java android 中的信号器连接事件句柄
signalr connection event handle in java android
我正在使用 SignalR 制作一款双人在线游戏。到目前为止,一切都很好。但是当其中一个玩家的互联网被切断(几毫秒内)或服务器因任何原因被切断时,该程序将不再运行。如何收到连接事件的通知?
如何自动重新连接?
研究了很久,一直没有找到解决办法
Asp.NET 核心 5.0 服务器端:
public class ChatHub : Hub
{
LogHelper logHelper = new LogHelper("hub");
public async Task Move(float x, float y)
{
await Clients.Others.SendAsync("ReceivePosition", x, y);
}
}
客户端javaandroid工作室:
public class MainActivity extends AppCompatActivity {
private static final String TAG = "qazwsx";
Button btnStart;
View viewMove;
HubConnection hubConnection;
@SuppressLint({"ClickableViewAccessibility", "CheckResult"})
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
hubConnection = HubConnectionBuilder.create("http://192.168.1.10:45455/notification").build();
hubConnection.on("ReceivePosition", (x, y) -> {
runOnUiThread(() -> {
viewMove.setX(x);
viewMove.setY(y);
Log.i(TAG, String.format("onCreate: x:%s y:%s", x, y));
});
}, Float.class, Float.class);
btnStart = findViewById(R.id.bntStart);
viewMove = findViewById(R.id.viewMove);
btnStart.setOnClickListener(view -> {
Log.i(TAG, "onCreate > btnStart.setOnClickListener > status:" + hubConnection.getConnectionState());
if (btnStart.getText().toString().equals("START")) {
if (hubConnection.getConnectionState() == HubConnectionState.DISCONNECTED) {
btnStart.setText("STOP");
}
} else {
if (hubConnection.getConnectionState() == HubConnectionState.CONNECTED) {
hubConnection.stop();
}
}
});
final float[] dX = new float[1];
final float[] dY = new float[1];
viewMove.setOnTouchListener((view, event) -> {
switch (event.getAction()) {
//this is your code
case MotionEvent.ACTION_DOWN:
dX[0] = view.getX() - event.getRawX();
dY[0] = view.getY() - event.getRawY();
break;
case MotionEvent.ACTION_MOVE:
view.animate()
.x(event.getRawX() + dX[0])
.y(event.getRawY() + dY[0])
.setDuration(0)
.start();
if (hubConnection.getConnectionState() == HubConnectionState.CONNECTED) {
hubConnection.send("Move", event.getRawX() + dX[0], event.getRawY() + dY[0]);
Log.i(TAG, String.format("move: x:%s y:%s", dX[0], dY[0]));
}
break;
default:
return false;
}
return true;
});
}
}
经过大量研究,我设法解决了这个问题。我正在写这个问题的解决方案。我希望它能解决别人的问题
处理信号器连接关闭:
hubConnection.onClosed(exception -> {
//do something
//attemt to connect
//note: exception is null when the user stop connection
});
处理信号器连接开始:
hubConnection.start()
.doOnError(throwable -> {
Log.e(TAG, "doInBackground > doOnError: ", throwable);
//start fail , try again
//note: the start function need try chach when we use this function
})
.doOnComplete(() -> {
Log.i(TAG, "doInBackground > doOnComplete.");
//start complated
})
.blockingAwait();//you must write this function ,else other function not worck
我正在使用 SignalR 制作一款双人在线游戏。到目前为止,一切都很好。但是当其中一个玩家的互联网被切断(几毫秒内)或服务器因任何原因被切断时,该程序将不再运行。如何收到连接事件的通知? 如何自动重新连接?
研究了很久,一直没有找到解决办法
Asp.NET 核心 5.0 服务器端:
public class ChatHub : Hub
{
LogHelper logHelper = new LogHelper("hub");
public async Task Move(float x, float y)
{
await Clients.Others.SendAsync("ReceivePosition", x, y);
}
}
客户端javaandroid工作室:
public class MainActivity extends AppCompatActivity {
private static final String TAG = "qazwsx";
Button btnStart;
View viewMove;
HubConnection hubConnection;
@SuppressLint({"ClickableViewAccessibility", "CheckResult"})
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
hubConnection = HubConnectionBuilder.create("http://192.168.1.10:45455/notification").build();
hubConnection.on("ReceivePosition", (x, y) -> {
runOnUiThread(() -> {
viewMove.setX(x);
viewMove.setY(y);
Log.i(TAG, String.format("onCreate: x:%s y:%s", x, y));
});
}, Float.class, Float.class);
btnStart = findViewById(R.id.bntStart);
viewMove = findViewById(R.id.viewMove);
btnStart.setOnClickListener(view -> {
Log.i(TAG, "onCreate > btnStart.setOnClickListener > status:" + hubConnection.getConnectionState());
if (btnStart.getText().toString().equals("START")) {
if (hubConnection.getConnectionState() == HubConnectionState.DISCONNECTED) {
btnStart.setText("STOP");
}
} else {
if (hubConnection.getConnectionState() == HubConnectionState.CONNECTED) {
hubConnection.stop();
}
}
});
final float[] dX = new float[1];
final float[] dY = new float[1];
viewMove.setOnTouchListener((view, event) -> {
switch (event.getAction()) {
//this is your code
case MotionEvent.ACTION_DOWN:
dX[0] = view.getX() - event.getRawX();
dY[0] = view.getY() - event.getRawY();
break;
case MotionEvent.ACTION_MOVE:
view.animate()
.x(event.getRawX() + dX[0])
.y(event.getRawY() + dY[0])
.setDuration(0)
.start();
if (hubConnection.getConnectionState() == HubConnectionState.CONNECTED) {
hubConnection.send("Move", event.getRawX() + dX[0], event.getRawY() + dY[0]);
Log.i(TAG, String.format("move: x:%s y:%s", dX[0], dY[0]));
}
break;
default:
return false;
}
return true;
});
}
}
经过大量研究,我设法解决了这个问题。我正在写这个问题的解决方案。我希望它能解决别人的问题
处理信号器连接关闭:
hubConnection.onClosed(exception -> {
//do something
//attemt to connect
//note: exception is null when the user stop connection
});
处理信号器连接开始:
hubConnection.start()
.doOnError(throwable -> {
Log.e(TAG, "doInBackground > doOnError: ", throwable);
//start fail , try again
//note: the start function need try chach when we use this function
})
.doOnComplete(() -> {
Log.i(TAG, "doInBackground > doOnComplete.");
//start complated
})
.blockingAwait();//you must write this function ,else other function not worck