在ajax响应中传递JavaScript函数

I'm trying to return a callback from an AJAX submitted form. The user submits a form, the server processes and returns the valid response, i.e. an error message and also a JavaScript function that could perform an action. I'm using Zepto.js faling back to jQuery depending on browser.

我试图从AJAX提交的表单返回回调。用户提交表单、服务器进程并返回有效响应，即错误消息和可以执行操作的JavaScript函数。我用Zepto。js通过浏览器返回到jQuery。

My ajax request is:

我的ajax请求是:

$.ajax({
    success: function(data, status, xhr) {
        data.callback();
    },
    url: form.attr('action'),
    data: form.serialize(),
    dataType: 'json'
});

On the server I want to return something like:

在服务器上，我想返回如下内容:

// PHP code
?>
{
    return: false,
    error: 'Sorry, we couldn’t find an account with that username or password.',
    callback: function() {
        console.log('this is the callback');
    }
}
<?php
// more PHP code

When returned to the browser callback function should fire. I want the server to return the callback so I can use the same JavaScript code and have it respond accordingly to the server response.

当返回到浏览器回调函数时，应该触发。我希望服务器返回回调，以便我可以使用相同的JavaScript代码，并让它相应地响应服务器响应。

Would I need to change the dataType to script? However I thought this was just for loading .js files, not blocks of code.

我需要将数据类型更改为脚本吗?但是，我认为这只是用于加载.js文件，而不是代码块。

Any help appreciated.

任何帮助表示赞赏。

2 个解决方案

#1

The general feeling here is I am approaching this in the wrong way. So revised code:

这里的普遍感觉是，我用了错误的方式来处理这个问题。所以修改后的代码:

$.ajax({
    success: function(data, status, xhr) {
        var callback = data['callback'];
        callback();
    },
    url: form.attr('action'), // in this example it's badLogin
    data: form.serialize(),
    dataType: 'json'
});
// callback specified in PHP
badLogin: function() {
    console.log('bad login');
}

And my PHP

和我的PHP

if (!$valid) {
?>
{
    "return": false,
    "error": "Sorry, we couldn’t find an account with that username or password.",
    "callback": "badLogin"
}
<?php
}

Thanks for pointing me in the right direction.

谢谢你给我指出了正确的方向。

#2

You can always return the code as a string and use eval() if you are absolutely sure that the string will always be correct and no code can be injected.

如果您绝对确信该字符串始终是正确的，并且不能注入任何代码，那么可以将代码作为字符串返回并使用eval()。

#1