Question

需要有关java + json的帮助。大家好！我需要按歌曲标题对数据进行排序，并提供特殊情况的处理。我需要编写对json文件中的数据进行排序的方法。例如，此方法还必须能够对具有相同字段的数据进行排序。该文件的解析有效。但是它的结构如下：

{
    "musicAlbum": [
        {
            "groupname": "twenty one pilots",
            "songduration": 3.27,
            "songname": "Heathens"
        },
        {
            "groupname": "twenty one pilots",
            "songduration": 4.4,
            "songname": "Car Radio"
        },
        {
            "groupname": "Linkin Park",
            "songduration": 3.06,
            "songname": "Numb"
        }
    ]
}

我现在不怎么从对象数组中获取任何元素。例如，当我尝试执行以下操作：.getSongName（）时，我得到“空”。

我有带字符串groupName的Record类；字符串songName;和double songDuration；所有获取器和设置器。另外JsonParser类具有解析方法。和MusicAlbum类一起使用List专辑。并与main一起上课：

import java.io.IOException;
import java.util.List;

public class App {
public static void main(String[] args) throws IOException // exception to be 
handled
{
    List<Record> album = JsonParser.parseJson();
    System.out.println(album); 

    for (int i = 0; i < album.size(); i++) {
        System.out.println(album.get(i));
    }
    Record songName = new Record();
    System.out.println(songName); 
    System.out.println(songName.getSongName());

}
}

我现在在控制台中拥有什么：

[MusicAlbum [album=[Record [groupName=twenty one pilots, songName=Heathens, songDuration=3.27] , Record [groupName=twenty one pilots, songName=Car Radio, songDuration=4.4] , Record [groupName=Linkin Park, songName=Numb, songDuration=3.06] , Record [groupName=Lana Del Rey, songName=Summertime sadness, songDuration=3.56] , Record [groupName=Imagine Dragons, songName=Thunder, songDuration=3.24] , Record [groupName=Three Days Grace, songName=Outsider, songDuration=2.43] , Record [groupName=ONUKA, songName=When I Met You, songDuration=4.04] , Record [groupName=Foster The People, songName=Best Friend, songDuration=4.25] , Record [groupName=Massive Attack, songName=Angel, songDuration=6.18] , Record [groupName=Florence + The Machine, songName=Big God, songDuration=4.28] , Record [groupName=Die antwoord, songName=banana brain, songDuration=7.12] , Record [groupName=Coldplay, songName=Hypnotised, songDuration=5.54] ]]]

MusicAlbum [album=[Record [groupName=twenty one pilots, songName=Heathens, songDuration=3.27] , Record [groupName=twenty one pilots, songName=Car Radio, songDuration=4.4] , Record [groupName=Linkin Park, songName=Numb, songDuration=3.06] , Record [groupName=Lana Del Rey, songName=Summertime sadness, songDuration=3.56] , Record [groupName=Imagine Dragons, songName=Thunder, songDuration=3.24] , Record [groupName=Three Days Grace, songName=Outsider, songDuration=2.43] , Record [groupName=ONUKA, songName=When I Met You, songDuration=4.04] , Record [groupName=Foster The People, songName=Best Friend, songDuration=4.25] , Record [groupName=Massive Attack, songName=Angel, songDuration=6.18] , Record [groupName=Florence + The Machine, songName=Big God, songDuration=4.28] , Record [groupName=Die antwoord, songName=banana brain, songDuration=7.12] , Record [groupName=Coldplay, songName=Hypnotised, songDuration=5.54] ]]

Record [groupName=null, songName=null, songDuration=0.0] 

null

我不知道要怎么做以及如何编写数据接收方法。

感谢您提供有关此方法的帮助！

Answer 1

您没有将任何值添加到songName对象：

Record songName = new Record(); // you should add values to "songName" object
System.out.println(songName); 
System.out.println(songName.getSongName());

因此，所有内容均为空。尝试做这样的事情：

将响应“ musicAlbum”添加到JSONArray对象，然后遍历该数组并获取每条记录。

List<Record> records = new ArrayList<>();
Record record = new Record();

for (int i=0;i<jsonArray.length();i++) {
    JSONObject jObj = jsonArray.getJSONObject(i);

    record.setGroupName = jObj.getString("groupname");
    record.setSongName = jObj.getString("songname");
    record.setSongDuration = jObj.getString("songduration");

    records.add(record);
}

Answer 2

好的，没有关于您的Record类和解析器的更多详细信息，我只能进行一些反向工程。

这就是您的数据的样子。

MusicAlbum [
    album= [
        Record [groupName=twenty one pilots, songName=Heathens, songDuration=3.27] ,
        Record [groupName=twenty one pilots, songName=Car Radio, songDuration=4.4] ,
        Record [groupName=Linkin Park, songName=Numb, songDuration=3.06] ,
        Record [groupName=Lana Del Rey, songName=Summertime sadness, songDuration=3.56] ,
        Record [groupName=Imagine Dragons, songName=Thunder, songDuration=3.24] ,
        Record [groupName=Three Days Grace, songName=Outsider, songDuration=2.43] ,
        Record [groupName=ONUKA, songName=When I Met You, songDuration=4.04] ,
        Record [groupName=Foster The People, songName=Best Friend, songDuration=4.25] ,
        Record [groupName=Massive Attack, songName=Angel, songDuration=6.18] ,
        Record [groupName=Florence + The Machine, songName=Big God, songDuration=4.28] ,
        Record [groupName=Die antwoord, songName=banana brain, songDuration=7.12] , Record [groupName=Coldplay, songName=Hypnotised, songDuration=5.54]
    ]
]

MusicAlbum有一个records数组，每个记录都有一些字段，songName是其中之一。

因此，获取记录详细信息的总体算法如下：

List<Album> albums = parseJson("{...}");
for (Album album : albums) {
    // 'album' is a bad name here, it's storing records, not albums
    List<Record> records = album.getAlbum();
    for (Record record : records) {
        System.out.println(record.getSonName());
    }
}

这基本上是伪代码，不会编译，只是给您一个想法。

java + json：如何从包含对象数组{[{}，{}]}

2 个答案: