解析JSON对象数组里面杰克逊 [英] Parsing JSON Object inside Array in Jackson

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本文介绍了解析JSON对象数组里面杰克逊的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我使用的杰克逊JSON处理器,以获得详细信息形成的JSON,我有以下

JSON字符串

  {
FIRST_NAME:FIRST_NAME
姓氏:最后一个,
城市:在某处,
user_user:[
   {
        ID:1
        域:http://google.com/
    },
    {
        ID:34,
        域:http://so.com/
    },
    {
        ID:42,
        域:http://ww.com/
    }
]}

的UserDetails

 包com.example.com;进口的java.util.List;公共类的UserDetails {
私人字符串的城市;
私人字符串FIRST_NAME;
私人字符串姓氏;
私人列表<> user_user;公共字符串getCity(){
    返回this.city;
}
公共无效setCity(字符串市){
    this.city =城市;
}
公共字符串getFirst_name(){
    返回this.first_name;
}
公共无效setFirst_name(字符串FIRST_NAME){
    this.first_name = FIRST_NAME;
}
公共字符串getLast_name(){
    返回this.last_name;
}
公共无效setLast_name(字符串姓氏){
    this.last_name =姓氏;
}
公开名单<> getUser_user(){
    返回this.user_user;
}
公共无效setUser_user(列表<> user_user){
    this.user_user = user_user;
}
}

User_user

 包com.example.com;进口的java.util.List;公共类User_user {
私人字符串域;
私人字符串ID;公共字符串getDomain(){
    返回this.domain;
}
公共无效setDomain(字符串域){
    this.domain =域;
}
公共字符串的getId(){
    返回this.id;
}
公共无效SETID(字符串ID){
    this.id = ID;
}
}

code

  ObjectMapper映射器=新ObjectMapper();
的UserDetails为userDetails = mapper.readValue(jsonString,UserDetails.class);的System.out.println(userDetails.getFirst_name());
的System.out.println(userDetails.getLast_name());User_user userF = mapper.readValue(jsonString,User_user.class);的for(int i = 0; I< userDetails.getUser_user()大小();我++)
{
的System.out.println(userF.getId());
}


  1. 从上面的code我无法从 User_user 获得编号。


  2. 我如何解析和 GET fav_colors 如果我的JSON有数组试



解决方案

我改的 列表&LT;&GT; user_user; 以列表&LT; <一href=\"https://github.com/VenomVendor/Jackson-JSON-Parsing/blob/master/src/vee/demo/jackson/json/parsing/User_user.java\"相对=nofollow> User_user &GT; user_user;在<一个href=\"https://github.com/VenomVendor/Jackson-JSON-Parsing/blob/master/src/vee/demo/jackson/json/parsing/UserDetails.java\"相对=nofollow>的UserDetails 结果
由<建议href=\"http://stackoverflow.com/questions/18614312/parsing-json-object-inside-array-in-jackson/#comment27417022_18614312\">Michał Ziober

之后,我能够轻松地检索所有的结果,这里是code,结果没有那些<大骨节病> SYSOUT 它几乎没有 3行

 的System.out.println(userDetails.getCity()); //市,名字,姓氏
    的for(int i = 0; I&LT; userDetails.getUser_user()大小();我++){
    的System.out.println(userDetails.getUser_user()获得(ⅰ).getId()); // ID,域名
    尝试{
        为(中间体K = 0; K&下; userDetails.getUser_user()获得(ⅰ).getFav_colors()大小();说明k ++){
            的System.out.println(userDetails.getUser_user()获得(ⅰ).getFav_colors()得到(k)的); // fav_color
        }
    }
    赶上(例外五){
        的System.out.println(no最爱颜色);
        }
    }

有关我已经加入GitHub的中中的项目。参照

Am using Jackson JSON Processor to get details form JSON, I have the following

JSON String

{
"first_name": "first_name",
"last_name": "last",
"city": "Somewhere",
"user_user": [
   {
        "id": "1",
        "domain": "http://google.com/"
    },
    {
        "id": "34",
        "domain": "http://so.com/"
    },
    {
        "id": "42",
        "domain": "http://ww.com/"
    }
]}

UserDetails

package com.example.com;

import java.util.List;

public class UserDetails{
private String city;
private String first_name;
private String last_name;
private List<?> user_user;

public String getCity(){
    return this.city;
}
public void setCity(String city){
    this.city = city;
}
public String getFirst_name(){
    return this.first_name;
}
public void setFirst_name(String first_name){
    this.first_name = first_name;
}
public String getLast_name(){
    return this.last_name;
}
public void setLast_name(String last_name){
    this.last_name = last_name;
}
public List<?> getUser_user(){
    return this.user_user;
}
public void setUser_user(List<?> user_user){
    this.user_user = user_user;
}
}

User_user

package com.example.com;

import java.util.List;

public class User_user{
private String domain;
private String id;

public String getDomain(){
    return this.domain;
}
public void setDomain(String domain){
    this.domain = domain;
}
public String getId(){
    return this.id;
}
public void setId(String id){
    this.id = id;
}
}

Code

ObjectMapper mapper = new ObjectMapper();
UserDetails userDetails = mapper.readValue(jsonString , UserDetails.class);

System.out.println(userDetails.getFirst_name());
System.out.println(userDetails.getLast_name());

User_user userF =   mapper.readValue(jsonString , User_user.class);

for(int i = 0; i < userDetails.getUser_user().size(); i++)
{
System.out.println(userF.getId());
}

  1. From the above code am not able to get Id from User_user.

  2. How do I parse and get fav_colors if my JSON had array again.

解决方案

I changed List<?> user_user; to List<User_user> user_user; in UserDetails
as suggested by Michał Ziober

After that I was able to retrieve all results easily, here is the code.
Without those Sysout it's hardly 3 lines.

    System.out.println(userDetails.getCity());      //City, FirstName, LastName 
    for(int i = 0; i < userDetails.getUser_user().size(); i++) {
    System.out.println(userDetails.getUser_user().get(i).getId());      // id, domain
    try {
        for(int k = 0; k < userDetails.getUser_user().get(i).getFav_colors().size(); k++) {
            System.out.println(userDetails.getUser_user().get(i).getFav_colors().get(k));   //fav_color
        }
    }
    catch (Exception e) {
        System.out.println("NO FAV COLORS");
        }
    }

For a reference I have added the project in GitHub.

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