使用AJAX将变量传递给php脚本

I'm not sure I understand how ajax works even though I read a lot about it. I want to run the following php if a button is clicked, without loading the page:

我不确定我是否理解ajax是如何工作的，尽管我读了很多关于它的文章。如果单击按钮，我想运行以下php，而不加载页面:

unset($checkout_fields['billing']['billing_postcode']);

So I put the following:

所以我写了如下内容:

jQuery(document).ready(function($) {
    $('.keep-buying-wrapper').click(function(){

        $.ajax({
            url: "url-to-the-script.php",
            method: "POST",
            data: {'checked': checked},
            success: alert('success!'),

        });
    });
});

And in my php script:

在我的php脚本中:

if( $_POST['checked'] == 'checked' ){
        unset($checkout_fields['billing']['billing_postcode']);
}

However nothing happen. even though the success alert is popping, POST['checked'] is null.

但是没有发生。即使成功警报是弹出的，POST['checked']也是空的。

Is the ajax supposes to trigger the php script?
What if I want to send some variable to functions.php?

ajax假设会触发php脚本吗?如果我想将某个变量发送到函数。php呢?

1 个解决方案

#1

The problem is that you need to serialize post data first.

问题是您需要首先序列化post数据。

HTML code (The id and name of checkbox is "billing_postcode"):

HTML代码(复选框的id和名称为“billing_postcode”):

<input type = "checkbox" id = "billing_postcode" name = "billing_postcode">

JS Code

JS代码

$(document).on('click', '#billing_postcode', function (event) {

    var data = $("#billing_postcode").serializeArray();
    $.ajax({
        url: "url-to-the-script.php",
        method: "POST",
        data: data,
        success: alert('success!'),
    })
});

You will get value in post array on server side and enter code here in php script:

您将在服务器端获取post数组的值，并在这里输入php脚本中的代码:

if($_POST['billing_postcode'])
    unset($checkout_fields['billing']['billing_postcode']);

#1