android解析json,Android解析json

android json解析成map格式

先看json数据

之前使用的是gson解析，把 discount 写成Object格式，但是通过解析之后转成String类型，成为

这种明显没有办法解析成map集合

使用原始的Json解析，一层一层的找到这个对象，转成String类型成为

这种情况下就可以了 之后是copy的解析方法

第一种方法是

第二种是

这样就可以获取到map集合了

相关问答

Q1: android中这样的JSON数据应该如何解析，

//json我的理解就是一系列键值对，只要KEY值对了，对应数据格式也对了，那就可以了！~

JsonObject j = ...//此处是获取json的方法，自定

boolean b = j.getBoolean("result");//取出“result”对应的值，这里的结果应该

//是"true";

JsonArray ja = j.getJsonArray("asks");//取出“asks”对应的值，因为asks对应的值

//中含有方括号，所以这是JsonArray型数据，需要取出进一步解析；JsonArray可以与数组

//进行比较理解，我个人是这样的，供参考

for(int i = 0; i ja.length(); i++){

JsonArray ja1 = ja.getJsonArray(i);//获取ja中的第一个元素，因为这个元素

//也是jsonArray,所以可以再进一步解析

double d1 = ja1.getDouble(0);//jsonArray中没有getFloat()这个方法

//，自行转型就可以了

double d2 = ja1.getDouble(1);//

} 以上，希望对你有所帮助

Q2: Android 解析json问题

Android 解析json的方式为：

1、首先，搭建一个服务器的工程：JsonProject这个项目

源代码：

Person.java

package com.json.domain;

public class Person {

private int id;

private String name;

private String address;

public Person() {

super();

}

public Person(int id, String name, String addrss) {

super();

this.id = id;

this.name = name;

this.address = addrss;

}

public int getId() {

return id;

}

public void setId(int id) {

this.id = id;

}

public String getName() {

return name;

}

public void setName(String name) {

this.name = name;

}

public String getAddress() {

return address;

}

public void setAddress(String address) {

this.address = address;

}

@Override

public String toString() {

return "Person [addrss=" + address + ", id=" + id + ", name=" + name

+ "]";

}

}

JsonService.java

package com.json.service;

import java.util.ArrayList;

import java.util.HashMap;

import java.util.List;

import java.util.Map;

import com.json.domain.Person;

public class JsonService {

public JsonService() {

}

public Person getPerson(){

Person person = new Person(1001,"jack","上海黄浦区");

return person;

}

public ListPerson getListPerson(){

ListPerson list = new ArrayListPerson();

Person person1 = new Person(1001,"jack","上海黄浦区");

Person person2 = new Person(1002,"rose","上海闵行区");

Person person3 = new Person(1003,"mick","上海黄浦区");

list.add(person1);

list.add(person2);

list.add(person3);

return list;

}

public ListString getListString(){

ListString list = new ArrayListString();

list.add("北京");

list.add("上海");

list.add("湖南");

return list;

}

public ListMapString,Object getListMaps(){

ListMapString,Object list = new ArrayListMapString,Object();

MapString,Object map1 = new HashMapString, Object();

MapString,Object map2 = new HashMapString, Object();

map1.put("id", 1001);

map1.put("name", "jack");

map1.put("address", "北京");

map2.put("id", 1001);

map2.put("name", "rose");

map2.put("address", "上海");

list.add(map1);

list.add(map2);

return list;

}

}

JsonServlet.java

package com.json.servlet;

import java.io.IOException;

import java.io.PrintWriter;

import javax.servlet.ServletException;

import javax.servlet.http.HttpServlet;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpServletResponse;

import com.json.service.JsonService;

import com.json.tools.JsonTools;

public class JsonServlet extends HttpServlet {

private JsonService service;

/**

* Constructor of the object.

*/

public JsonServlet() {

super();

}

/**

* Destruction of the servlet. br

*/

public void destroy() {

super.destroy(); // Just puts "destroy" string in log

// Put your code here

}

/**

* The doGet method of the servlet. br

*

* This method is called when a form has its tag value method equals to get.

*

* @param request the request send by the client to the server

* @param response the response send by the server to the client

* @throws ServletException if an error occurred

* @throws IOException if an error occurred

*/

public void doGet(HttpServletRequest request, HttpServletResponse response)

throws ServletException, IOException {

this.doPost(request, response);

}

/**

* The doPost method of the servlet. br

*

* This method is called when a form has its tag value method equals to post.

*

* @param request the request send by the client to the server

* @param response the response send by the server to the client

* @throws ServletException if an error occurred

* @throws IOException if an error occurred

*/

public void doPost(HttpServletRequest request, HttpServletResponse response)

throws ServletException, IOException {

response.setContentType("text/html;charset=utf-8");

request.setCharacterEncoding("utf-8");

response.setCharacterEncoding("utf-8");

PrintWriter out = response.getWriter();

String jsonString = "";

String action_flag = request.getParameter("action_flag");

if(action_flag.equals("person")){

jsonString = JsonTools.createJsonString("person", service.getPerson());

}else if(action_flag.equals("persons")){

jsonString = JsonTools.createJsonString("persons", service.getListPerson());

}else if(action_flag.equals("listString")){

jsonString = JsonTools.createJsonString("listString", service.getListString());

}else if(action_flag.equals("listMap")){

jsonString = JsonTools.createJsonString("listMap", service.getListMaps());

}

out.println(jsonString);

out.flush();

out.close();

}

/**

* Initialization of the servlet. br

*

* @throws ServletException if an error occurs

*/

public void init() throws ServletException {

service = new JsonService();

}

}

2、通过浏览器

访问地址一：

输出以下结果：

{"person":{"address":"上海黄浦区","id":1001,"name":"jack"}

访问地址二：

输出以下结果：

{"persons":[{"address":"上海黄浦区","id":1001,"name":"jack"},{"addrss":"上海闵行区","id":1002,"name":"rose"},{"address":"上海黄浦区","id":1003,"name":"mick"}]}

访问地址三：

输出以下结果：

{"persons":["北京","上海","湖南"]}

访问地址四：

输出以下结果：

{"persons":[{"id":1001,"address":"北京","name":"jack"},{"id":1001,"address":"上海","name":"rose"}]}

Q3: android json解析三种方式哪种效率最高

用org.json以及谷歌提供gson来解析json数据的方式更好一些。

安卓下通常采用以下几种方式解析json数据：

1、org.json包（已经集成到android.jar中了）

2、google提供的gson库

3、阿里巴巴的fastjson库

4、json-lib

以Google出品的Gson为例，具体步骤为：

1、首先，从 code.google.com/p/google-gson/downloads/list下载GsonAPI：

google-gson-1.7.1-release.zip 把gson-1.7.jar copy到libs(项目根目录新建一个libs文件夹)中。 可以使用以下两种方法解析JSON数据，通过获取JsonReader对象解析JSON数据。

代码如下:

String jsonData = "[{\"username\":\"arthinking\",\"userId\":001},{\"username\":\"Jason\",\"userId\":002}]";

try{

JsonReader reader = new JsonReader(new StringReader(jsonData));

reader.beginArray();

while(reader.hasNext()){

reader.beginObject();

while(reader.hasNext()){

String tagName = reader.nextName();

if(tagName.equals("username")){

System.out.println(reader.nextString());

}

else if(tagName.equals("userId")){

System.out.println(reader.nextString());

}

}

reader.endObject();

}

reader.endArray();

}

catch(Exception e){

e.printStackTrace();

}

2、使用Gson对象获取User对象数据进行相应的操作：

代码如下：

Type listType = new TypeTokenLinkedListUser(){}.getType();

Gson gson = new Gson();

LinkedListUser users = gson.fromJson(jsonData, listType);

for (Iterator iterator = users.iterator(); iterator.hasNext();) {

User user = (User) iterator.next();

System.out.println(user.getUsername());

System.out.println(user.getUserId());

}

3、如果要处理的JSON字符串只包含一个JSON对象，则可以直接使用fromJson获取一个User对象：

代码如下：

String jsonData = "{\"username\":\"arthinking\",\"userId\":001}";

Gson gson = new Gson();

User user = gson.fromJson(jsonData, User.class);

System.out.println(user.getUsername());

System.out.println(user.getUserId());