,"user_posts");

 
$i++;

}

 
echo "$posts";

This returns

PHP Code:

  Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/local/psa/home/vhosts/darkfx.co.uk/httpdocs/studios/test.php on line 13

moondog · 09-12-2005, 11:51 AM

I think the problem is that you are trying to execute the query twice.

PHP Code:

   $query = mysql_query("SELECT * FROM FORUM_users WHERE username='$username'");

The above line actually takes the sql statement, executes it, and assigns the result to $query. In the next line:

PHP Code:

  $result=mysql_query($query); 

????: NamePros.com http://www.namepros.com/showthread.php?t=123073

You try to do a query again, but this time, you are using the variable $query, which is already a result set from the previous line.

I think the solution might be obvious now. This should work:

PHP Code:

   
$query = "SELECT * FROM FORUM_users WHERE username='$username'";

 
$result=mysql_query($query);

Happy coding.

-Bob

Encenta.com · 09-12-2005, 12:46 PM

This man, he is a legend. I thank you so much

moondog · 09-12-2005, 02:56 PM

Originally Posted by miseria

This man, he is a legend. I thank you so much

HAHA!

09-12-2005, 10:37 AM	THREAD STARTER #1 (permalink)
Encenta.com SQLdumpster.com Join Date: Jun 2005 Location: West Sussex, UK Posts: 573 Follow @MiseriaSK	Quick problem Hey, Can anybody tell me what's wrong with the following? I'd be very grateful PHP Code: `$query = mysql_query("SELECT * FROM FORUM_users WHERE username='$sessionusername'"); ????: NamePros.com http://www.namepros.com/programming/123073-quick-problem.html $sql = mysql_query($query) or die(mysql_error()); $obj = mysql_fetch_object($sql); $postcount = $obj->user_posts; echo "$postcount";` Returns PHP Code: `You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #5' at line 1`

09-12-2005, 11:08 AM	THREAD STARTER #3 (permalink)
Encenta.com SQLdumpster.com Join Date: Jun 2005 Location: West Sussex, UK Posts: 573 Follow @MiseriaSK	Thanks for the reply. I have retyped te code in the following way where $username is the username for the session: PHP Code: `$query = mysql_query("SELECT * FROM FORUM_users WHERE username='$username'"); ????: NamePros.com http://www.namepros.com/showthread.php?t=123073 $result=mysql_query($query); ????: NamePros.com http://www.namepros.com/showthread.php?t=123073 $num=mysql_num_rows($result); $i=0; while ($i < $num) { $posts=mysql_result($result,$i,"user_posts"); $i++; } echo "$posts";` This returns PHP Code: `Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/local/psa/home/vhosts/darkfx.co.uk/httpdocs/studios/test.php on line 13` Last edited by Encenta.com; 09-12-2005 at 11:12 AM.

09-12-2005, 11:51 AM	#4 (permalink)
moondog NamePros Regular Join Date: Jun 2004 Posts: 587	I think the problem is that you are trying to execute the query twice. PHP Code: `$query = mysql_query("SELECT * FROM FORUM_users WHERE username='$username'");` The above line actually takes the sql statement, executes it, and assigns the result to $query. In the next line: PHP Code: `$result=mysql_query($query); ????: NamePros.com http://www.namepros.com/showthread.php?t=123073` You try to do a query again, but this time, you are using the variable $query, which is already a result set from the previous line. I think the solution might be obvious now. This should work: PHP Code: `$query = "SELECT * FROM FORUM_users WHERE username='$username'"; $result=mysql_query($query);` Happy coding. -Bob __________________ Can't wait to be out of this forsaken business. Getting close! :)

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09-12-2005, 10:55 AM	#2 (permalink)
dgaussin Senior Member Join Date: May 2004 Location: France Posts: 1,226 Follow @dgaussin	I think it's because $sessionusername contains 'Resource id #5' and # should be escaped. Anyway I'm not sure it's normal you have $sessionusernam set to that... Try to display the value before execute the query...

09-12-2005, 12:46 PM	THREAD STARTER #5 (permalink)
Encenta.com SQLdumpster.com Join Date: Jun 2005 Location: West Sussex, UK Posts: 573 Follow @MiseriaSK	This man, he is a legend. I thank you so much

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