php - 我基于 php 的客户端 IP 跟踪器并不总是触发，为什么？

Question

更新 没关系，它工作正常，我写它是为了如果同一个网站被多次查看，它不会再次添加，而是递增...哈哈，我不知道我自己的代码

我在这里写了这个脚本，因为它写入了两个不同的表，所以它有点多余，但唯一的区别是一个计算浏览量，另一个计算新的 ips。我想我可以重做它们。

问题是跟踪器并不总是被触发。为什么会这样？

<?php

mysqli_report(MYSQLI_REPORT_OFF);
error_reporting(E_ALL);
error_reporting(-1);
ini_set('display_errors',true);

function track() {

$ip = "my ip";

$client_ip = $_SERVER['REMOTE_ADDR'];

require_once(dirname(__FILE__) . DIRECTORY_SEPARATOR.'dbconnect.php');
$link = new mysqli("$servername", "$username", "$password", "$dbname");

$actual_link = "http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]";

if($client_ip!=$ip){

$stmt = $link->prepare('SELECT website_url FROM website_views where website_url=?');
$stmt->bind_param('s',$actual_link);

if($stmt->execute())
{
    // means it exists
    $stmt->bind_result($website_url_from_db);
    if($stmt->fetch())
    {
    $link = new mysqli("$servername", "$username", "$password", "$dbname");
    $id = "";
    $stmt = mysqli_prepare($link, "UPDATE website_views SET views=views+1 where website_url=?");
    $stmt->bind_param('s',$actual_link);
    $stmt->execute();

    $link = new mysqli("$servername", "$username", "$password", "$dbname");
    $stmt = $link->prepare('SELECT client_ip FROM views where client_ip=?');
    $stmt->bind_param('s',$client_ip);
    if($stmt->execute()){

    $link = new mysqli("$servername", "$username", "$password", "$dbname");
    $stmt = mysqli_prepare($link, "UPDATE views SET num_times=num_times+1 where client_ip=? and website_url=?");
    $stmt->bind_param('ss',$client_ip,$actual_link);
    $stmt->execute();

    }else{

    $num_times = 1;
    $link = new mysqli("$servername", "$username", "$password", "$dbname");
    $stmt = mysqli_prepare($link, "INSERT INTO views VALUES (?,?,?,?)");
    $stmt->bind_param('issi',$id,$actual_link,$client_ip,$num_times);
    $stmt->execute();
    }
    }else{
$link = new mysqli("$servername", "$username", "$password", "$dbname");
$id = "";
$website_url = $actual_link;
$views = 1;
$ip = $client_ip;
$num_times=1;

$stmt = mysqli_prepare($link, "INSERT INTO website_views VALUES (?,?,?)");
$stmt->bind_param('isi',$id,$website_url,$views);
$stmt->execute();
$link = new mysqli("$servername", "$username", "$password", "$dbname");
$stmt = mysqli_prepare($link, "INSERT INTO views VALUES (?,?,?,?)");
$stmt->bind_param('issi',$id,$actual_link,$client_ip,$num_times);
$stmt->execute();
}

}
}
}

track();

?>

Answer 1

尝试

$ipAddress = $_SERVER['REMOTE_ADDR'];
if (array_key_exists('HTTP_X_FORWARDED_FOR', $_SERVER)) {
    $ipAddress = array_pop(explode(',', $_SERVER['HTTP_X_FORWARDED_FOR']));
}

另外，检查 print_r( $_SERVER ) 以确保它提供您所期望的。