jsonpCallback函数不起作用 [英] jsonpCallback function not working

问题描述

更新1:

我刚刚从jquery 1.4.4升级到1.6.1.这对原始问题中的脚本有何影响?

I've just upgraded from jquery 1.4.4 to 1.6.1. How does that effect the script in the original question?

原始问题:

就像我测试的那样，我做到了:

Just as I test, I did:

$(document).ready(function() {
    get_jsonp_feed();

    function get_jsonp_feed() {
        $.ajax({
            url: 'http://www.remote_host.co.uk/feed.php',
            type: 'GET',
            dataType: 'jsonp',
            jsonp: 'callback',
            jsonpCallback: 'jsonpCallback',
            error: function(xhr, status, error) {
                alert("error");
            },
            success: function(jsonp) { 
                alert("success");
            }
        });
    }

    function jsonpCallback(data){
        alert("jsonpCallback");
    }
});

我希望收到2条警报，第一个显示success，第二个显示jsonpCallback.但是我只得到第一个警报success.为什么第二个警报没有显示?

I was expecting to get 2 alerts, the first showing success and the second showing jsonpCallback. But I am only getting the first alert success. Why is the second alert not showing up?

推荐答案

George是正确的，将jsonp参数设置为false-从jQuery 1.5开始(因此，设置方式取决于jQuery版本).我不相信您提供的回调名称是作为一个函数调用的(而是在提供给服务器的URL中提供的名称).如果您获得成功，那么您已经收到了数据.好奇:您是否为开发人员设置了主机条目，因为我尝试进行一些测试，并且 http ://www.remote_host.co.uk/feed.php 不能为我解决.

George is correct, set the jsonp param to false -- as of jQuery 1.5 (so, how you set this up is jQuery version dependent). I don't believe that your supplied callback name is invoked as a function (rather, it is the name provided in the URL presented to the server). If you are getting success, then you have received the data. Curious: do you have a hosts entry set up for dev, because I tried to do some testing, and http://www.remote_host.co.uk/feed.php does not resolve for me.

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