java - 将 ArrayList 转换为 json 对象

Question

我在数据库中有一个表testemployee，其中包含三列eid，ename，esalary。我想在json对象中显示表的所有行(稍后我想使用 kendo UI 网格给 json obj 一个 html View )我使用 arraylist 来存储所有数据行，然后将列表转换为 json。但是当我运行该文件时，它显示 HTTP 状态 405 - HTTP 方法 GET 不受支持这个网址
下面是我的代码

  package com.JSONoflist;

  import com.google.gson.Gson;

  import java.io.IOException;
import java.io.PrintWriter;
import java.sql.Connection;

import java.sql.ResultSet;
import java.sql.SQLException;

import java.sql.Statement;

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 java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import net.sf.json.JSONObject;

@WebServlet(urlPatterns = {"/list2json"})

public class list2json extends HttpServlet {

protected void processRequest(HttpServletRequest request,     HttpServletResponse response) throws IOException {
       response.setContentType("text/html;charset=UTF-8");

    PrintWriter out = response.getWriter();

    Statement stmt = null;

     ResultSet rs = null;
    dbconn Obj = new dbconn();
    Connection connection = Obj.Open();
       try {

      String qry = "SELECT * FROM testemployee";
        stmt = connection.createStatement();
        rs = stmt.executeQuery(qry);

      JSONObject jObj = new JSONObject();
         ArrayList<Employee> list = new ArrayList<Employee>();
         Employee sPojo = null;
         while (rs.next()) {
          sPojo = new Employee();
          sPojo.setId(rs.getString("eid"));
         sPojo.setEname(rs.getString("ename"));
          sPojo.setSalary(rs.getString("esalary"));
          list.add(sPojo);
       }
       //  String json = new Gson().toJson(list);
        System.out.println(list);
        jObj.put("testemployee", list);
       System.out.println(jObj.toString());
    } catch (Exception ex) {
      ex.printStackTrace();
   }
}

}

我也尝试过在 Arraylist 中使用静态值，但出现 http 错误。我是否遗漏了什么？请帮助我。由于这个错误，我什至无法检查 json 是否在 url 中显示数据库值。

@WebServlet(urlPatterns = {"/list2json"})
public class list2json extends HttpServlet 
{

    protected void processRequest(HttpServletRequest request, HttpServletResponse response) throws IOException {
    response.setContentType("text/html;charset=UTF-8");
    PrintWriter out = response.getWriter();

   try {
      List<String> foo = new ArrayList<String>();
      foo.add("A");
       foo.add("B");
      foo.add("C");

      String json = new Gson().toJson(foo);

    } catch (Exception ex) {
      ex.printStackTrace();
   }
}

}

Answer 1

只需将您的 servlet 代码重写为

@WebServlet(urlPatterns = {"/list2json"})
public class list2json extends HttpServlet 
{

   @Override
public void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    response.setContentType("text/html;charset=UTF-8");
    PrintWriter out = response.getWriter();

   try {
      List<String> foo = new ArrayList<String>();
      foo.add("A");
       foo.add("B");
      foo.add("C");

      String json = new Gson().toJson(foo);

    } catch (Exception ex) {
      ex.printStackTrace();
   }
}