如何使用Jackson将JSON字符串解析为数组

Question

我有一个值为以下内容的String：

[
  {
    "key1": "value11",
    "key2": "value12"
  },
  {
    "key1": "value21",
    "key2": "value22"
  }
]

并且以下是该类：

public class SomeClass {
    private String key1;
    private String key2;
    /* ... getters and setters omitted ...*/
}

我想将它解析为一个List<SomeClass>或者SomeClass[]，使用Jackson的ObjectMapper，最简单的方式是什么？

- mmutilva

---

我曾经遇到过类似的情况，但是我决定将其反序列化为JSONArray对象，而不是反序列化为一个类，因为这样可以避免未来出现类序列化问题。我是否忽略了任何优势？ - Arun

Answer 1

我终于明白了：

ObjectMapper objectMapper = new ObjectMapper();
TypeFactory typeFactory = objectMapper.getTypeFactory();
List<SomeClass> someClassList = objectMapper.readValue(jsonString, typeFactory.constructCollectionType(List.class, SomeClass.class));

- mmutilva

---

什么是ObjectMapper？你能具体说明一下吗？ - hitesh141

@hitesh141 ObjectMapper 是 com.fasterxml.jackson.databind.ObjectMapper 的成员。 - Chu

Answer 2

另一个答案是正确的，但为了完整起见，这里提供其他方法：

List<SomeClass> list = mapper.readValue(jsonString, new TypeReference<List<SomeClass>>() { });
SomeClass[] array = mapper.readValue(jsonString, SomeClass[].class);

- StaxMan

Answer 3

一个带有数组的完整示例。 将“constructArrayType()”替换为“constructCollectionType()”或任何您需要的其他类型。

import java.io.IOException;

import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.TypeFactory;

public class Sorting {

    private String property;

    private String direction;

    public Sorting() {

    }

    public Sorting(String property, String direction) {
        this.property = property;
        this.direction = direction;
    }

    public String getProperty() {
        return property;
    }

    public void setProperty(String property) {
        this.property = property;
    }

    public String getDirection() {
        return direction;
    }

    public void setDirection(String direction) {
        this.direction = direction;
    }

    public static void main(String[] args) throws JsonParseException, IOException {
        final String json = "[{\"property\":\"title1\", \"direction\":\"ASC\"}, {\"property\":\"title2\", \"direction\":\"DESC\"}]";
        ObjectMapper mapper = new ObjectMapper();
        Sorting[] sortings = mapper.readValue(json, TypeFactory.defaultInstance().constructArrayType(Sorting.class));
        System.out.println(sortings);
    }
}

- Didasko

Answer 4

我通过在JSONLint.com上验证JSON，然后使用Jackson解决了这个问题。以下是相同的代码。

 Main Class:-

String jsonStr = "[{\r\n" + "       \"name\": \"John\",\r\n" + "        \"city\": \"Berlin\",\r\n"
                + "         \"cars\": [\r\n" + "            \"FIAT\",\r\n" + "          \"Toyata\"\r\n"
                + "     ],\r\n" + "     \"job\": \"Teacher\"\r\n" + "   },\r\n" + " {\r\n"
                + "     \"name\": \"Mark\",\r\n" + "        \"city\": \"Oslo\",\r\n" + "        \"cars\": [\r\n"
                + "         \"VW\",\r\n" + "            \"Toyata\"\r\n" + "     ],\r\n"
                + "     \"job\": \"Doctor\"\r\n" + "    }\r\n" + "]";

        ObjectMapper mapper = new ObjectMapper();

        MyPojo jsonObj[] = mapper.readValue(jsonStr, MyPojo[].class);

        for (MyPojo itr : jsonObj) {

            System.out.println("Val of getName is: " + itr.getName());
            System.out.println("Val of getCity is: " + itr.getCity());
            System.out.println("Val of getJob is: " + itr.getJob());
            System.out.println("Val of getCars is: " + itr.getCars() + "\n");

        }

POJO:

public class MyPojo {

private List<String> cars = new ArrayList<String>();

private String name;

private String job;

private String city;

public List<String> getCars() {
    return cars;
}

public void setCars(List<String> cars) {
    this.cars = cars;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getJob() {
    return job;
}

public void setJob(String job) {
    this.job = job;
}

public String getCity() {
    return city;
}

public void setCity(String city) {
    this.city = city;
} }

  RESULT:-
         Val of getName is: John
         Val of getCity is: Berlin
         Val of getJob is: Teacher
         Val of getCars is: [FIAT, Toyata]

          Val of getName is: Mark
          Val of getCity is: Oslo
          Val of getJob is: Doctor
          Val of getCars is: [VW, Toyata]

- Atul KS