未捕获的引用错误:未定义新货币

Question

为什么没有定义它，尽管我已经在顶部定义了newMoney，这是我的代码：

             var deduct = 100;
             var newMoney = {{user.e_money}} - deduct;
             var username = getUserName(); //get the current user
              // send a message to the server that the e-money value has changed
             socket.emit('update e-money', username, newMoney);
             console.log("Emitting the data to the server side - emoney: " + newMoney + " with the name money of : " + username);
            //end 
            clearTimeout(interval);
            //send the data to the server
            socket.emit('chat message', getUser());
            var interval = setTimeout(function(){
                $('.'+getUser()).fadeIn();
            },5000);
        });
             socket.on('update e-money response', function (data) {
             alert("Your money is: "+ data.newMoney);
             console.log("Your money is: "+ data.newMoney);
             });

在我的服务器端，这是我的代码：

 //emoney
socket.on('update e-money', function (data) {
 var userName = data.username;
 var newMoney = data.newMoney;
//var query = {"name": userName};
  // update the entry on the database
  User.findOneAndUpdate({"username":userName}, {"$set":{"e_money": "300" }}, { upsert: true, returnOriginal:false }, function (err, doc) {
if (err) {
  console.log("There was an error: " + err);
  console.log(userName);
  io.emit('update e-money error', { error: err });
} else {
    io.emit('update e-money response', { newMoney: newMoney });
    console.log(newMoney);
}
  });
  });

我不知道为什么它是未定义的.伙计们，你能帮帮我吗.任何人。

Answer 1

问题就在这里

  var deduct = 100;
  var newMoney = {{user.e_money}} - deduct;
  var username = getUserName(); //get the current user
   // send a message to the server that the e-money value has changed
  socket.emit('update e-money', username, newMoney);

我应该这样做

  var deduct = 100;
  var newMoney = {{user.e_money}} - deduct;
  var username = getUserName();
  var compile = {newMoney, username};

然后在我的服务器端使用它:)

那我就这样发出来

 socket.emit('update e-money', compile);