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具有无效过滤器的filter_input

[英]filter_input with invalid filter

I just came across this snippet in our code base: 我刚刚在我们的代码库中遇到了以下片段:

$token = filter_input(INPUT_GET, 'token', FILTER_VALIDATE_STRING);
if ($token === false || $token === null) {
    die('invalid token');
}

FILTER_VALIDATE_STRING is not a valid filter type. FILTER_VALIDATE_STRING不是有效的过滤器类型。 Does that mean it would just revert to FILTER_DEFAULT , as an unrecognized filter has been passed in? 这是否意味着由于FILTER_DEFAULT了无法识别的过滤器而将其恢复为FILTER_DEFAULT

You'll get a warning that an undefined constant is being used, and PHP will go ahead and turn it into a string literal. 您将收到一条警告,指出正在使用未定义的常量,PHP会继续将其转换为字符串文字。 The best thing to do is use a constant that is defined in the documentation . 最好的办法是使用文档中定义的常量。

All GET and POST vars are strings, and as you have noted FILTER_VALIDATE_STRING is not a defined constant. 所有GET和POST变量都是字符串,并且您已经注意到FILTER_VALIDATE_STRING不是定义的常量。 If you enable error reporting you will see: 如果启用错误报告,您将看到:

Notice: Use of undefined constant FILTER_VALIDATE_STRING - assumed 'FILTER_VALIDATE_STRING' 注意:使用未定义的常量FILTER_VALIDATE_STRING-假定为'FILTER_VALIDATE_STRING'

Warning: filter_input() expects parameter 3 to be long, string given 警告:filter_input()期望参数3长,并给出字符串

So filter_input() will return NULL just like any other function that is not passed required arguments. 因此filter_input()将返回NULL ,就像未传递必需参数的任何其他函数一样。

This is assuming that whoever wrote this did not also define FILTER_VALIDATE_STRING . 假设撰写此内容的人也未定义FILTER_VALIDATE_STRING

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