how do I fillter chars like ` & ^ %
Filltering
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Ok here is what I got
PHP Code:$text = preg_replace('/[^[:alnum:]\-\.\_\^]/', '', $username);
if($text == true)
{
die ("Unsupported character(s)");
}
Last edited by ripkk2tfk; 09.01.10, 13:14.Visit my site
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to prevent mysql injection use: mysql_real_escape_string
PHP: mysql_real_escape_string - Manual
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one other way
put this in your confg.php or ini.php
PHP Code:function scharin($word)
{
$chars = "abcdefghijklmnopqrstuvwxyz0123456789";
for($i=0;$i<strlen($word);$i++)
{
$ch = substr($word,$i,1);
$nol = substr_count($chars,$ch);
if($nol==0)
{
return true;
}
}
return false;
}
put this line in your register.php and you replace $login to your user name etc
PHP Code:if(scharin($login) == true) $error=$error.'Invalid Characters in Nick!<br/>';
PHP Code:if(strlen($login)<3) $error=$error.'Nick should be Minimum of 3 words!<br/>';
if you need more and strong filter send me pm will tell youLast edited by GiLL; 09.01.10, 16:19.left wap stuff
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Originally posted by GiLL View Postput this in your confg.php or ini.php
PHP Code:function scharin($word)
{
$chars = "abcdefghijklmnopqrstuvwxyz0123456789";
for($i=0;$i<strlen($word);$i++)
{
$ch = substr($word,$i,1);
$nol = substr_count($chars,$ch);
if($nol==0)
{
return true;
}
}
return false;
}
put this line in your register.php and you replace $login to your user name etc
PHP Code:if(scharin($login) == true) $error=$error.'Invalid Characters in Nick!<br/>';
PHP Code:if(strlen($login)<3) $error=$error.'Nick should be Minimum of 3 words!<br/>';
if you need more and strong filter send me pm will tell you
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