2014-06-18 61 views
0

我嵌套了POJO,如下所示。所有的POJO都在同一個包中。請參閱學生的名字POJO和所有其他POJO都在裏面從嵌套POJOS形成Json

class Student{ 

    String firstName; 
    String lastName; 

    List <Activities> activites; 

    public String getFirstName() { 
     return firstName; 
    } 

    public void setFirstName(String firstName) { 
     this.firstName = firstName; 
    } 

    public String getLastName() { 
     return lastName; 
    } 

    public void setLastName(String lastName) { 
     this.lastName = lastName; 
    } 

    public List<Activities> getActivites() { 
     return activites; 
    } 

    public void setActivites(List<Activities> activites) { 
     this.activites = activites; 
    } 

    } 

    class Activites{ 

    List<Quipments> quipments; 

    String time; 

    public List<Quipments> getQuipments() { 
     return quipments; 
    } 

    public void setQuipments(List<Quipments> quipments) { 
     this.quipments = quipments; 
    } 

    public String getTime() { 
     return time; 
    } 

    public void setTime(String time) { 
     this.time = time; 
    } 



    } 

    class Quipments{ 

    String Type; 

    public String getType() { 
     return Type; 
    } 

    public void setType(String type) { 
     Type = type; 
    } 

    } 

我想在JSON以上學生POJO,將有其他的POJO的所有值轉換。

我們不假設在api下面使用。我知道下面的作品。

import com.sun.jersey.api.json.JSONJAXBContext; 
import com.sun.jersey.api.json.JSONMarshaller; 

想用類似下面

import org.json.simple.JSONArray; 
import org.json.simple.JSONObject; 
import org.json.simple.JSONValue; 

請指點

+0

對於所需的庫,請參閱[鏈接](HTTPS ://code.google.com/p/json-simple/) –

+0

json-simple似乎太簡單了:-)我想你會有人使用它重新發明輪子的工作。你爲什麼不嘗試flexjson,gson或jackson呢? – Leo

回答

0

使用這個非常流行的庫https://github.com/google/gson

String sample = "{firstName: \"mardar\", lastName: \"pandit\"}"; 
com.google.gson.Gson gson = new com.google.gson.Gson(); 
Student student = gson.fromJson(sample, Student.class); 
System.out.println(gson.toJson(student));