If you have already done the method that returns a String, just put it in JSON format. 
An alternative would be to generate a JSON, from an object, see below:
If you are using Maven in your project, you can include these two libraries as a dependency:
<dependency>
   <groupId>com.fasterxml.jackson.core</groupId>
   <artifactId>jackson-core</artifactId>
   <version>2.6.1</version>
</dependency>
<dependency>
   <groupId>com.fasterxml.jackson.core</groupId>
   <artifactId>jackson-databind</artifactId>
   <version>2.6.1</version>
</dependency>
Or simply download them and include in your project.
And the code would be something like this: 
A class with attributes
public class Person{
    private String name;
    private Integer age;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public Integer getAge() {
        return age;
    }
    public void setAge(Integer age) {
        this.age = age;
    }
}
And his method:
ObjectMapper mapper = new ObjectMapper();
Person person = new Person();
person.setName("Name");
person.setAge(12);
try {
      String out = mapper.writeValueAsString(person);
      System.out.println(out);
} catch (JsonProcessingException e) {
      e.printStackTrace();
}
							
							
						 
Post your code so we can try to help you.
– Renan
If you’ve already done the hardest part, which is to do JSON (string), then creating the file is a piece of cake - open it for writing and write the string on it! A JSON file is a text file, and only (but I think the encoding has to be UTF-8, I’m not sure). Nothing special. (or got it wrong, and you still haven’t set up JSON?)
– mgibsonbr
Thanks I did what you told me I took the string already with the json format and wrote a file solving the problem.
– gabrielguedes