PHP - 检查字符串中是否包含非法字符

Question

In JS you can do:

 var chs = "[](){}";
 var str = "hello[asd]}";
 if (str.indexOf(chs) != -1) {
    alert("The string can't contain the following characters: " + chs.split("").join(", "));
 }

How can you do this in PHP (replacing alert with echo)?

I do not want to use a regex for the simplicity of what I think.

EDIT:

What I've tried:

 <?php
    $chs = /[\[\]\(\)\{\}]/;
    $str = "hella[asd]}";
    if (preg_match(chs, str)) {
       echo ("The string can't contain the following characters: " . $chs);
    }
 ?>

Which obviously doesn't work and idk how to do it without regex.

Answer 1

In php you should do this:

$string = "Sometring[inside]";

if(preg_match("/(?:\[|\]|\(|\)|\{|\})+/", $string) === FALSE)
{
     echo "it does not contain.";
}
else
{
     echo "it contains";
]

The regex says check to see any of the characters are inside the string. you can read more about it here:

http://en.wikipedia.org/wiki/Regular_expression

And about PHP preg_match() :

http://php.net/manual/en/function.preg-match.php

Update:

I have written an updated regex for this, which captures the letters inside:

$rule = "/(?:(?:\[([\s\da-zA-Z]+)\])|\{([\d\sa-zA-Z]+)\})|\(([\d\sa-zA-Z]+)\)+/"
$matches = array();
if(preg_match($rule, $string, $matches) === true)
{
   echo "It contains: " . $matches[0];
}

It returnes something like this:

It contains: [inside]

I have changed the regex only which becomes:

$rule = "/(?:(?:(\[)(?:[\s\da-zA-Z]+)(\]))|(\{)(?:[\d\sa-zA-Z]+)(\}))|(\()(?:[\d\sa-zA-Z]+)(\))+/";

// it returns an array of occurred illegal characters

It now returns [] for this "I am [good]"