JSON数据和Java对象的相互转换

JSON解析器：

• 常见的解析器：Jsonlib，Gson，Fastjson，Jackson

JSON转为Java对象

1. 导入jackson的相关jar包
2. 创建Jackson核心对象 ObjectMapper
3. 调用ObjectMapper的readValue()方法进行转换

@Test
public void test() throws IOException {
    //初始化JSON字符串
    String json = "{\"name\":\"Mr.Q\",\"age\":22,\"address\":\"Inner Mongolia\",\"birthday\":\"2020-03-16\"}";
    ObjectMapper mapper = new ObjectMapper();
    //转换为Java对象
    Person per = mapper.readValue(json, Person.class);
    System.out.println(per);
}

Person{name='Mr.Q', age=22, address='Inner Mongolia', birthday=Mon Mar 16 08:00:00 CST 2020}

Java对象转换JSON

转换步骤：

1. 导入Jackson的相关jar包

2. 创建Jackson核心对象ObjectMapper

3. 调用ObjectMapper的相关方法进行转换

转换方法：

Java对象转为JSON

writeValue(参数，obj)

参数：

• File：将obj对象转换为JSON字符串，并保存到指定的文件中
• Writer：将obj对象转换为JSON宇符串，并将JSON数据填充到字符输岀流中
• OutputStream：将obj对象转换为JSON字符串，并将JSON数据填充到字节输岀流中

writeValueAsString(obj)：将对象转为json字符串

---

I. 新建JavaBean对象

@Data
@ToString
public class Person {
    private String name;
    private int age;
    private String address;
}

II. 单元测试类

@Test
public void test1() throws Exception {
    Person per = new Person();
    per.setName("Mr.Q");;
    per.setAge(22);
    per.setAddress("Inner Mongolia");
    //创建Jackson对象ObjectMapper
    ObjectMapper mapper = new ObjectMapper();
    String json = mapper.writeValueAsString(per);
    //{"name":"Mr.Q","age":22,"address":"Inner Mongolia"}
    //System.out.println(json);

    //writeValue:将数据写到文件中
    mapper.writeValue(new File("F://Json.txt"),per);

    //writeValue:将数据写到输出流中
    mapper.writeValue(new FileWriter("F://JsonStream.txt"),per);
    mapper.writeValue(new FileOutputStream("F://JsonTest.txt"),per);
}

生成的文件：

【注解】

1. @JsonIgnore：排除属性
2. @JsonFormat：属性值得格式化

如果在Person中添加属性birthday

则上面的对象属性解析为：

{"name":"Mr.Q","age":22,"address":"Inner Mongolia","birthday":1584346372533}

其中 "birthday":1584346372533 为时间戳

• 如果我们不想转换这个属性，就用@JsonIgnore来排除此birthday属性

• 如果格式化显示则用@JsonFormat

{"name":"Mr.Q","age":22,"address":"Inner Mongolia","birthday":"2020-03-16"}

复杂java对象转换

1. List：数组
2. Map：对象格式一致

List

@Test
public void testList() throws JsonProcessingException {
    Person per = new Person();
    per.setName("Mr.Q");;
    per.setAge(22);
    per.setAddress("Inner Mongolia");
    per.setBirthday(new Date());

    Person per1 = new Person();
    per1.setName("Mr.Q");;
    per1.setAge(22);
    per1.setAddress("Inner Mongolia");
    per1.setBirthday(new Date());

    Person per2 = new Person();
    per2.setName("Mr.Q");;
    per2.setAge(22);
    per2.setAddress("Inner Mongolia");
    per2.setBirthday(new Date());

    List<Person> list = new ArrayList<Person>();
    list.add(per);
    list.add(per1);
    list.add(per2);
    ObjectMapper mapper = new ObjectMapper();
    String json = mapper.writeValueAsString(list);
    System.out.println(json);
}

Map

@Test
public void testMap() throws JsonProcessingException {

    Map<String,Object> map = new HashMap<String, Object> ();

    map.put("name", "Mr.Q");

    map.put("age", 22);

    map.put("address" , "Xian");

    ObjectMapper mapper = new ObjectMapper();

    String json = mapper.writeValueAsString(map);

    //{"address":"Xian","name":"Mr.Q","age":22}

    System.out.println(json);

}

{"address":"Xian","name":"Mr.Q","age":22}