现在,当Gson创建这个类的实例时,我正以非常规的方式创建单例实例.见下文：

public class ApplicationModel {

    private static ApplicationModel instance;

    private GeneralVO general;

    protected ApplicationModel() {
        instance = this;
    }

    public static ApplicationModel getInstance() {
        return instance;
    }

    public String getVersionDate() {
        return general.getVersionDate();
    }
}

这是我创建它然后在应用程序中重用它的方式：

InputStreamReader reader = new InputStreamReader(is);
ApplicationModel model1 = new Gson().fromJson(reader,ApplicationModel.class);

Log.i("mytag","InputStream1 = "+model1.toString());
Log.i("mytag","Date: "+model1.getVersionDate());
ApplicationModel model2 = ApplicationModel.getInstance();
Log.i("mytag","InputStream1 = "+model2.toString());
Log.i("mytag","Date: "+model2.getVersionDate());

这有效,因为getInstance()返回相同的模型,但不知何故,这似乎不正确.

我的问题是“这是一个很好的办法,还是有更好的解决方案？”

编辑

做单例的更好方法是使用带有一个INSTANCE元素的枚举.

See this post for an explanation

public class ApplicationModel {

    private static ApplicationModel instance; //= new ApplicationModel(); 
    //instantiating here is called an "Eagerly Instantiated"

    private GeneralVO general;

    private ApplicationModel() { 
    }

    public static ApplicationModel getInstance() {
        //instantiating here is called "Lazily Instantiated",using : 
        //if (instance==null) {                  --> Check whether ''instance'' is instantiated,or null
        //    instance = new ApplicationModel(); --> Instantiate if null
        //}
        return instance;  //return the single static instance
    }

    public String getVersionDate() {
        return general.getVersionDate();
    }
}

通过将构造函数设置为private,可以防止对象被另一个类重新实例化,要使用该对象,您必须使用ApplicationModel.getInstance()调用该对象.

因此,如果要设置值,请调用ApplicationModel.getInstance().setterMethod(value),为什么这有用？如果要跟踪更改,则只需跟踪setter方法.如果使用了构造函数,则还必须跟踪构造函数.

示例：

// To SET the value:
// instead of ApplicationModel model1 = new Gson().fromJson(reader,ApplicationModel.class);
ApplicationModel.getInstance.setValue(new Gson().fromJson(reader,ApplicationModel.class);

// To GET the value :
ApplicationModel.getInstance.getValue();

“渴望实例化”与“懒惰实例化”：

>如果您想要一种简单的方法来处理,那么Eager Instantiation非常有用
主题
> Lazy Instantiation具有更好的内存占用

除此之外,您可以谷歌搜索更多信息,但我认为这对您来说应该已经足够了.

希望这有帮助,祝你好运^^

问候,

里德

第一个IParser.java

public interface IParser<T> {
    public T parse(String json);
}

第二个Parser.java

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.reflect.Typetoken;

public class Parser<T> implements IParser<T> {

    @Override
    public T parse(String json) {
        GsonBuilder builder = new GsonBuilder();
        Gson gson = builder.enableComplexMapKeySerialization().create();
        MyJson<T> jsonParsed = gson.fromJson(json,new Typetoken<MyJson<T>>() {
        }.getType());

        System.out.println(jsonParsed.toString());
        return null;
    }

}

第三个MyJson.java

public class MyJson<T> {
    private int status;
    private T data;

    public int getStatus() {
        return status;
    }

    public void setStatus(int status) {
        this.status = status;
    }

    public T getData() {
        return data;
    }

    public void setData(T data) {
        this.data = data;
    }
}

和User.java

import java.io.Serializable;

public class User implements Serializable {
    private static final long serialVersionUID = 1L;
    private long usrId;
    private String usrNm;
    private String usrPwd;
    private String usrEml;

    public User() {

    }

    public long getUsrId() {
        return usrId;
    }

    public void setUsrId(long usrId) {
        this.usrId = usrId;
    }

    public String getUsrNm() {
        return usrNm;
    }

    public void setUsrNm(String usrNm) {
        this.usrNm = usrNm;
    }

    public String getUsrPwd() {
        return usrPwd;
    }

    public void setUsrPwd(String usrPwd) {
        this.usrPwd = usrPwd;
    }

    public String getUsrEml() {
        return usrEml;
    }

    public void setUsrEml(String usrEml) {
        this.usrEml = usrEml;
    }

    @Override
    public String toString() {
        return "User [usrId=" + usrId + ",usrNm=" + usrNm + ",usrPwd="
                + usrPwd + ",usrEml=" + usrEml + "]";
    }
}

我的JSON字符串：

{
    status: 200,data: {
        usrId: 2,usrNm: "John",usrPwd: "123",usrEml: "john@test.com"
    }
}

我想在jsonString之上解析为MyJson< T>宾语.

我做：

MyJson<T> jsonParsed = gson.fromJson(json,new Typetoken<MyJson<T>>() {
            }.getType());

=> Gson将MyJson中的用户解析为LinkedTreeMap

它的意思是：Object data = jsonParsed.getData(); => LinkedTreeMap的数据实例

我想要的：用户的数据实例

但是,当我尝试：

MyJson<User> jsonParsed = gson.fromJson(json,new Typetoken<MyJson<User>>() {
            }.getType());

gson工作得很好.

所以我想知道我怎么能用第一种方式使gson工作正常(使用T而不是User),因为我想通过使用Parser解析其他对象

谢谢你的帮助.

@Override
public T parse(String json,Type type) {
    GsonBuilder builder = new GsonBuilder();
    Gson gson = builder.enableComplexMapKeySerialization().create();
    MyJson<T> jsonParsed = gson.fromJson(json,type);

    System.out.println(jsonParsed.toString());
    return null;
}

使用：

User user = myJson.parse(jsonString,new Typetoken<User>() {
        }.getType());

[
  [
    "Passport Number","NATIONALITY","REASONS"
  ],[
    "SHAIS100","INDIA",""
  ],[
    "","","Agent ID is not matched."
  ],""
  ]
]

我想将其填充到ArrayList< String []>,请告诉我该怎么办？

空字符串不应转换为空.

1.-首先创建Gson对象：

Gson gson = new Gson();

2.-然后获取您的列表的通讯录类型< String []> (请注意,由于Java的type erasure,您无法像List< String []> .class这样做)：

Type type = new Typetoken<List<String[]>>() {}.getType();

3.-最后将JSON解析为类型类型的结构：

List<String[]> yourList = gson.fromJson(yourjsonString,type);