我正在开发一个Android聊天应用程序,使用Node Js和Redis存储消息和用户信息。我正在使用套接字io进行通信,并使用Room将消息存储在本地数据库中。当用户离线时,我希望他们再次在线时收到他们的消息。我的问题是,当用户A离线时,用户B向他发送许多消息(例如5条消息),而当用户A再次在线时,他仅收到第一条消息,而最后一条消息却收到4次。这就是我正在做的事情,一旦用户收到一条消息,我会将Redis中的消息状态从“已发送”更新为“已发送”。在用户脱机的情况下,我将其消息存储在Redis中,其状态为消息“已发送”,并且再次联机时,我检查例如从用户B收到的消息,如果其状态为“已发送”,则我进行传递它发送给用户,然后将其更新为“已交付”,如下面的代码所示:
//On this event, we update the socket ID of the sender in Redis so they can
receive private messages from their contacts
socket.on('sender', (sender, destinat) =>{
tempId = socket.id;
senderId = sender;
users[sender] = sender;
users [destinat] = destinat;
//We also update the user status: online
client.hset(senderId, 'lastSeen', 'Now', function(reply){
console.log( senderId + reply);
});
//Stocking to the user socket id
client.hset(users[sender], 'tempId', tempId, function(){
console.log("Welcome " + sender);
console.log("Welcome " + tempId);
});
//Getting all the messages of the sender from users
//If the sender has any messages that hasn't received yet, they'll be sent
here
//the id of each message is compsed of two parts: the phone number of the
receiver, and the id of the message itself
(receiverPhoneNumber:idMessage)
client.keys(users [sender] + ':*', function(err, results) {
results.forEach(function(key) {
client.hgetall(key, function(err, reply){
if(err)
console.log(err);
else if(reply){
//Compare the message status: if not sent, deliver it to receiver once online
if('Sent'.localeCompare(reply.status) == 0 && users
[destinat].localeCompare(reply.fromUser) == 0) {
io.to(tempId).emit('message', reply);
}
}
});
});
});
});
从服务器接收到消息后,我使用Async将它们存储在Room Database中,然后将其显示给用户,如以下代码所示
这是AsyncTask类:
class AddMessage extends AsyncTask<Void, Void, Void> {
@Override
protected Void doInBackground(Void... voids) {
//Creating a user account
m = new Message();
m.setContent( message );
m.setTime( time );
m.setUrl( url );
m.setStatus( status );
m.setFromUser( fromUser );
m.setToUser( toUser );
m.setUsername( receiver.getUsername() );
//adding to database
DatabaseClient.getInstance(getContext()).getAppDatabase()
.messageDao()
.insert(m);
return null;
}
@Override
protected void onPostExecute(Void aVoid) {
super.onPostExecute(aVoid);
Toast.makeText( getContext(), "Added!", Toast.LENGTH_SHORT ).show();
}
}
我检查了是否已正确从服务器收到消息到android应用程序(一旦将消息重新发送到服务器后再重新发送到服务器)。我相信问题与AsyncTask有关,但我想不通,非常感谢您的帮助,非常感谢。
//When receving a message
socket.on("message", new Emitter.Listener() {
@Override
public void call(final Object... args) {
if(getActivity() != null){
getActivity().runOnUiThread(new Runnable() {
@Override
public void run() {
JSONObject data = (JSONObject) args[0];
try {
//extract data from fired event
idMessage = data.getString( "idMessage" );
message = data.getString("message");
fromUser = data.getString( "fromUser" );
toUser = data.getString( "toUser" );
time = data.getString( "time" );
status = data.getString( "status" );
url = data.getString( "url" );
//Here we call asyncTask to Add it to Database
addMessage = new AddMessage();
addMessage.execute( );
//We emit this event to update the status of
the message to delivered
socket.emit( "sent", idMessage, userID );
} catch (JSONException e) {
e.printStackTrace();
}
}
});
}
}
});
答案 0 :(得分:0)
我通过切换到RxJava而不是AsyncTask解决了这个问题。该问题与AsyncTask有关,因为它有时会影响数据链,而this link中提到的RxJava则不是这样:“ AsyncTasks的另一个问题是如果您一次运行多个。您无法保证它们将以什么顺序完成,从而导致检查所有任务何时完成的逻辑复杂,更糟糕的是假设一个任务要比另一个任务先完成,直到遇到使第一次呼叫变慢的极端情况为止,这会使它们以错误的顺序完成并产生不良结果。”