我扩展了本机实现FirebaseMessagingService
等待本机推送通知android
.
当用户单击推送通知时,我需要启动我的 flutter 应用程序,所以..如何将数据发送到我的 flutter 应用程序?
在颤振中
void main() {
runApp(MyApp());
}
class MyApp extends StatelessWidget{
@override
Widget build(BuildContext context) {
return MaterialApp(
home: ScreenPage(),
);
}
}
class ScreenPage extends StatefulWidget {
@override
_ScreenPageState createState() => _ScreenPageState();
}
class _ScreenPageState extends State<ScreenPage> {
static const platform = const MethodChannel("myChannel");
@override
void initState() {
platform.setMethodCallHandler(nativeMethodCallHandler);
super.initState();
}
Future<dynamic> nativeMethodCallHandler(MethodCall methodCall) async {
print('Native call!');
switch (methodCall.method) {
case "methodNameItz" :
return "This data from flutter.....";
break;
default:
return "Nothing";
break;
}
}
@override
Widget build(BuildContext context) {
//return ();
}
}
In Java
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;
import io.flutter.app.FlutterActivity;
import io.flutter.plugin.common.MethodChannel;
//import io.flutter.view.FlutterNativeView;
public class MyJavaFile extends FlutterActivity {
Button clickMeButton;
MethodChannel channel;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
channel = new MethodChannel(getFlutterView(), "myChannel");
setContentView(R.layout.home_activity);
clickMeButton = findViewById(R.id.clickMeButton);
clickMeButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
channel.invokeMethod("methodNameItz", null, new MethodChannel.Result() {
@Override
public void success(Object o) {
Log.d("Results", o.toString());
}
@Override
public void error(String s, String s1, Object o) {
}
@Override
public void notImplemented() {
}
});
}
});
}
}
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