从 servlet 发送 json答案

【问题标题】：Sending json from servlet从 servlet 发送 json
【发布时间】：2016-11-17 13:12:17
【问题描述】：

我写了这个 servlet：

protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {

    response.setContentType("application/json");

    BufferedReader br = new BufferedReader(new InputStreamReader(request.getInputStream()));
    String json = "";
    if(br != null){
        json = br.readLine();
    }

    PrintWriter out = response.getWriter();

    out.write(json);
    out.flush();
}

通过 ajax 请求发送数据，servlet 我返回它们。问题是，当我尝试让我返回在 servlet 中创建的 JSONObject 时，会出现错误 500。

给我这个错误的代码：

protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {




    response.setContentType("application/json");

    BufferedReader br = new BufferedReader(new InputStreamReader(request.getInputStream()));

    JSONObject json = new JSONObject();

    json.put("name", "federico");

    PrintWriter out = response.getWriter();

    out.print(json);
    out.flush();
}

我的 requestAjax 是：

$.ajax({
                url: "ValidServlet",
                type: 'POST',
                data : JSON.stringify(submitFormData),
                success: function (data) {
                    alert(data.name);
                },
                error: function (data) {
                    alert('error');
                }
            });

我的错误是：

Failed to load resource: the server responded with a status of 500 (Internal Server Error) (17:33:34:492 | error, network)at http://localhost:8080/ProgettoProsa/ValidServlet>

也许我错误地返回所以 json？

帮我找出失败的地方。

谢谢大家。

【问题讨论】：

能否说明您收到的错误以及您发送的请求？

标签： java json ajax servlets jsonobject

【解决方案1】：

试试这段代码用http发送json数据：

包 com.nl.services.ws.sincronizacao；

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import com.google.gson.Gson;

@WebServlet(value = "/TestGsonHttp")
public class TestGsonHttp extends HttpServlet {

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {

        // Your class
        MyJson m = new MyJson();
        m.setName("frederico");

        Gson g = new Gson();
        String jsonResult = g.toJson(m);

        PrintWriter writer = resp.getWriter();
        writer.print(jsonResult);
        writer.flush();
        writer.close();
    }

    class MyJson {

        private String  name;

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }
    }
}

【讨论】：