javascript - Parse.com 云代码 : function called simultaneously by 2 users

标签 javascript parse-platform parse-cloud-code

我正在审查为我的应用程序编写的所有云代码,并且我已经发现了一些需要纠正的内容。我有一个充满用户和“ session ”对象的解析数据库。每个用户都可以创建或查看并接受 session 。当用户想要接受 session 时,将调用下一个函数。如果所需人数等于接受 session 并确认的人数,否则该 session 仍可供其他用户使用。今天我尝试同时接受 2 个客户的 session ,但 session 没有得到确认。当然,所需的人数是2。这是我正在调用的函数。我该如何纠正这种行为?

// accept meeting 
Parse.Cloud.define("acceptMeeting", function(request, response) {
	Parse.Cloud.useMasterKey();
	var userAcceptingTheMeeting = request.user;
	var meetingId = request.params.meetingId;
	var meetingToAccept;
	var userCreatorOfMeeting;
	var changedObjects = [];
	var queryForMeeting = new Parse.Query("MeetingObject");
	queryForMeeting.get(meetingId).then(function(meeting) {
		meetingToAccept = meeting;
		userCreatorOfMeeting = meeting.get("user");
		// incrementing the "acceptedMeetings" number field on the database for the user that accepted the meeting
		userAcceptingTheMeeting.increment("acceptedMeetings", +1);
		changedObjects.push(userAcceptingTheMeeting);
		return changedObjects;
	}).then(function(changedObjects) {
		meetingToAccept.add("participantsObjectId", userAcceptingTheMeeting.id);
		meetingToAccept.add("participantsName", userAcceptingTheMeeting.get("username"));
		// if the length of the array containing all the participants is equal to the number required "meetingNumberOfPersons" then set "isAvailable" to false (the meeting is confirmed)
		if (meetingToAccept.get("participantsObjectId").length === meetingToAccept.get("meetingNumberOfPersons")) {
			meetingToAccept.set("isAvailable", false);
		}
		changedObjects.push(meetingToAccept);
		console.log(changedObjects.length);
		return changedObjects;
	}).then(function(saveChangedObjects) {
		return Parse.Object.saveAll(changedObjects);	
	}).then(function(push) {
		// check if the meeting is still available
		if (meetingToAccept.get("isAvailable") === true) {
			
			// the meeting is still available, send a notification only to the creator of the meeting
			// push to the creator of the meeting 
			
		} else if (meetingToAccept.get("isAvailable") === false) {
			
			// the meeting is confirmed, send notifications to everyone (creator and participants)
			// push to the creator of the meeting 
			
			var participantsArray = [];
			participantsArray = meetingToAccept.get("participantsObjectId");
			participantsArray.splice(participantsArray.indexOf(userAcceptingTheMeeting.id), 1 );
			for (var i = 0; i < participantsArray.length; i++) {
				var participant = new Parse.User({
					id: participantsArray[i]
				});
				
				// push to the other participants 
				
			}
		}	
		return changedObjects;	
	}).then(function(savedObjects) {
		if (meetingToAccept.get("isAvailable") === true) {
			response.success("unconfirmed");
		} else {
			response.success("confirmed");
		}
	}, function(error) {
		response.error("Failed to accept the meeting");
	});
});

最佳答案

我认为您应该在 .add() 之后使用 .save() meetingToAccept.add("participantsObjectId", userAcceptingTheMeeting.id)

考虑以下事件序列:

{call 1}acceptMeeting//开始通话1,partnersObjectId = [](空数组)

{call 2}acceptMeeting//开始通话 2,participantsObjectId = []

{call 1} meetToAccept.add("participantsObjectId", userAcceptingTheMeeting.id)//participantsObjectId = [user1]

{call 2} meetToAccept.add("participantsObjectId", userAcceptingTheMeeting.id)//t = 2participantsObjectId = [user2]

{call 1} meetToAccept.get("participantsObjectId").length 检查返回 1//participantsObjectId = [user2]

{call 2} meetToAccept.get("participantsObjectId").length 检查返回 1

{call 1} Parse.Object.saveAll(changedObjects)//结果partitionsObjectId = [user1]

{call 2} Parse.Object.saveAll(changedObjects)//结果partitionsObjectId = [user2] 覆盖partnersObjectId = [user1]

另外,关于您的代码的评论:如果您在代码中留出空间,它会变得更具可读性。这样它就不会那么稠密。另外,我建议您在每个“然后”处对“然后”的作用发表评论。

关于javascript - Parse.com 云代码 : function called simultaneously by 2 users,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30243176/

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