填充以下用户猫鼬

Question

让我花点时间解释一下从头到尾发生的事情。

序言:

一个用户a关注了另外10个人。当用户 A 登录时，这 10 人中每人的 X 条帖子都会被拉入视图。

我不知道这样做是否正确，希望有更好的方法。但是，我想试一试，但还是不行。

关注模特:

let mongoose = require('mongoose');
let Schema = mongoose.Schema;

let FollowSchema = new Schema({
  user: {
    type: Schema.Types.ObjectId,
    ref: 'User'
  },
  followers: [{
    type: Schema.Types.ObjectId,
    ref: 'Card'
  }],
  following: [{
    type: Schema.Types.ObjectId,
    ref: 'Card'
  }]
});

module.exports = mongoose.model('Follow', FollowSchema);

卡片型号

let mongoose = require('mongoose');
let Schema = mongoose.Schema;

let CardSchema = new Schema({
  title: String,
  content: String,
  createdById: {
    type: Schema.Types.ObjectId,
    ref: 'User'
  },
  createdBy: {
    type: String
  }
});

module.exports = mongoose.model('Card', CardSchema);

遵循逻辑

当用户A关注用户B时，做两件事：

将 B 的 user_id 推送到字段 'following' 上的用户 A 文档（A 在 B 之后）

将 A 的 user_id 推送到字段 'followers' 上的用户 B 文档（B 后跟 A）

router.post('/follow', utils.loginRequired, function(req, res) {
const user_id = req.user._id;
const follow = req.body.follow_id;

let bulk = Follow.collection.initializeUnorderedBulkOp();

bulk.find({ 'user': Types.ObjectId(user_id) }).upsert().updateOne({
    $addToSet: {
        following: Types.ObjectId(follow)
    }
});

bulk.find({ 'user': Types.ObjectId(follow) }).upsert().updateOne({
    $addToSet: {
        followers: Types.ObjectId(user_id)
    }
})

bulk.execute(function(err, doc) {
    if (err) {
        return res.json({
            'state': false,
            'msg': err
        })
    }
    res.json({
        'state': true,
        'msg': 'Followed'
    })
})

})

实际数据库值

> db.follows.find().pretty()
{
    "_id" : ObjectId("59e3e27dace1f14e0a70862d"),
    "user" : ObjectId("59e2194177cae833894c9956"),
    "following" : [
        ObjectId("59e3e618ace1f14e0a708713")
    ]
}
{
    "_id" : ObjectId("59e3e27dace1f14e0a70862e"),
    "user" : ObjectId("59e13b2dca5652efc4ca2cf5"),
    "followers" : [
        ObjectId("59e2194177cae833894c9956"),
        ObjectId("59e13b2d27cfed535928c0e7"),
        ObjectId("59e3e617149f0a3f1281e849")
    ]
}
{
    "_id" : ObjectId("59e3e71face1f14e0a708770"),
    "user" : ObjectId("59e13b2d27cfed535928c0e7"),
    "following" : [
        ObjectId("59e3e618ace1f14e0a708713"),
        ObjectId("59e13b2dca5652efc4ca2cf5"),
        ObjectId("59e21942ca5652efc4ca30ab")
    ]
}
{
    "_id" : ObjectId("59e3e71face1f14e0a708771"),
    "user" : ObjectId("59e3e618ace1f14e0a708713"),
    "followers" : [
        ObjectId("59e13b2d27cfed535928c0e7"),
        ObjectId("59e2194177cae833894c9956")
    ]
}
{
    "_id" : ObjectId("59e3e72bace1f14e0a708779"),
    "user" : ObjectId("59e21942ca5652efc4ca30ab"),
    "followers" : [
        ObjectId("59e13b2d27cfed535928c0e7"),
        ObjectId("59e2194177cae833894c9956"),
        ObjectId("59e3e617149f0a3f1281e849")
    ]
}
{
    "_id" : ObjectId("59f0eef155ee5a5897e1a66d"),
    "user" : ObjectId("59e3e617149f0a3f1281e849"),
    "following" : [
        ObjectId("59e21942ca5652efc4ca30ab"),
        ObjectId("59e13b2dca5652efc4ca2cf5")
    ]
}
>

有了上面的数据库结果，这是我的查询：

查询

router.get('/follow/list', utils.loginRequired, function(req, res) {
    const user_id = req.user._id;

    Follow.findOne({ 'user': Types.ObjectId(user_id) })
      .populate('following')
      .exec(function(err, doc) {
          if (err) {
              return res.json({
                  'state': false,
                  'msg': err
              })
          };

          console.log(doc.username);

          res.json({
              'state': true,
              'msg': 'Follow list',
              'doc': doc
          })
      })
});

通过上面的查询，根据我对 Mongoose populate 的一点了解，我希望从 following 数组中的每个用户那里得到卡片。

我的理解和期望可能是错误的，但是以这样的最终目标，这种填充方法可以吗？还是我在尝试解决人口聚合任务？

更新：

感谢您的回答。非常接近，但 followingCards 数组仍然不包含任何结果。这是我当前 Follow 模型的内容：

> db.follows.find().pretty()
{
    "_id" : ObjectId("59f24c0555ee5a5897e1b23d"),
    "user" : ObjectId("59f24bda1d048d1edad4bda8"),
    "following" : [
        ObjectId("59f24b3a55ee5a5897e1b1ec"),
        ObjectId("59f24bda55ee5a5897e1b22c")
    ]
}
{
    "_id" : ObjectId("59f24c0555ee5a5897e1b23e"),
    "user" : ObjectId("59f24b3a55ee5a5897e1b1ec"),
    "followers" : [
        ObjectId("59f24bda1d048d1edad4bda8")
    ]
}
{
    "_id" : ObjectId("59f24c8855ee5a5897e1b292"),
    "user" : ObjectId("59f24bda55ee5a5897e1b22c"),
    "followers" : [
        ObjectId("59f24bda1d048d1edad4bda8")
    ]
}
>

以下是我从 Card 模特那里获得的所有当前内容：

> db.cards.find().pretty()
{
    "_id" : ObjectId("59f24bc01d048d1edad4bda6"),
    "title" : "A day or two with Hubtel's HTTP API",
    "content" : "a day or two",
    "external" : "",
    "slug" : "a-day-or-two-with-hubtels-http-api-df77056d",
    "createdBy" : "seanmavley",
    "createdById" : ObjectId("59f24b391d048d1edad4bda5"),
    "createdAt" : ISODate("2017-10-26T20:55:28.293Z"),
    "__v" : 0
}
{
    "_id" : ObjectId("59f24c5f1d048d1edad4bda9"),
    "title" : "US couple stole goods worth .2m from Amazon",
    "content" : "for what",
    "external" : "https://bbc.com",
    "slug" : "us-couple-stole-goods-worth-dollar12m-from-amazon-49b0a524",
    "createdBy" : "nkansahrexford",
    "createdById" : ObjectId("59f24bda1d048d1edad4bda8"),
    "createdAt" : ISODate("2017-10-26T20:58:07.793Z"),
    "__v" : 0
}

使用您 (@Veeram) 的 Populate Virtual 示例，这是我得到的响应：

{"state":true,"msg":"Follow list","doc":{"_id":"59f24c0555ee5a5897e1b23d","user":"59f24bda1d048d1edad4bda8","following":["59f24b3a55ee5a5897e1b1ec","59f24bda55ee5a5897e1b22c"],"followers":[],"id":"59f24c0555ee5a5897e1b23d","followingCards":[]}}

followingCards 数组为空。

另一方面，使用 $lookup 查询只需 returns []

我可能遗漏了什么？

Answer 1

var mongoose = require('mongoose'), Schema = mongoose.Schema

var eventSchema = Schema({
title     : String,
location  : String,
startDate : Date,
endDate   : Date
});

var personSchema = Schema({
firstname: String,
lastname: String,
email: String,
dob: Date,
city: String,
eventsAttended: [{ type: Schema.Types.ObjectId, ref: 'Event' }]
});

var Event  = mongoose.model('Event', eventSchema);
var Person = mongoose.model('Person', personSchema);

为了展示如何使用 populate，首先创建一个人对象，

   aaron = new Person({firstname: 'Aaron'}) and an event object, 
   event1 = new Event({title: 'Hackathon', location: 'foo'}):

   aaron.eventsAttended.push(event1);
   aaron.save(callback);

然后，当您进行查询时，您可以像这样填充引用：

 Person
.findOne({ firstname: 'Aaron' })
.populate('eventsAttended') .exec(function(err, person) {
if (err) return handleError(err);
console.log(person);
 });

// 只有当我们将 refs 推到 person.eventsAttended

时才有效

Answer 2

您可以在聚合管道中使用虚拟填充或 $lookup 运算符。

使用虚拟填充

FollowSchema.virtual('followingCards', {
    ref: 'Card',
    localField: 'following',
    foreignField: 'createdById'
});

Follow.findOne({ 
   'user': Types.ObjectId(user_id) })
   .populate('followingCards')
   .exec(function(err, doc) {      
      console.log(JSON.stringify(doc));
});

使用$lookup聚合

Follow.aggregate([
  {
    "$match": {
      "user": Types.ObjectId(user_id) 
    }
  },
  {
    "$lookup": {
      "from": "cards",
      "localField": "following",
      "foreignField": "createdById",
      "as": "followingCards"
    }
  }
]).exec(function (err, doc) {
     console.log(JSON.stringify(doc));
})

Answer 3

注意：将Activity.find改为Card.find

const { ObjectID } = require("mongodb");
// import Follow and Activity(Card) schema

const userId = req.tokenData.userId; // edit this too...
Follow.aggregate([
  {
    $match: {
      user: ObjectID(userId)
    }
  }
])
.then(data => {
  // console.log(data)
  var dataUsers = data[0].following.map(function(item) {
    return item._id;
  });
  // console.log(dataUsers)
  Activity.find(
    { createdById: { $in: dataUsers } },
    {
      _id: 1,
      title: 1,
      content: 1,
      createdBy: 1,
      creatorAvatar: 1,
      activityType: 1,
      createdAt: 1
    }
  )
    // .sort({createdAt:-1)
    .then(posts => res.send({ posts }));
});

填充以下用户猫鼬

populate following users mongoose

mongoose

mongodb

mongoose-populate