javascript - PHP 到 Ajax Post 使用 JQuery，全局事件有效，但 POST 似乎从未正确设置

Question

我正在尝试使用 $.ajax 调用将数据发布到表单。当输入表单的值发生更改时会触发该事件。为了简单起见，我将所有内容都放在 C9.io 上，以便可以看到我的所有代码。

http://c9.io/noman2000/php-ajax

这是启动该函数的脚本:

<?php 
    // Strictly for a demo, this isn't how it is actually written.
    ?>
    <form>
    <input type="text" id="user-name">
    </form>

    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
    <script> $jQ = jQuery.noConflict();</script>
    <script type="text/javascript" src="ajaxGlobaltests.js"></script>
    <script type="text/javascript">

            $jQ(document).ready(function() {

                /* 
                The DOMNodeInserted is a wonky event at best.  This should be refactored to a function that is exited upon
                completion and that uses SetInterval to prevent it from firing too often and killing page speed.
                But this is just a demo.
                */
                $jQ(document).on('DOMNodeInserted', function(e) {
                        var element = e.target;


                        var userName = $jQ('#user-name');
                        console.log(userName.val());
                        if(userName){
                            userName.on('change', function(){
                                $jQ(this).css('background-color', '#EEE');
                                $jQ.ajax({
                                url: 'submit.php',
                                type: 'post',
                                data: {action: userName.val()}
                                // .serialize() is used for multiple values, but I think this will work for singular.
                                }).done(function(){
                                    console.log('It has been done.');
                                }); // end ajax call
                                return false;
                            });//End change
                        }//End if
                }); //End DOM Node Insert
            }); // End on Ready.
                            </script>

用于捕获此AJAX事件的submit.php脚本:

<?php
    if($_POST['action'])
    {
        $user_info['username'] = 'test data';
          ?>
    <script type="text/javascript">
    // C9 registers and Error pushing between javascript and php, but this works fine in production.
    console.log(<?php echo "The post data is  {$_POST['action']} and the userID is {$user_info['username']}"; ?>);
    </script>


    <?php

        if(class_exists('dbde_mysql')){
            $uName_db = new dbde_mysql;
             $uName_connect = mysqli_connect($uName_db->host, $uName_db->user, $uName_db->pass, $uName_db->db) or die("Error " . mysqli_error($uName_db));
             $uName_str = "UPDATE users SET {$_POST['action']} WHERE id = {$user_id}";
             $sql_res = mysqli_query($uName_db, $uName_str) or die("sql error: $uName_str\n");
     }else{
         // Future proofing.  This ensures that a new database can be created if the class doesn't exist.
         // uName is connected, not a problem.  On this demo, yes, it obviously doesn't work.
         ?>

         <?php
     }


    } // Post action field.
    ?>

我的全局 Ajax 事件寄存器都显示代码开始和执行，没有错误。 .done 事件也会触发控制台消息。然而，submit.php 中似乎没有任何作用。那么我的 $_POST['action'] 变量正确还是我在这里遗漏了其他内容？

Answer 1

尝试将其添加到 PHP 文件的顶部以查看您收到的内容:

//if HTML recd, otherwise handle for type Eg. die(json_encode($_POST['action']));
die( $_POST['action'] );

然后，在您的 .done() 函数中，显示您收到的内容:

.done(function(retd){
    //If HTML received, otherwise handle for type
    console.log('Received: ' + retd);
});

这至少可以让您看到 PHP 端收到的内容。

<小时/>

您可以在下面添加:

$num_rows = mysqli_num_rows($sql_res);
die(' Received rows: ' . $num_rows);