流星訂閱回調運行時訂閱包含以前的訂閱導致

Question

我是相當新的流星，而我正在與訂閱回調一個奇怪的問題。我有一個包含課程和評論的數據庫。我在評論上使用發佈/訂閱模型來返回僅與所選類相關的評論，並且每次單擊新類時都希望更改評論。我想打印所有評論並編輯一些關於評論的指標（平均質量，難度評分）。使用下面的代碼，以訂閱，更新發送到客戶端的評論，印刷點評（這是從一個幫手搶下）返回正確，但指標（這是抓住上onReady回調助手）是不準確的。當onReady函數運行時，即使評論本身正確打印，本地評論集合的當前結果也包含點擊的類和先前點擊的類的聯合。

我也使用自動跟蹤試過，但我得到了同樣的結果。在更新之前是否有辦法清除以前的訂閱結果？

發佈：

Meteor.publish('reviews', function validReviews(courseId, visiblity) { 
 
\t \t console.log(courseId); 
 
\t \t console.log(visiblity); 
 
\t \t var ret = null 
 
\t \t //show valid reviews for this course 
 
\t \t if (courseId != undefined && courseId != "" && visiblity == 1) { 
 
\t \t \t console.log("checked reviews for a class"); 
 
\t \t \t ret = Reviews.find({class : courseId, visible : 1}, {limit: 700}); 
 
\t \t } else if (courseId != undefined && courseId != "" && visiblity == 0) { //invalidated reviews for a class 
 
\t \t \t console.log("unchecked reviews for a class"); 
 
\t \t \t ret = Reviews.find({class : courseId, visible : 0}, 
 
\t \t \t {limit: 700}); 
 
\t \t } else if (visiblity == 0) { //all invalidated reviews 
 
\t \t \t console.log("all unchecked reviews"); 
 
\t \t \t ret = Reviews.find({visible : 0}, {limit: 700}); 
 
\t \t } else { //no reviews 
 
\t \t \t console.log("no reviews"); 
 
\t \t \t //will always be empty because visible is 0 or 1. allows meteor to still send the ready 
 
\t \t \t //flag when a new publication is sent 
 
\t \t \t ret = Reviews.find({visible : 10}); 
 
\t \t } 
 
\t \t //console.log(ret.fetch()) 
 
\t \t return ret 
 
    \t });

訂閱：

this.helpers({ 
 
     reviews() { 
 
     return Reviews.find({}); 
 
     }   
 
    });

和訂閱調用，在構造與助手：

constructor($scope) { 
 
    $scope.viewModel(this); 
 

 
    //when a new class is selected, update the reviews that are returned by the database and update the gauges 
 
    this.subscribe('reviews',() => [(this.getReactively('selectedClass'))._id, 1], { 
 
     //callback function, should only run once the reveiws collection updates, BUT ISNT 
 
     //seems to be combining the previously clicked class's reviews into the collection 
 
     onReady: function() { 
 
     console.log("class is: ", this.selectedClass); 
 
     if (this.isClassSelected == true) { //will later need to check that the side window is open 
 
      //create initial variables 
 
      var countGrade = 0; 
 
      var countDiff = 0; 
 
      var countQual = 0; 
 
      var count = 0; 
 

 
      //table to translate grades from numerical value 
 
      var gradeTranslation = ["C-", "C", "C+", "B-", "B", "B-", "A-", "A", "A+"]; 
 

 
      //get all current reviews, which will now have only this class's reviews because of the subscribe. 
 
      var allReviews = Reviews.find({}); 
 
      console.log(allReviews.fetch()); 
 
      console.log("len is " + allReviews.fetch().length) 
 
      if (allReviews.fetch().length != 0) { 
 
      console.log("exist") 
 
      allReviews.forEach(function(review) { 
 
       count++; 
 
       countGrade = countGrade + Number(review["grade"]); 
 
       countDiff = countDiff + review["difficulty"]; 
 
       countQual = countQual + review["quality"]; 
 
      }); 
 

 
      this.qual = (countQual/count).toFixed(1); 
 
      this.diff = (countDiff/count).toFixed(1); 
 
      this.grade = gradeTranslation[Math.floor(countGrade/count) - 1]; 
 
      } else { 
 
      console.log("first else"); 
 
      this.qual = 0; 
 
      this.diff = 0; 
 
      this.grade = "-"; 
 
      } 
 
     } else { 
 
      console.log("second else"); 
 
      this.qual = 0; 
 
      this.diff = 0; 
 
      this.grade = "-"; 
 
     } 
 
     } 
 
    })

Answer 1

當使用發佈 - 訂閱客戶端上的minimongo數據庫將包含訂閱的工會，除非他們明確地清除。因此，您希望重複客戶端發佈中的查詢，以便按照相同的方式進行過濾和排序。 Minimongo在客戶端的速度非常快，通常你的數據少得多，所以不用擔心性能。

在你constructor您有：

var allReviews = Reviews.find({}); 

改用：

另一個側面提示：JavaScript是quite clever約truthy和falsy值。

if (courseId != undefined && courseId != "" && visibility == 1) 

可以簡化爲：

if (courseId && visibility) 

假設你使用visibility == 1表示true和visibility == 0表示false