doucheng1891 2018-10-26 20:48
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为什么这个preg_replace调用返回NULL?

Why does this call return NULL? Is the regex wrong? With the test input it doesn't return NULL. The docs say NULL indicates an error but what error could it be?

$s = hex2bin('5b5d202073205b0d0a0d0a0d0a0d0a20202020202020203a');
// $s = 'test';
$s = preg_replace('/\[\](\s|.)*\]/s', '', $s);
var_dump($s);

// PHP 7.2.10-1+0~20181001133118.7+stretch~1.gbpb6e829 (cli) (built: Oct  1 2018 13:31:18) ( NTS )
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  • drk49438 2018-10-26 20:58
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    Your regex is causing catastrophic backtracking and causing PHP regex engine to fail. You can use preg_last_error() function to check this.

    $r = preg_replace("/\[\](\s|.)*\]/s", "", $s);
    if (preg_last_error() == PREG_BACKTRACK_LIMIT_ERROR) {
        print 'Backtrack limit was exhausted!';
    }
    

    Output:

    Backtrack limit was exhausted!
    

    You are getting NULL return value from preg_replace due to this error. As per PHP doc of preg_replace:

    If matches are found, the new subject will be returned, otherwise subject will be returned unchanged or NULL if an error occurred.


    Fix: You don't need (\s|.) when using s modifier (DOTALL). since dot matches any character including newline when using s modifier.

    You should just use this regex:

    $r = preg_replace('/\[\].*?\]/s', "", $s);
    echo preg_last_error();
    //=> 0
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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