Quick problem

[SiKing]

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

$query = mysql_query("SELECT * FROM FORUM_users WHERE username='$sessionusername'");
$sql = mysql_query($query) or die(mysql_error());
    $obj = mysql_fetch_object($sql);
    $postcount = $obj->user_posts; 

echo "$postcount";

Returns

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]

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...

 

[SiKing]

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

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

$i=0;
while ($i < $num) {

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

$i++;
}

echo "$posts";

This returns

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]

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

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

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

$result=mysql_query($query);

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:

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

$result=mysql_query($query);

Happy coding.

-Bob

 

[SiKing]

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

 

[moondog]

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

HAHA! :bingo: