如何在Bundle中传递自定义对象？

Question

我试图将自己的自定义对象传递到一个包中:

Bundle b = new Bundle();
                STMessage message = (STMessage)a.getAdapter().getItem(position);
                b.putObject("message",message);

我收到错误:

The method putObject(String, Object) is undefined for the type Bundle

Answer 1

一种方法是让您的自定义对象实现Parcelable接口并使用 Bundle.putParcelable/Bundle.getParcelable

Answer 2

模型类

package com.sam.bundleobjectpass;

import java.io.Serializable;

/**
 * Created by Samir on 31.8.2016.
 */
public class Model implements Serializable {
    private String name;
    private String surName;
    private int age;

    public Model(String name, String surName, int age) {
        this.name = name;
        this.surName = surName;
        this.age = age;
    }

    public String getName() {
        return name;
    }


    public String getSurName() {
        return surName;
    }


    public int getAge() {
        return age;
    }


}

主要活动

public class MainActivity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);


        Model model = new Model("Sam", "Sami",32);

        Intent i = new Intent(MainActivity.this, ReceiverActivity.class);
        i.putExtra("Editing", model); // sending our object. In Kotlin is the same
        startActivity(i);

    }
}

ReceiverActivity

public class ReceiverActivity extends Activity {

    TextView txt_name;
    TextView txt_surname;
    TextView txt_age;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.second);

        txt_name = (TextView)findViewById(R.id.txt_name);
        txt_surname = (TextView)findViewById(R.id.txt_surname);
        txt_age = (TextView)findViewById(R.id.txt_age);
        // receiving our object
        Model model = (Model) getIntent().getSerializableExtra("Editing");

        txt_name.setText(model.getName());
        txt_surname.setText(model.getSurName());
        txt_age.setText(""+model.getAge());


    }
}

// Kotlin

val model: ProgramModel = intent.getSerializableExtra("Editing") as ProgramModel
        model.let { // means if not null or empty
            txt_act_daily_topic.text = model.title
        }

Answer 3

使您的自定义对象Parcelable或Serializable然后使用putParcelable或putSerializable.

根据对象的复杂程度,一个或另一个可能更容易.

Answer 4

由于使用Parsable是为某些评论中提到的高性能IPC传输而设计的,我尝试使用不同的方法.

我的方法是谷歌使用GSON库.

例

public class Person{
    private String name;
    private int age;
    // Getter and Setters omitted
}

您可以在实用程序类中使用返回Gson实例的方法,这是为了清洁代码和组织.我将使用GsonBuilder来为某人注册自定义适配器.

public class Utils {
    private static Gson gson;

    private static Gson getGsonParser() {
        if(null == gson) {
            GsonBuilder builder = new GsonBuilder();
            gson = builder.create();
        }
        return gson;
    }
}

片刻的真相!

放

Bundle args = new Bundle();
String personJsonString = Utils.getGsonParser().toJson(person);
args.putString(PERSON_KEY, personJsonString);

得到

Bundle args = getArguments();
String personJsonString = args.get(PERSON_KEY);
Person person= Utils.getGsonParser().fromJson(str, Person.class);

目前我不知道这种方法的性能限制.但它的工作正常