javascript - Jquery/JavaScript - 将 Ajax jSONP 响应存储到变量中

Question

我得到了使用 JSONP 的 ajax 请求的结果，没有任何问题。这是我的代码

    function TestJSONP()
    {
    $.ajax({
        url: "https://www.sample.com/api/users.json?account_api_key=0000&unique_install_id=0000&email_address=test@test.com&locale_id=en-US&operating_system_version=6.1.7601.65536&operating_system_architecture=64&outlook_version=2013&version=0.0.5.0",

        // the name of the callback parameter, as specified by the YQL service
        jsonp: "callback",

        // tell jQuery we're expecting JSONP
        dataType: "jsonp",

        // tell YQL what we want and that we want JSON
        data: {
            q: "select title,abstract,url from search.news where query=\"cat\"",
            format: "json"
        },

        // work with the response
        success: function (response) {
            console.log(response); // server response
        }
    });
}

我需要将响应数据设置为我可以在该请求之外访问的变量。请给我建议。 (我读了一些类似的问题，但我无法将他们的解决方案应用到我的身上。因为我认为我的响应数据结构有点不同)请查看以下 block 以查看 console.log(response); 的结果。

{
 account: 
 {
 id: "sadasdd4234",
 name: "Sample Development",
 support_email_address: "test1@sample.com",
 report_threat_button_text: "text1",
 successful_report_text: "text2",
 false_report_text: "text3",
 },
 current_plugin_version: "0.0.1",
 id: "trt45rety",
 status: "ok",
 type: "api_response",
 user: 
 {
 id: "erwrretV0",
 language: "en",
 first_name: "Robert",
 last_name: "Croos",
 email_address: "test2@sample.net"
 }
}

提前致谢。库山兰迪玛

Answer 1

试试这个例子:

只需在函数外声明一个全局变量，并在 ajax 响应后将响应变量分配给该全局变量。

   var jsonData;
   function TestJSONP()
    {
    $.ajax({
        url: "https://www.sample.com/api/users.json?account_api_key=0000&unique_install_id=0000&email_address=test@test.com&locale_id=en-US&operating_system_version=6.1.7601.65536&operating_system_architecture=64&outlook_version=2013&version=0.0.5.0",

        // the name of the callback parameter, as specified by the YQL service
        jsonp: "callback",

        // tell jQuery we're expecting JSONP
        dataType: "jsonp",

        // tell YQL what we want and that we want JSON
        data: {
            q: "select title,abstract,url from search.news where query=\"cat\"",
            format: "json"
        },

        // work with the response
        success: function (response) {
            console.log(response); // server response
            jsonData = response; // you can use jsonData variable in outside of the function
        }
    });
}