[Php] [Mysql] dziwny problem.. Opcje

[raxon]

Mam oto taki problem podczas próby rejestracji dostaje taki oto błąd..

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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 'key='37267459315397548983', status='0', ban='0', points='0', rang1='0', rang2='' at line 3
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niby wszystko działa a jednak nie (IMG:style_emoticons/default/sad.gif)

może coś tu pomoże

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@mysql_query("insert into members set username='$username', password='$password', email='$email', nickwmc='$nickwmc', 
sex='$sex', dud='$dud', dum='$dum', dur='$dur', gadugadu='$gadugadu', skype='$skype', steam='$steam', youtube='$youtube',
website='$website', ip='$ip', dateregister='$dateregister', key='$key', status='0', ban='0', points='0', 
rang1='0', rang2='0', rang3='0', rang4='0', rang5='0', avatar='upload/avatars/avatar.gif'")or die(mysql_error());
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a linia 3 to connection();

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<?php
require "config.php";
connection();
?>
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[raxon]

są tylko te co w tej części skryptu..
próbowałem je też zmieniać ale nic.

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$key = "".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."".rand(0,9)."";
 
$wiadomosc = "register.php?step=3&key=".$key;
 
$zaporaLoginQuery = mysql_query("select * from members where username='$username'")or die(mysql_errror());
$zaporaEmailQuery = mysql_query("select * from members where email='$email'")or die(mysql_errror());
 
 
if(mysql_num_rows($zaporaLoginQuery) <= 0){
if(mysql_num_rows($zaporaEmailQuery) <= 0){
 
 
@mysql_query("insert into members set username='$username', password='$password', email='$email', nickwmc='$nickwmc', 
sex='$sex', dud='$dud', dum='$dum', dur='$dur', gadugadu='$gadugadu', skype='$skype', steam='$steam', youtube='$youtube',
website='$website', ip='$ip', dateregister='$dateregister', key='$key', status='0', ban='0', points='0', 
rang1='0', rang2='0', rang3='0', rang4='0', rang5='0', avatar='upload/avatars/avatar.gif'")or die(mysql_error());
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Ten post edytował raxon 27.07.2011, 11:16:14