遍历JSON并将列表数组添加到对象/数组

2024-04-27 • 问答

我有以下两个JSON数组：

{ 
  "Person": {
    "Info": [
      "name": "Becky","age": 14
    ]
   },"Fruits": [
    {
      "name": "avocado","organic": true
    },{
      "name": "mango","organic": true
    }
  ],"Vegetables": [
    {
      "name": "brocoli",{
      "name": "lettuce","organic": true
    }
  ]
}

我想做的是使用Jackson和Gson库使所有内容看起来都很漂亮。

类似这样的事情。 Gson可以正常工作。所以我想要的输出是：

{ 
  "Person": {
    "Info": [
      "name":"Becky","age": 14
    ]
  },"FruitsList": {
    "Fruits": [
      {
        "name": "avocado","organic": true
      },{
        "name": "mango","organic": true
      }
    ]
  },"VegetablesList": {
    "Vegetables": [
      {
        "name": "brocoli",{
        "name": "lettuce","organic": true
      }
    ]
  }
}

我将班级设置为：

class Person{
   private List<Info> InfoList;
   //Set and get were set
}

class Info{
   private String name;
   private int age;
   //Set and get were set
}

class Fruits{
   private String name;
   private boolean organic;
   //Set and get were set
   public String toString(){
            return "Fruits:{" +
            "name:'" + name+ '\'' +
            ",organic:" + organic+'\''+
            '}';
   }
 }

 class Vegetables{
   private String name;
   private boolean;
   //Set and get were set
   public String toString(){
            return "Fruits:[" +
            "name:'" + name+ '\'' +
            ",organic:" + organic+'\''+
            ']';
   }
 }

class rootFinal{
    private List<Fruits> fruitList;
    private List<Vegetables> vegetablesList;
    private List<Person> personList;
    //Set and get were set
}

class mainJson{
   final InputStream fileData = ..("testPVF.json");

   ObjectMapper map = new Ob..();
   rootFinal root = map.readValue(fileData,rootFinal.class);
   // I can access each class with 
   System.out.printl(root.getHeaderList.get(0));
}

这将输出...

[Fruit{name:'avocado',organic:true},Fruit{name:'mango',organic:true}]

但这不是我想要的。

我正在尝试对JSON文件进行迭代，或者尝试以某种方式检查数组是否存在。向其中添加其他对象/数组。

如果我发现Veg或Fruit，我想以某种方式添加VegList和FruitList，如图所示。由于它位于"Person": {}符号中，因此应忽略{}。

是否可以使用Gson来做到这一点？

zmshenzheng 回答：遍历JSON并将列表数组添加到对象/数组

如果我对您的理解正确，那么您想使用JSON Array节点包装每个JSON Object节点。为此，您无需使用POJO模型，可以将JSON的有效载荷读取为ObjectNode，并使用其API对其进行更新。

Jackson

带有Jackson库的简单示例：

import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.databind.node.ObjectNode;

import java.io.File;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.Map;

public class JsonObjectApp {

    public static void main(String[] args) throws Exception {
        File jsonFile = new File("./resource/test.json").getAbsoluteFile();

        ObjectMapper mapper = new ObjectMapper();
        mapper.enable(SerializationFeature.INDENT_OUTPUT);

        ObjectNode root = (ObjectNode) mapper.readTree(jsonFile);

        Map<String,JsonNode> valuesToAdd = new LinkedHashMap<>();

        // create fields iterator
        Iterator<Map.Entry<String,JsonNode>> fieldsIterator = root.fields();
        while (fieldsIterator.hasNext()) {
            Map.Entry<String,JsonNode> entry = fieldsIterator.next();

            // if entry represents array
            if (entry.getValue().isArray()) {
                // create wrapper object
                ObjectNode arrayWrapper = mapper.getNodeFactory().objectNode();
                arrayWrapper.set(entry.getKey(),root.get(entry.getKey()));

                valuesToAdd.put(entry.getKey(),arrayWrapper);

                // remove it from object.
                fieldsIterator.remove();
            }
        }

        valuesToAdd.forEach((k,v) -> root.set(k + "List",v));

        System.out.println(mapper.writeValueAsString(root));
    }
}

为您的JSON打印以上代码：

{
  "Person" : {
    "Info" : [ {
      "name" : "Becky","age" : 14
    } ]
  },"FruitsList" : {
    "Fruits" : [ {
      "name" : "avocado","organic" : true
    },{
      "name" : "mango","organic" : true
    } ]
  },"VegetablesList" : {
    "Vegetables" : [ {
      "name" : "brocoli",{
      "name" : "lettuce","organic" : true
    } ]
  }
}

Gson

我们可以使用Gson库实现非常相似的解决方案：

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonElement;
import com.google.gson.JsonObject;

import java.io.File;
import java.io.FileReader;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.Map;

public class GsonApp {

    public static void main(String[] args) throws Exception {
        File jsonFile = new File("./resource/test.json").getAbsoluteFile();

        Gson gson = new GsonBuilder()
                .setPrettyPrinting()
                .create();

        try (FileReader reader = new FileReader(jsonFile)) {
            JsonObject root = gson.fromJson(reader,JsonObject.class);

            Map<String,JsonElement> valuesToAdd = new LinkedHashMap<>();

            // create fields iterator
            Iterator<Map.Entry<String,JsonElement>> fieldsIterator = root.entrySet().iterator();
            while (fieldsIterator.hasNext()) {
                Map.Entry<String,JsonElement> entry = fieldsIterator.next();
                // if entry represents array
                if (entry.getValue().isJsonArray()) {
                    // create wrapper object
                    JsonObject arrayWrapper = new JsonObject();
                    arrayWrapper.add(entry.getKey(),root.get(entry.getKey()));

                    valuesToAdd.put(entry.getKey(),arrayWrapper);

                    // remove it from object.
                    fieldsIterator.remove();
                }
            }

            valuesToAdd.forEach((k,v) -> root.add(k + "List",v));

            System.out.println(gson.toJson(root));
        }
    }
}

输出相同。

选中Get Pretty Printer

还要检查specific APIs for Pretty Printing

示例代码：

//This example is using Input as JSONNode
//One can serialize POJOs to JSONNode using ObjectMapper.
ObjectMapper mapper = new ObjectMapper();
mapper.writerWithDefaultPrettyPrinter().writeValueAsString(
					inputNode)

deserialization gson java json json-deserialization

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遍历JSON并将列表数组添加到对象/数组

zmshenzheng 回答：遍历JSON并将列表数组添加到对象/数组

Jackson

Gson

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