如何使用Jackson迭代JSON对象？

Question

我最终想要使用我的JSON数据集创建反向索引。 我知道如何解析一个JSON对象但是如何迭代很多？ 这是我的工作：

文件1：

{
    "doc_id": "2324jos",
    "screen_name": "twitter_user101",
    "tweet_text": "Its a beautiful day to be productive",
    "hashtags": "[]",
    "links": "[]",
    "place_type": "city",
    "place_name": "Evergreen Park",
    "created_at": "2019-02-08 22:24:03"
}

我的代码：

public class ParseJson {
    public static void main(String[] args) throws Exception {
        // this is the key object to convert JSON to Java
        Tweet tweet;
        ObjectMapper mapper = new ObjectMapper();
        try {
            File json = new File("test.json");
            tweet = mapper.readValue(json, Tweet.class);
            System.out.println("Java object created from JSON String :");
            System.out.println(tweet);
        } catch (IOException ex) {
            ex.printStackTrace();
        }
    }
}

public class Tweet {
    public String doc_id;
    public String screen_name;
    public String tweet_text;
    public String hashtags;
    public String links;
    public String place_type;
    public String place_name;
    public String created_at;

    public Tweet() {

    }

    public Tweet(String doc_id, String screen_name, String tweet_text, String hashtags, String links, String place_type, String place_name, String created_at) {
        this.doc_id = doc_id;
        this.screen_name = screen_name;
        this.tweet_text = tweet_text;
        this.hashtags = hashtags;
        this.links = links;
        this.place_name = place_name;
        this.place_type = place_type;
        this.created_at = created_at;
    }

    @Override
    public String toString() {
        return doc_id + screen_name + tweet_text;
    }
}

现在，我想遍历这个在数组中有2个JSON对象的JSON文件：

文件2：

[
    {
        "doc_id": "2324jos",
        "screen_name": "b'LIBBYRULZ'",
        "tweet_text": "@ABC ya'll be lying",
        "hashtags": "[]",
        "links": "[]",
        "place_type": "city",
        "place_name": "Evergreen Park",
        "created_at": "2019-02-08 22:24:03"
    },
    {
        "doc_id": "8982hol",
        "screen_name": "b'eddylee_1'",
        "tweet_text": "Hungry for money",
        "hashtags": "[]",
        "links": "[]",
        "place_type": "city",
        "place_name": "Manhattan",
        "created_at": "2/7/2019 17:01"
    }
]     

如果doc_id是唯一键，我如何使用Jackson调整上面的代码呢？ 我希望能够为每个doc_id返回每个JSON对象中的所有数据。

Answer 1

使用Jackson解析JSON对象数组：

Tweet[] tweets = mapper.readValue(json, Tweet[].class);

应该做的伎俩。 见下文：

public class ParseJson {
    public static void main(String[] args) throws Exception {

        Tweet[] tweets;
        ObjectMapper mapper = new ObjectMapper();
        try {
            File json = new File("test.json");
            tweets = mapper.readValue(json, Tweet[].class);
            System.out.println("Java object created from JSON String :");
            Arrays.asList(tweets).forEach(System.out::println); // Prints each element in the array
        } catch (IOException ex) {
            ex.printStackTrace();
        }
    }
}

Answer 2

你可以尝试将它放在一个列表中，这样你就可以迭代它：

List<Tweet> data = mapper.readValue(json, new TypeReference<List<Tweet>>(){});

Answer 3

我建议使用TypeFactory创建一个CollectionType并使用它将JSON解析为List<Tweet> 。

CollectionType tweetListType = mapper.getTypeFactory().constructCollectionType(ArrayList.class, Tweet.class);
List<Tweet> tweets = mapper.readValue(json, tweetListType);
tweets.forEach(System.out::println);

以下是您分享的完整示例：

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.CollectionType;

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

public class ParseJson {
    public static void main(String[] args) {
        // this is the key object to convert JSON to Java
        ObjectMapper mapper = new ObjectMapper();
        try {
            File json = new File("test.json");
            CollectionType tweetListType = mapper.getTypeFactory().constructCollectionType(ArrayList.class, Tweet.class);
            List<Tweet> tweets = mapper.readValue(json, tweetListType);
            System.out.println("Java objects created from JSON String:");
            tweets.forEach(System.out::println);
        } catch (IOException ex) {
            ex.printStackTrace();
        }
    }
}