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Java 当我使用;答复.数据“;要打印JSON数据,它只打印整个响应,而不是JSON_Java_Angularjs_Json - Fatal编程技术网

Java 当我使用;答复.数据“;要打印JSON数据,它只打印整个响应,而不是JSON

Java 当我使用;答复.数据“;要打印JSON数据,它只打印整个响应,而不是JSON,java,angularjs,json,Java,Angularjs,Json,我将从angularjs 1.5x客户端向Java HTTP方法发送get请求,并在客户端打印包含一些JSON数据的响应 // method from an angular service that sends ajax requests this.getTask = function(taskName) { var url = 'http://localhost:9998/tasks/' + taskName; var config = {

我将从angularjs 1.5x客户端向Java HTTP方法发送get请求,并在客户端打印包含一些JSON数据的响应

// method from an angular service that sends ajax requests
this.getTask = function(taskName) {
        var url = 'http://localhost:9998/tasks/' + taskName;
        var config = {
            headers: {
                'Content-Type': 'application/json'
            }
        };

        $http.get(url, config)
            .then(function successCallback(response) {
               console.log(response.data);
            }, function errorCallback(response) {
               console.log(response);
            });
    };
请求成功执行,但语句
response.data
不返回JSON数据,而是返回JSON数据


Object{data:Object,status:302,headers:headersGetter/我猜它使用的不是successCallback,而是输出响应的errorCallback。请尝试调整代码,使其看起来像
then(function(response){console.log(response.data})。catch(function(err){console.error(err)};
var data = JSON.parse(apiResponse);

var name = data.response;