流星订阅回调运行时订阅包含以前的订阅导致

Question

我是相当新的流星，而我正在与订阅回调一个奇怪的问题。我有一个包含课程和评论的数据库。我在评论上使用发布/订阅模型来返回仅与所选类相关的评论，并且每次单击新类时都希望更改评论。我想打印所有评论并编辑一些关于评论的指标（平均质量，难度评分）。使用下面的代码，以订阅，更新发送到客户端的评论，印刷点评（这是从一个帮手抢下）返回正确，但指标（这是抓住上onReady回调助手）是不准确的。当onReady函数运行时，即使评论本身正确打印，本地评论集合的当前结果也包含点击的类和先前点击的类的联合。

我也使用自动跟踪试过，但我得到了同样的结果。在更新之前是否有办法清除以前的订阅结果？

发布：

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 });

订阅：

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

和订阅调用，在构造与助手：

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

当使用发布 - 订阅客户端上的minimongo数据库将包含订阅的工会，除非他们明确地清除。因此，您希望重复客户端发布中的查询，以便按照相同的方式进行过滤和排序。 Minimongo在客户端的速度非常快，通常你的数据少得多，所以不用担心性能。

在你constructor您有：

var allReviews = Reviews.find({}); 

改用：

另一个侧面提示：JavaScript是quite clever约truthy和falsy值。

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

可以简化为：

if (courseId && visibility) 

假设你使用visibility == 1表示true和visibility == 0表示false