javascript - ajax中success回调函数里如何输出json数据？

回复内容：

麻烦仔细看你的数据类型，低级错误好不好

results是一个数组，所以你在调用时，应该写：

js.results[0].currentCity
//或者
js.results[0]["currentCity"]

你的success应该改为下面的形式；然后直接把alert替换成你自己的回调函数就可以啦！

 
    $.ajax({
        url: window.location.href,  //the endpoint, commonly same url
        type: "POST", //http method
        data: { csrfmiddlewaretoken : csrftoken, 
                email: email,
                password: password,
                },  // data sent with the post request

    //handle a successful response
        success : function(json){
            console.log(json);  // another sanity check
            //On success show thr data posted to server as a message
            alert('Hi '+json['email']+'!.'+' You have entered password:'+json['password']+json["rsp"]);
                },
    //handle a non-success response
            error: function(xhr,errmsg,err){
            console.log(xhr.status + ": " + xhr.responseText); // provide a bit more info about error to the console
            }
        });