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jQuery 发布时间：2022-04-19 发布网站：大佬教程 code.js-code.com

大佬教程收集整理的这篇文章主要介绍了如何通过jQuery.ajax将JavaScript数组发送到PHP？，大佬教程大佬觉得挺不错的，现在分享给大家，也给大家做个参考。

参见英文答案 > Convert js Array() to JSon object for use with JQuery .ajax 5个
我正在尝试通过jQuery.ajax将JavaScript数组发送到PHP 页面,但该数组仅使用空值发送.

如果我在Chrome上打开F12控制台并检查JS对象,它就在那里.全部填满了.但是当我使用ChromePHP工具记录PHP变量时,它只显示空白值(同样,如果我遍历PHP-array,回显它的值,我得到所有空白).

我在这里很困惑.

这是我的示例代码：

<?PHP 
include 'ChromePHP.PHP';
if (isset($_GET['newUsers'])) {
    $newUsers = $_GET['newUsers'];
    ChromePHP::log($newUsers);
} else { ?>

<html>
<body>

    <script src="js/jquery-2.0.3.min.js"></script>
    <script type="text/javascript">
        var newUsers = [];

        newUser = [];
        newUser['nome'] = 'alvaro';
        newUser['idade'] = '34';
        newUsers.push(newUser);

        newUser1 = [];
        newUser1['nome'] = 'bia';
        newUser1['idade'] = '7';
        newUsers.push(newUser1);

        newUser2 = [];
        newUser2['nome'] = 'alice';
        newUser2['idade'] = '2';
        newUsers.push(newUser2);

        $.ajax({
            url: "TESTCookie.PHP",type: "GET",data: {
                'newUsers[]': newUsers
            }
        });

    </script>
</body>
</html>
<?PHP } ?>

根据第一条评论更新.现在我得传递对象,但不知道如何读取它的属性.已经尝试了$user [‘nome’]没有结果.

<?PHP 
include 'ChromePHP.PHP';

if (isset($_POST['newUsers'])) {

    $newUsers = $_POST['newUsers'];

    foreach ($newUsers as $user) {
        # code...
        # HOW DO I READ THE nome AND idade VALUES HERE?
    }

} else { ?>

<html>
<body>

    <script src="js/jquery-2.0.3.min.js"></script>
    <script type="text/javascript">
        //var newUsersObj = {};
        var newUsers = [];

        newUser = {};
        newUser['nome'] = 'alvaro';
        newUser['idade'] = '34';
        newUsers.push(newUser);

        newUser1 = {};
        newUser1['nome'] = 'bia';
        newUser1['idade'] = '7';
        newUsers.push(newUser1);

        newUser2 = {};
        newUser2['nome'] = 'alice';
        newUser2['idade'] = '2';
        newUsers.push(newUser2);

        $.ajax({
            url: "TESTCookie.PHP",type: "POST",data: {
                'newUsers[]': newUsers
            },success: function () {

            },error: function () {

            }
        });

    </script>
</body>
</html>
<?PHP } ?>