Question

我正在使用socket.IO在NodeJS和Android中构建测验应用程序，

我从服务器发出事件quizzoStatus时遇到问题，该事件第一次触发一次，第二次触发两次，依此类推。

我在这里附上我的代码段

///server side: NodeJS
socket.on('sendQuizzoAnsPoints', async (data)=>{
          try {
              const obj = JSON.parse(data);
              const game = await QuizzoPlayModel.findOne({_id:obj.gameId});
              const player = await UserModel.findOne({_id: game.playerId});
              const opponent = await UserModel.findOne({_id: game.opponentId});
              if(obj.userId == game.opponentId){
                  let update = {
                      opponentPoints: game.opponentPoints + obj.points || 0,
                      opponentWA: game.opponentWA + obj.wrongAns || 0,
                  };
                  await QuizzoPlayModel.findByIdAndUpdate(obj.gameId, update).lean().exec();
                  userNamespace.to(player.socketId).emit('quizzoStatus', {
                      fullName: opponent.fullName,
                      points: game.playerPoints + obj.points,
                      wrongAns: obj.wrongAns,
                      gameId: obj.gameId
                  });
              }
              if(obj.userId == game.playerId) {
                  let update = {
                      playerPoints: game.playerPoints + obj.points || 0,
                      playerWA: game.playerWA + obj.wrongAns || 0,
                  };
                  await QuizzoPlayModel.findByIdAndUpdate(obj.gameId, update).lean().exec();
                  userNamespace.to(opponent.socketId).emit('quizzoStatus', {
                      fullName: player.fullName,
                      points: game.playerPoints+ obj.points,
                      wrongAns: obj.wrongAns,
                      gameId: obj.gameId
                  });
              }
          } catch (e) {
              console.log(e);
          }
        });

在这里，我收听一个名为sendQuizzoAnsPoints的事件，然后在另一个名为quizzoStatus的事件中向玩家或对手发出一个事件。

quizzoStatus事件从服务器到Android多次触发。在这里我附上了android代码

/// Android code
socket.emit("sendQuizzoAnsPoints", new Gson().toJson(quizzoStatusRequestDto)); 
socket.on("quizzoStatus", new Emitter.Listener(){
           @Override
           public void call(Object... args){
               runOnUiThread(new Runnable(){
                   @Override
                   public void run(){
                       Log.e("opponet point", opponentPoints + " " + quizzoStatusResponseDto.getPoints());
                   }
               });
           }
       });

Answer 1

问题出在$html = file_get_contents("log.html"); // Split up your html into header, data row, and the bottom $html_pieces = explode("", $html, 3); // Show the table header echo $html_pieces[0]; $path = './visitors.txt'; $visitor_log = fopen($path, 'r'); while (($line = fgets($visitor_log)) !== false) { $string_split = explode("--", $line, 3); // Generate a new line of $output from the data row of your $html_pieces $output = str_replace("---date---", $string_split[0], $html_pieces[1]); $output = str_replace("---browser---", $string_split[1], $output); $output = str_replace("---ip---", $string_split[2], $output); // And display, line-by-line echo $output; } // All done, finish off your table echo $html_pieces[2];中。您每次都在分配新的侦听器而不删除它。您需要为该Android侦听器创建一个变量，并在完成工作后将其Emmiter或其他地方删除：

onDestroy

希望这会起作用

套接字IO事件多次触发NodeJS

1 个答案: