如何在Java中定制JSON字符串

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.12.4</version>
</dependency>

public class Person {
    private String name;
    private int age;

    // 省略构造方法、getter和setter
}

import com.fasterxml.jackson.databind.ObjectMapper;

public class Main {
    public static void main(String[] args) throws Exception {
        Person person = new Person("John", 25);

        ObjectMapper objectMapper = new ObjectMapper();
        String jsonString = objectMapper.writeValueAsString(person);

        System.out.println(jsonString);
    }
}

{"name":"John","age":25}