在发送活动中尝试以下代码

   //EXTRA_MESSAGE is our key and it's value is 'packagename.MESSAGE'
    public static final String EXTRA_MESSAGE = "packageName.MESSAGE";

    @Override
    protected void onCreate(Bundle savedInstanceState) {
       ....

        //Here we declare our send button
        Button sendButton = (Button) findViewById(R.id.send_button);
        sendButton.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                //declare our intent object which takes two parameters, the context and the new activity name

                // the name of the receiving activity is declared in the Intent Constructor
                Intent intent = new Intent(getApplicationContext(), NameOfReceivingActivity.class);

                String sendMessage = "hello world"
                //put the text inside the intent and send it to another Activity
                intent.putExtra(EXTRA_MESSAGE, sendMessage);
                //start the activity
                startActivity(intent);

            }

从接收活动尝试以下代码:

   protected void onCreate(Bundle savedInstanceState) {
 //use the getIntent()method to receive the data from another activity
 Intent intent = getIntent();

//extract the string, with the getStringExtra method
String message = intent.getStringExtra(NewActivityName.EXTRA_MESSAGE);

然后将以下代码添加到AndroidManifest.xml文件中

  android:name="packagename.NameOfTheReceivingActivity"
  android:label="Title of the Activity"
  android:parentActivityName="packagename.NameOfSendingActivity"

启动新活动的方法是广播一个意图，您可以使用特定类型的意图将数据从一个活动传递到另一个活动。我的建议是你去看看Android开发者文档中与intent相关的内容;这是一个丰富的信息的主题，并有例子。

在发送活动中尝试以下代码

   //EXTRA_MESSAGE is our key and it's value is 'packagename.MESSAGE'
    public static final String EXTRA_MESSAGE = "packageName.MESSAGE";

    @Override
    protected void onCreate(Bundle savedInstanceState) {
       ....

        //Here we declare our send button
        Button sendButton = (Button) findViewById(R.id.send_button);
        sendButton.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                //declare our intent object which takes two parameters, the context and the new activity name

                // the name of the receiving activity is declared in the Intent Constructor
                Intent intent = new Intent(getApplicationContext(), NameOfReceivingActivity.class);

                String sendMessage = "hello world"
                //put the text inside the intent and send it to another Activity
                intent.putExtra(EXTRA_MESSAGE, sendMessage);
                //start the activity
                startActivity(intent);

            }

从接收活动尝试以下代码:

   protected void onCreate(Bundle savedInstanceState) {
 //use the getIntent()method to receive the data from another activity
 Intent intent = getIntent();

//extract the string, with the getStringExtra method
String message = intent.getStringExtra(NewActivityName.EXTRA_MESSAGE);

然后将以下代码添加到AndroidManifest.xml文件中

  android:name="packagename.NameOfTheReceivingActivity"
  android:label="Title of the Activity"
  android:parentActivityName="packagename.NameOfSendingActivity"

一个老问题，但如果目标是切换显示的页面，我只有一个活动，当我想切换页面时调用setContentView()(通常是响应用户单击按钮)。这允许我简单地从一个页面的内容调用到另一个页面。没有意图疯狂的额外包裹捆绑和任何试图来回传递数据。

我像往常一样在分辨率/布局中做了一堆页面，但没有为每个页面做一个活动。只需使用setContentView()来根据需要切换它们。

所以我的一个和唯一的onCreate()有:

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    LayoutInflater layoutInflater = getLayoutInflater();

    final View mainPage = layoutInflater.inflate(R.layout.activity_main, null);
    setContentView (mainPage);
    Button openMenuButton = findViewById(R.id.openMenuButton);

    final View menuPage = layoutInflatter.inflate(R.layout.menu_page, null);
    Button someMenuButton = menuPage.findViewById(R.id.someMenuButton);

    openMenuButton.setOnClickListener(new View.OnClickListener() {
        public void onClick(View v) {
            setContentView(menuPage);
        }
    });

    someMenuButton.setOnClickListener(new View.OnClickListener() {
        public void onClick(View v) {
            do-something-interesting;
            setContentView(mainPage);
        }
    }
}

如果你想要返回按钮在退出应用程序之前返回内部页面，只需包装setContentView()将页面保存在一个小的页面堆栈中，并在onBackPressed()处理程序中弹出这些页面。

试试这个简单的方法。

startActivity(new Intent(MainActivity.this, SecondActivity.class));

Intent i = new Intent(firstactivity.this, secondactivity.class);
startActivity(i);