如何使用一个ajax请求从java servlet返回多个json对象

时间:2011-05-07 19:54:06

标签: java jquery json servlets

我正在使用jsp和servlet构建Web应用程序,我从jsp发送ajax请求,我想从servlet返回两个json对象。我尝试执行以下操作但代码不起作用。

//在jquery中我写了这段代码

        var id = $(this).attr('id');

        var paramenters = {"param":id};

        $.getJSON("MyServlet", paramenters, function (data1,data2){

            $("h3#name").text(data1["name"]);

            $("span#level").text(data1["level"]);

            $("span#college").text(data2["college"]);

            $("span#department").text(data2["department"]);

        });

//在servlet中我编写了这段代码

    String json1 = new Gson().toJson(object1);

    String json2 = new Gson().toJson(object2);

    response.setContentType("application/json");

    response.setCharacterEncoding("utf-8");

    response.getWriter().write(json1);

    response.getWriter().write(json2);
有人可以帮助我吗?

5 个答案:

答案 0 :(得分:20)

你应该这样做:

服务器端:

String json1 = new Gson().toJson(object1); 
String json2 = new Gson().toJson(object2); 
response.setContentType("application/json"); 
response.setCharacterEncoding("utf-8"); 
String bothJson = "["+json1+","+json2+"]"; //Put both objects in an array of 2 elements
response.getWriter().write(bothJson);

客户方:

$.getJSON("MyServlet", paramenters, function (data){ 
   var data1=data[0], data2=data[1]; //We get both data1 and data2 from the array
   $("h3#name").text(data1["name"]); 
   $("span#level").text(data1["level"]); 
   $("span#college").text(data2["college"]); 
   $("span#department").text(data2["department"]);
});

希望这会有所帮助。干杯

答案 1 :(得分:2)

将它们包装在JSON数组中:

[ {..}, {..}, {..}]

或者将它们包装在另一个对象中:

{ "result1":{..}, "result2":{..} }

答案 2 :(得分:1)

您可以返回一个JSON数组,其中两个对象都是数组的元素。让您的servlet返回具有如下结构的JSON:

[{"name": "object1"}, {"name": "object2"}]

然后你的javascript代码可能是这样的:

$.getJSON("MyServlet", paramenters, function (data){
        var data1 = data[0];
        var data2 = data[1];

        $("h3#name").text(data1["name"]);

        $("span#level").text(data1["level"]);

        $("span#college").text(data2["college"]);

        $("span#department").text(data2["department"]);

    });

答案 3 :(得分:1)

你需要将两者放入一个像这样的单个json字符串

response.getWriter().write("[");
response.getWriter().write(json1);
response.getWriter().write(",");
response.getWriter().write(json2);
response.getWriter().write("]");

这将它们放在json数组中

您也可以将它们放在json对象中

response.getWriter().write("{\"object1\":");
response.getWriter().write(json1);
response.getWriter().write(",\"object2\":");
response.getWriter().write(json2);
response.getWriter().write("}");

答案 4 :(得分:0)

@Edgar的回答对我有用。但我认为我们应该避免自己组建阵列,所以我建议使用一个列表。代码将是这样的:

protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
    ...
    resp.setContentType("application/json");
    resp.setCharacterEncoding("utf-8");
    ArrayList<Object> obj_arr = new ArrayList<Object>();
    obj_arr.add(obj1);
    obj_arr.add(obj2);
    Gson gson = new Gson();
    String tmp = gson.toJson(obj_arr);
    resp.getWriter().write(tmp);
}

在前端,对于我们获得的数据,我们可以使用data[0]来检索obj1data[1]来检索obj2。代码将是这样的(我在这里使用ajax):

$('#facts_form').submit(function (e) {
    e.preventDefault();
    var data = new FormData(this);
    var url = 'import';
    $.ajax({
        url: url,
        type: "POST",
        data: data,
        processData: false,  
        contentType: false,   
        async: false,
        cache: false,
        success: function (data) {                
            for (var i = 1; i < data.length; i++){
               //do whatever 
            }
        },
        error: function (xhr, status, error) {
            alert(status + "\n" + error);
        }
    });
});