通过JQuery JSONP调用Google Ajax Search API

I know this has been asked a zillion times, but I still can't get my code to work. I am trying to a simple JSONP call from my Javascript application. The cod fragment looks like:

我知道这已被问过无数次，但我仍然无法让我的代码工作。我正在尝试从我的Javascript应用程序中进行简单的JSONP调用。鳕鱼片段看起来像：

url="http://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=AAA&callback=?";

and then either:

然后是：

$.getJSON(url, function(data) {
    alert('hello 1');
  });

or:

要么：

$.ajax({url: url,
    datatype: 'jsonp',
    success: function(data) { alert("hello 2"); },
    error: function(j, t, e) {  alert(t);}
});

Neither approach works. The second approach results in the alert of "error". The first does not return success either. What am I doing wrong? Many, many thanks!!

这两种方法都不奏效。第二种方法导致“错误”的警报。第一个也没有成功。我究竟做错了什么？非常感谢！！

UPDATE: I think I found at least one problem. Let me look more into this.

更新：我想我至少发现了一个问题。让我看看这个。

UPDATE 2: Sorry, this code actually works, at least the first approach. There was a subtle error around this code fragment that resulted in the code not working, but overall this works just fine. Asynchronous calls are sometimes a little tricky :-)

更新2：对不起，这段代码实际上是有效的，至少是第一种方法。这个代码片段周围有一个微妙的错误导致代码不能正常工作，但整体而言这很好。异步调用有时候有点棘手:-)

2 个解决方案

#1

Check this out JsFIddleDemo

看看JsFIddleDemo

    /*
     * create callbak function for jsonP
     * @params
     * data is response from http://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=AAA&callback=myjsonpfunction
     */
      function myjsonpfunction(data){
           alert(data.responseData.results) //showing results data
           $.each(data.responseData.results,function(i,rows){
              alert(rows.url); //showing  results url
           });
      }

    //request data using jsonP
    $(function(){
        $.ajax({
        url:'http://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=AAA&callback=myjsonpfunction',
        type:"GET",
        dataType: 'jsonp',
        jsonp: 'myjsonpfunction',
        async:'true',
        success:function (data) {
            //alert("success");
          }
        });
      });

you need write a callback parameter and callback function,the google ajax apis will be return only json if you don't set of callback.

你需要写一个回调参数和回调函数，如果你没有设置回调，google ajax apis将只返回json。

if you set url like this

如果你设置这样的URL

http://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=AAA&callback=?(another)

the response is

反应是

{"responseData": null, "responseDetails": "bad or missing callback or context", "responseStatus": 400}

{“responseData”：null，“responseDetails”：“错误或缺少回调或上下文”，“responseStatus”：400}

#2

Looks like the method you are using is deprecated: https://developers.google.com/web-search/docs/reference

您的使用方法看起来已弃用：https：//developers.google.com/web-search/docs/reference

And has moved on to: http://code.google.com/apis/customsearch/v1/overview.html

并转到：http：//code.google.com/apis/customsearch/v1/overview.html

#1