如何在投票被分组并在同一聚合函数中求和之前检查当前用户的投票

Question

var PostSchema = new mongoose.Schema({
    item: {
    type: mongoose.Schema.ObjectId,
        ref: 'item',
        required: true
    },
  user: {
    type: mongoose.Schema.ObjectId,
    ref: 'User',
    required: true
  },
  vote: {
    type: Number,
    default: 0
  },
  total: {
    type: Number,
    default: 0
  },
  awsPostKey: {type: String},
  picture: {type: String, required: true}
});

---

var data = function(){
return Post
.find({})
.then(function(post){
    return post;
 })
};


var userId = //mongo objectId for current user

//postVote schema:
var PostVoteSchema = new mongoose.Schema({
  post: {
    type: mongoose.Schema.ObjectId,
        ref: 'Post',
        required: true
    },
  user: {
    type: mongoose.Schema.ObjectId,
    ref: 'User',
    required: true
  },
  vote: {
    type: Number,
    default: 0
  }
});

---

//pass data from Post query to PostVote sum function:

PostVoteSchema.statics.sum = function (data, userId) {

 var postIds = data.map(function (a) {
    return a._id;
  });

return PostVote
.aggregate(
    [
   { $match: { 'post': {$in: postIds}}},
   { $group: { _id:'$post' ,vote:{$sum:'$vote'}}}
 ])
.execAsync()
.then(function(votes){

    return votes;

   //desired output to client, _id is for specific post
   {_id: 5802ea4bc00cb0beca1972cc, vote: 3, currentUserVote: -1}

 });
};

我成功地获得了具有相同 postId 的所有选票的总和。 现在，我想查看当前用户 (userId) 是否也对给定的帖子进行了投票，然后返回他们的投票方式（+1 或 -1）以及对特定帖子的所有投票的总和.

是否可以这样做，或者我必须在我的聚合管道之外执行此操作 - 在第二个查询中？不得不再次查询该集合似乎很费力。

Answer 1

是的，这是可能的。在 $group 管道中，您可以使用 $cond 运算符作为输入 $sum 累加器运算符的逻辑。例如：

return PostVote.aggregate([
   { "$match": { "post": { "$in": postIds } } },
   { 
        "$group": {
            "_id": "$post",
            "votes": { "$sum": "$vote" },                 
            "userVotes": {
                "$sum": {
                    "$cond": [
                        { "$eq": ["$user", userId] },
                        "$vote",
                        0
                    ]
                }
            }
        }
    }
 ]).execAsync().then(function(votes){
    return votes;
});