Parse.com Cloud Code：由2个用户同时调用的函数

  • Z时代
  • 2024-01-10
  • 分类：问答

我在审核我为我的应用程序编写的所有云代码，并且我已找到需要更正的内容。我有一个充满用户和“会议”对象的解析数据库。每个用户都可以创建或查看和接受会议。当用户想要接受会议时，下一个功能会被调用。如果所需人数等于接受会议的人数已确认，否则它可供其他用户使用。今天，我试着从两位客户那里同时接受一次会议，会议没有得到证实。当然，所需的人数是2.这是我打电话的功能。我怎样才能纠正这种行为？Parse.com Cloud Code：由2个用户同时调用的函数

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

回答：

，我认为你应该使用.save()的.add()meetingToAccept.add("participantsObjectId", userAcceptingTheMeeting.id)

后考虑这一系列事件：

{调用1} acceptMeeting //开始通话1，participantsObjectId = []（空阵列）

{调用2} acceptMeeting //开始呼叫2，participantsObjectId = []

{调用1} meetingToAccept.add（ “participantsObjectId”，userAcceptingTheMeeting.id）// participantsObjectId = [USER1]

{调用2 } meetingToAccept.add（“participantsObjectId”，userAcceptingTheMeeting.id）// t = 2 participantsObjectId = [user2]

{call 1} meetingToAccept.get（“participantsObjectId”）。length check returns 1 // participantsObjectId = [user2 ]

{call 2} meetingToAccept.get（“participantsObjectId”）。length check返回1

{调用1} Parse.Object.saveAll（changedObjects）//导致participantsObjectId = [USER1]

{调用2} Parse.Object.saveAll（changedObjects）//导致participantsObjectId = [USER2] OVERRIDING participaObjectId = [user1]

此外，有关您的代码的评论：如果您将代码放在空间中，它会变得更具可读性。所以它不那么密集。另外，我建议在每个“当时”你对“那时”做什么做评论。

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