We can send data one Activty1 to Activity2 with multiple ways like.
1- Intent
2- bundle
3- create an object and send through intent
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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");

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。

最简单的方法是在item是字符串的情况下使用以下代码:

intent.putextra("selected_item",item)

接收:

String name = data.getStringExtra("selected_item");

Start another activity from this activity pass parameters via Bundle Object

Intent intent = new Intent(this, YourActivity.class);
Intent.putExtra(AppConstants.EXTRAS.MODEL, cModel);
startActivity(intent);
Retrieve on another activity (YourActivity)

ContentResultData cModel = getIntent().getParcelableExtra(AppConstants.EXTRAS.MODEL);

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。

你可以使用putExtra(Serializable..)和getSerializableExtra()方法来传递和检索你的类类型的对象;你必须将你的类标记为Serializable，并确保你所有的成员变量也是Serializable…