谢谢你的帮助，但我发现了一个可选的解决方案

 public class getsetclass implements Serializable {
        private int dt = 10;
    //pass any object, drwabale 
        public int getDt() {
            return dt;
        }

        public void setDt(int dt) {
            this.dt = dt;
        }
    }

活动一

getsetclass d = new getsetclass ();
                d.setDt(50);
                LinkedHashMap<String, Object> obj = new LinkedHashMap<String, Object>();
                obj.put("hashmapkey", d);
            Intent inew = new Intent(SgParceLableSampelActivity.this,
                    ActivityNext.class);
            Bundle b = new Bundle();
            b.putSerializable("bundleobj", obj);
            inew.putExtras(b);
            startActivity(inew);

在活动2中获取数据

 try {  setContentView(R.layout.main);
            Bundle bn = new Bundle();
            bn = getIntent().getExtras();
            HashMap<String, Object> getobj = new HashMap<String, Object>();
            getobj = (HashMap<String, Object>) bn.getSerializable("bundleobj");
            getsetclass  d = (getsetclass) getobj.get("hashmapkey");
        } catch (Exception e) {
            Log.e("Err", e.getMessage());
        }

如果你有一个单例类(fx Service)作为你的模型层的网关，它可以通过在该类中有一个带有getter和setter的变量来解决。

活动一:

Intent intent = new Intent(getApplicationContext(), Activity2.class);
service.setSavedOrder(order);
startActivity(intent);

活动二:

private Service service;
private Order order;

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

    service = Service.getInstance();
    order = service.getSavedOrder();
    service.setSavedOrder(null) //If you don't want to save it for the entire session of the app.
}

在服务:

private static Service instance;

private Service()
{
    //Constructor content
}

public static Service getInstance()
{
    if(instance == null)
    {
        instance = new Service();
    }
    return instance;
}
private Order savedOrder;

public Order getSavedOrder()
{
    return savedOrder;
}

public void setSavedOrder(Order order)
{
    this.savedOrder = order;
}

此解决方案不需要对相关对象进行任何序列化或其他“打包”。但是，只有在使用这种架构时才会有好处。

您需要将对象序列化为某种字符串表示形式。一种可能的字符串表示是JSON，如果你问我，在android中序列化JSON最简单的方法之一是通过谷歌GSON。

在这种情况下，你只需要输入字符串返回值from (new Gson()).toJson(myObject);并检索字符串值并使用fromJson将其转换回您的对象。

但是，如果您的对象不是非常复杂，那么可能不值得这样做，您可以考虑传递对象的单独值。

我使用Gson及其强大而简单的api在活动之间发送对象，

例子

// This is the object to be sent, can be any object
public class AndroidPacket {

    public String CustomerName;

   //constructor
   public AndroidPacket(String cName){
       CustomerName = cName;
   }   
   // other fields ....


    // You can add those functions as LiveTemplate !
    public String toJson() {
        Gson gson = new Gson();
        return gson.toJson(this);
    }

    public static AndroidPacket fromJson(String json) {
        Gson gson = new Gson();
        return gson.fromJson(json, AndroidPacket.class);
    }
}

2个函数，你将它们添加到你想要发送的对象中

使用

将对象从A发送到B

    // Convert the object to string using Gson
    AndroidPacket androidPacket = new AndroidPacket("Ahmad");
    String objAsJson = androidPacket.toJson();

    Intent intent = new Intent(A.this, B.class);
    intent.putExtra("my_obj", objAsJson);
    startActivity(intent);

在B接收

@Override
protected void onCreate(Bundle savedInstanceState) {        
    Bundle bundle = getIntent().getExtras();
    String objAsJson = bundle.getString("my_obj");
    AndroidPacket androidPacket = AndroidPacket.fromJson(objAsJson);

    // Here you can use your Object
    Log.d("Gson", androidPacket.CustomerName);
}

我几乎在我做的每个项目中都使用它，我没有任何性能问题。

We can send data one Activty1 to Activity2 with multiple ways like.
1- Intent
2- bundle
3- create an object and send through intent
.................................................
1 - Using intent
Pass the data through intent
Intent intentActivity1 = new Intent(Activity1.this, Activity2.class);
intentActivity1.putExtra("name", "Android");
startActivity(intentActivity1);
Get the data in Activity2 calss
Intent intent = getIntent();
if(intent.hasExtra("name")){
     String userName = getIntent().getStringExtra("name");
}
..................................................
2- Using Bundle
Intent intentActivity1 = new Intent(Activity1.this, Activity2.class);
Bundle bundle = new Bundle();
bundle.putExtra("name", "Android");
intentActivity1.putExtra(bundle);
startActivity(bundle);
Get the data in Activity2 calss
Intent intent = getIntent();
if(intent.hasExtra("name")){
     String userName = getIntent().getStringExtra("name");
}
..................................................
3-  Put your Object into Intent
Intent intentActivity1 = new Intent(Activity1.this, Activity2.class);
            intentActivity1.putExtra("myobject", myObject);
            startActivity(intentActivity1); 
 Receive object in the Activity2 Class
Intent intent = getIntent();
    Myobject obj  = (Myobject) intent.getSerializableExtra("myobject");

迄今为止最简单的方法IMHO包裹对象。只需在希望可打包的对象上方添加注释标记。

下面是该库的一个示例https://github.com/johncarl81/parceler

@Parcel
public class Example {
    String name;
    int age;

    public Example(){ /*Required empty bean constructor*/ }

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

    public String getName() { return name; }

    public int getAge() { return age; }
}