Question

我想从Intent方法中提取第一个参数。我怎么能这样做？

示例，BroadcastCodes.LOGIN_BROADCAST只是字符串参数。

Utilities.sendLocalBroadcastActivity(new Intent(BroadcastCodes.LOGIN_BROADCAST, null,
                  ServerConnectionManager.this, LoginActivity.class));

在MainActivity类中将是：

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

    progressInfo = findViewById(R.id.splash_screen_progress_info);

    Utilities.startService(this, StickyService.class);

        localBroadcastManager = LocalBroadcastManager.getInstance(getApplicationContext());

        broadcastReceiver = new BroadcastReceiver() {
        @Override
        public void onReceive(Context context, Intent intent) {

        // How should I extract the first parameter to compare and call other function.

        String broadcast = intent.getStringExtra("?");

        logger.info("Received broadcast: " + broadcast);
        if(?)
        {
          handleConnectBroadcast(broadcast);
        }
      }
    };
    connectionMessagesHandler = new ConnectionMessagesHandler();
  }

Answer 1

如果我理解，你想获取第一个参数，即BroadcastCodes.LOGIN_BROADCAST这是动作参数，

使用intent.getAction();

希望它有所帮助！

Answer 2

这是一个如何提取意图数据的例子：

Intent intent = getIntent();
String age= intent.getStringExtra("age");
String name = intent.getStringExtra("name");

在您的代码中，它将是：

public void onReceive(Context context, Intent intent) {
        String broadcast = intent.getStringExtra("YOUR_EXTRA_NAME");

在您的评论案例中，您应该添加：

if (intent.getAction().equals(Constants.LOGIN_BROADCAST)){
//code
}

Answer 3

Intent intent = new Intent(ABC.this,XYZ.class);
intent.putExtra("any_key_name","any_value");
startActivity(intent)

如何从Intent中提取第一个参数？

3 个答案: