Question

我正在使用JSONP获取ajax请求的结果而没有任何问题。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。请指教。（我读了一些类似的问题，我无法将它们的解决方案应用于我的。因为我认为我的响应数据结构有点不同）请参阅下面的块以查看console.log的结果（响应）;

{
 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。 Kushan Randima

Answer 1

试试这个例子：

只需在函数外部声明一个全局变量，并在ajax响应之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
        }
    });
}

Answer 2

我尝试验证json响应并且它似乎无效，这可能就是你无法将其设置为变量的原因。您可以在http://jsonlint.com/验证json响应。

一旦纠正了json响应，就可以定义函数范围之外的变量，并可以将响应分配给变量。确保在函数之前定义变量。

var responseObject ;
function TestJSONP(){
 .....
 .....

 // work with the response
   success: function (response) {
      responseObject = JSON.parse(response);
}

希望这会有所帮助。

Answer 3

艾米的回答是正确的。做得好！我将用更多细节重新编写它。这对初学者很有帮助。

var jasonData;
   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

        //Save Account Data
        account_id = response.account.id;
        name = response.account.name;
        support_email_address = response.account.support_email_address;
        report_threat_button_text = response.account.report_threat_button_text;
        successful_report_text = response.account.successful_report_text;
        false_report_text = response.account.false_report_text;

        //Main Object Data
        current_plugin_version = response.current_plugin_version;
        id = response.id;
        status = response.status;
        type = response.type;           

        //Save User Data
        user_id = response.user.id;
        language = response.user.language;
        first_name = response.user.first_name;
        last_name = response.user.last_name;
        email_address = response.user.email_address;
    }
});
}

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

3 个答案: