用于排除@＃$％^＆*的PHP代码

How can I exclude @ # $%^&* from a given string?

Try this:

$str= preg_replace("/[^a-zA-Z0-9_\-\/=]+/", "", 'your string here');

This allows only common acceptable chars. Excludes the chars you mentioned.

Or you can try this too:

$str = '@ # $%^&*';
$new_str = str_replace(array('@', '#', '$', '%', '^', '&', '*'), '', $str);
print $new_str;

You can specify multiple characters to replace using str_replace:

$s= 'what a @bad #$%^ string *';
$s= str_replace(array('@', '#', '$', '%', '^', '%', '*'), '', $s);
echo($s);

This will ouput:

what a bad  string

$thisIsaVeryBadStringIndeed = "@wh#at %a b^a&d @#$%^&*string";
$unWantedBadCharacters = "@#$%^&*";

$chars = preg_split('//',$unWantedBadCharacters);

for ($i=0;$i<strlen($unWantedBadCharacters);++$i)
    $pairs[$unWantedBadCharacters{$i}] = '';

$stringWithoutBadCharacters = strtr($thisIsaVeryBadStringIndeed,$pairs);

This is one of the faster methods. If you only create the pairs array once.

A simple regular expression would be one way to do it:

$str = preg_replace('/[@#$%^&*]/', '', $str);