PHP in_array函数在服务器中不起作用

Question

I use this function and Its Fine on Local means when $hexIpStr exist function return false otherwise return true , But when I Deploy my Project on Server this Function only Return True, $arr is 2D array.

function ipChecker($arr, $hexIp)
{
    $hexIpStr = "HEX=" . $hexIp;
    foreach ($arr as $members) {
        if (in_array($hexIpStr, $members)) {
            return false;
        } else {
            return true;
        }
    }
}

The function is for example called with these values:

$hexIp = 'f528764d624db129b32c21fbca0cb8d6';
$arr = array(
  0 => [
    0 => 'FullName=mehdi',
    1 => 'Email=test@gmail.com',
    2 => 'IP=127.0.0.1',
    3 => 'HEX=f528764d624db129b32c21fbca0cb8d6',
    4 => '=>',
    5 => 'Opinion=1 ',
  ],
  1 => [
    0 => '',
  ],
);

no its not answer because its fine in the server i get this

FullName=asd Email=kingblack5230@gmail.com IP=192.119.160.197 HEX=dd029394f038b0775138a23df8d9cddd => Opinion=1 and its right and i use this function to get user ip:

function GetRealIp()
{
    if (!empty($_SERVER['HTTP_CLIENT_IP'])) //check ip from share internet
    {
        $ip = $_SERVER['HTTP_CLIENT_IP'];
    } elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) //to check ip is pass from proxy
    {
        $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
    } else {
        $ip = $_SERVER['REMOTE_ADDR'];
 }
    return $ip;}

Answer 1

f528764d624db129b32c21fbca0cb8d6 ( # ) is md5('127.0.0.1') , this is likely not the value (IP) you expect / that is in $members . You likely expect a public IP.

If you rely on $_SERVER['REMOTE_ADDR'] this can be the issue. If your server is proxying the request to your PHP worker on the same machine (rather than using eg mod_php directly), REMOTE_ADDR will be the IP of the proxy ( 127.0.0.1 / localhost ) rather than the visitor.

In this case, you should dump $_SERVER and check for a variable HTTP_X_FORWARDED_FOR / use that instead of REMOTE_ADDR , the proxy server will likely add this to include the original IP.

Answer 2

Your function "returns early" (and thus breaks the foreach ) after checking the first child for your HEX=... string. This means you only ever check the first member.

Given this $arr , your member is in there, just not at the first position:

array(
  0 => [
    0 => 'FullName=blub',
    1 => 'Email=blub@gmail.com',
    2 => 'IP=1.2.3.4',
    3 => 'HEX=6465ec74397c9126916786bbcd6d7601',
  ],
  1 => [
    0 => 'FullName=mehdi',
    1 => 'Email=test@gmail.com',
    2 => 'IP=127.0.0.1',
    3 => 'HEX=f528764d624db129b32c21fbca0cb8d6',
  ],
)

It would return true , because only in_array('HEX=f528764d624db129b32c21fbca0cb8d6', ['HEX=6465ec74397c9126916786bbcd6d7601']) is checked.

You should restructure your function so it can only " return early" if it found your member, and if not, continues to check all members, and returns the other case only in the end:

function ipChecker($arr, $hexIp)
{
    $hexIpStr = "HEX=" . $hexIp;
    foreach ($arr as $members) {
        if (in_array($hexIpStr, $members)) {
            return false;
        }
    }
    return true;
}