php - 简单的ajax POST不执行php函数

Question

我尝试使用ajax执行php函数，但该函数没有响应， 如果我使用 chrome 查看网络结果，我会得到这个

编辑:我使用 WAMP 运行此本地

manageusers.php
/geopedia/core
POST
200
OK
text/html
jquery-1.10.1.min.js:6
Script
850 B
625 B
7 ms
4 ms

那么这不应该起作用吗？

Ajax函数

    $(document).ready(function() {
    $('#facebook').click(function() { 
        var first_name = $('#first_name').val();
        var last_name = $('#last_name').val();
        var username = $('#username').val();

        $.ajax({ 
            url: 'core/manageusers.php',
            type: 'POST',
            data: {
                facebook:first_name,
                last_name:last_name,
                username:username

            },
        });
    });
}); 

事件处理程序

if(isset($_POST['facebook'])){ 
echo("inside");
include_once('core/class.users.php');
$users = new Users();

$first_name = $_POST['first_name'];
$last_name  = $_POST['last_name'];
$username   = $_POST['username'];

$fb_user = $users->fb_user($first_name,$last_name,$username);
}

Answer 1

这可能会帮助您找到错误:

        $.ajax({ 
            url: 'core/manageusers.php',
            type: 'POST',
            data: {
                facebook:first_name,
                last_name:last_name,
                username:username,

            },
            success: function(mydata) {
                alert(mydata);
            }
        });
    });
}); 

事件处理程序

<?php
//if(isset($_POST['facebook'])){ 
//echo("inside");
//include_once('core/class.users.php');
//$users = new Users();

$first_name = $_POST['first_name'];
$last_name  = $_POST['last_name'];
$username   = $_POST['username'];

echo 'You sent: First: [' .$first_name. ']<br>'Last: [' .'$last_name. '] etc.';

//$fb_user = $users->fb_user($first_name,$last_name,$username);