相关疑难解决方法(0)

Java中的内部类和静态嵌套类之间的主要区别是什么？设计/实施是否在选择其中一个方面发挥作用？

我正在使用spring 3.1.2,我需要将一个json对象解析为POJO.这是我需要解析的json:

{
"Person" : {
    "id" : "2"
 },
"Dog" : {
    "dateOfBirth" : "2012-08-20 00:00:00",
    "price" : "10.00"
    }
}

我需要将这个json对象(由两个对象组合)转换为一个POJO,这里是:

public class MyClass{
     public MyClass(){}
     public MyClass(String personsId, TimeStamp dogsDateOfBirth, BigDecimal dogsPrice){
     .... // assign each parameter to the appropriate field
     }
     private String personsId;
     private TimeStamp dogsDateOfBirth;
     private BigDecimal dogsPrice;
     //... Getters and Setters for each field
}

为此,我使用了ObjectMapper mapper = new ObjectMapper(); Now,因为我有几个json对象,我的代码看起来像这样:

    String json = ... ;// A json with several objects as …

我按如下方式创建JSON:

    var manager = {
        username: "admin",
        password: "admin"
    };
    var userToSubscribe = {
        username: "newuser",
        password: "newpassword",
        email: "user@1and1.es"
    };

    var openid = "myopenid";

    var subscription = {
            manager: manager,
            userToSubscribe : userToSubscribe,
            openid : openid
    };

    $.ajax({
        url: '/myapp/rest/subscribeUser.json',
        type: 'POST',
        dataType: 'json',
        contentType: 'application/json',
        mimeType: 'application/json',
        data: JSON.stringify({subscription : subscription})   
    });

这是发送的JSON:

{"subscription":{"manager":{"username":"admin","password":"admin"},"userToSubscribe":{"username":"newuser","password":"newpassword","email":"user@1and1.es"},"openid":"myopenid"}}  

我想将这个JSON映射到Wrapper类.这是包装器:

private class Subscription{
    private User manager;
    private User userToSubscribe;
    private String openid;
    public User getManager() {
        return manager;
    }
    public void setManager(User manager) …