removing characters from string if it's not in the regex

basically i want to remove unwanted characters from string - i have a list of valid characters in a regex or map ( or whatever that is )

基本上我想从字符串中删除不需要的字符 - 我有一个正则表达式或地图中的有效字符列表（或任何其他）

$permitted_uri_chars = ' ) ( ( ) a-z 0-9~%.+:_\-  δ ο κ ι μ ή χ ό ν';

right now i'm using this code which seems slow and messy and above all i have to write every single character ( i cant do a-z 0-9 )

现在我正在使用这个看似缓慢而凌乱的代码，最重要的是我必须编写每一个字符（我不能做a-z 0-9）

 $string = "this is a test";
 $permitted_uri_chars = ' ) ( ( ) a b c d e z 0 1 2 3 4 9 _ - δ ο κ ι μ ή χ ό ν'; 
 $permitted_uri_chars = explode(' ' , $permitted_uri_chars );
 $unwanted = array();

 for($i = 0 ; $i < strlen($string) ; $i++)
 {
     if(!in_array($string[$i] , $permitted_uri_chars ))
     $unwanted[] = $string[$i] ;
 }
 $string = str_replace($unwanted, '-' , $string);
 echo $string;

2 个解决方案

#1

preg_replace() is probably the best tool for the job:

preg_replace（）可能是这项工作的最佳工具：

$string = preg_replace('/[^\da-z~%\.\+:_\-δοκιμήχόν]/i', '', $string);

#2

Use preg_replace() here instead, using a negated character class.

在这里使用preg_replace（），使用否定的字符类。

Note: Not clear if parentheses are permitted, but you can remove them if you need to. I included them since you have multiple in $permitted_uri_chars.

注意：不清楚是否允许使用括号，但如果需要，可以删除它们。我把它们包括在内，因为你在$ allowed_uri_chars中有多个。

$string = preg_replace('/[^a-z0-9δοκιμήχόν_()%~.:+-]/i', '', $string);

#1