mongodb - 在不使用聚合的情况下简化深层嵌套的 mongo 查询
问题描述
我正在尝试为我的应用程序创建一种新闻源功能。我试图了解如何使用 Mongo/Mongoose 中的嵌套查询而不是使用聚合来做到这一点。(应用程序的其余部分使用嵌套查询),如果我不需要,我宁愿不必使用香草 javascript 修改查询以获得完美的对象。
我想获取当前用户的朋友,对于每个朋友,获取他们所有的帖子,然后按日期对其进行排序。
我想让我当前的查询更有效率,因为有一个额外的步骤
我在猫鼬中的查询
User.findOne({ _id: req.userId }, 'friends.user -_id')
.populate({
path: 'friends.user',
model: 'User',
select: 'posts -_id',
populate: {
path: 'posts',
model: 'Post',
select: 'date author user desc -_id',
options: { sort: { date: -1 } },
populate: {
path: 'author user',
model: 'User',
select: 'profile.firstname profile.lastname profile.avatar username'
},
},
})
结果
{
"newsfeed": [
{
"user": {
"posts": [
{
"user": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "kenne"
},
"author": {
"profile": {
"firstname": "user1",
"lastname": "user1",
"avatar": "9e7b60e534cf761f41d6afe3d97295c9.jpg"
},
"_id": "5b562382a16cde19638e4bee",
"username": "user1"
},
"date": "2018-07-24T06:40:37.413Z",
"desc": "sdfsdf"
},
{
"user": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "user2"
},
"author": {
"profile": {
"firstname": "user1",
"lastname": "user1",
"avatar": "9e7b60e534cf761f41d6afe3d97295c9.jpg"
},
"_id": "5b562382a16cde19638e4bee",
"username": "user1"
},
"date": "2018-07-24T06:40:17.180Z"
},
{
"user": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "user2"
},
"author": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "user2"
},
"date": "2018-07-23T22:20:15.246Z",
"desc": "Gibberish"
}
]
}
},
{
"user": {
"posts": [
{
"user": {
"profile": {
"firstname": "user3",
"lastname": "user3",
"avatar": "132df94df5733efd41609681bc7f71f9.jpeg"
},
"_id": "5b5382f3661a8d7023e1ae65",
"username": "user3"
},
"author": {
"profile": {
"firstname": "user3",
"lastname": "user3",
"avatar": "132df94df5733efd41609681bc7f71f9.jpeg"
},
"_id": "5b5382f3661a8d7023e1ae65",
"username": "user3"
},
"date": "2018-07-21T19:09:45.543Z",
"desc": "Gibberish"
}
]
}
}
]
}
因为我需要首先获取用户的朋友,所以当前查询创建了两个用户对象,每个对象都有自己的帖子和作者以及其中的用户对象。
eg. {"newsfeed": [{"user": {"bios": [...]}}], [{"user": {"bios": [...]}}] }
我想要的是这样的
{
"newsfeed": [{
"posts": [
{
"user": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "user2"
},
"author": {
"profile": {
"firstname": "user1",
"lastname": "user1",
"avatar": "9e7b60e534cf761f41d6afe3d97295c9.jpg"
},
"_id": "5b562382a16cde19638e4bee",
"username": "user1"
},
"date": "2018-07-24T06:40:37.413Z",
"desc": "sdfsdf"
},
{
"user": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "user2"
},
"author": {
"profile": {
"firstname": "user1",
"lastname": "user1",
"avatar": "9e7b60e534cf761f41d6afe3d97295c9.jpg"
},
"_id": "5b562382a16cde19638e4bee",
"username": "user1"
},
"date": "2018-07-24T06:40:17.180Z"
},
{
"user": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "user2"
},
"author": {
"profile": {
"firstname": "user2",
"lastname": "user2",
"avatar": "c66620a0ef057908a1663725956ac03a.jpg"
},
"_id": "5b56549fcba9231e5d1e848d",
"username": "user2"
},
"date": "2018-07-23T22:20:15.246Z",
"desc": "Gibberish"
},
{
"user": {
"profile": {
"firstname": "user3",
"lastname": "user3",
"avatar": "132df94df5733efd41609681bc7f71f9.jpeg"
},
"_id": "5b5382f3661a8d7023e1ae65",
"username": "user3"
},
"author": {
"profile": {
"firstname": "user3",
"lastname": "user3",
"avatar": "132df94df5733efd41609681bc7f71f9.jpeg"
},
"_id": "5b5382f3661a8d7023e1ae65",
"username": "user3"
},
"date": "2018-07-21T19:09:45.543Z",
"desc": "Gibberish"
}
]}
我怎么能不使用聚合来做到这一点?得到更像这样的东西
eg. {"newsfeed": [{"posts": [...]}] }
附加信息
用户
const UserSchema = new Schema({
username : String,
friends : [{ user: { type: Schema.Types.ObjectId, ref: 'User'}, status: String }],
profile: {
name : String,
firstname : String,
lastname : String,
avatar : String
}),
posts : [{ type: Schema.Types.ObjectId, ref: 'Post' }]
});
帖子
const PostsSchema = new Schema({
user : { type: Schema.Types.ObjectId, ref: 'User' },
author : { type: Schema.Types.ObjectId, ref: 'User' },
date : Date,
desc : String
});
示例数据:用户
{
"_id" : ObjectId("5b51fd2a3f4ec33546a06648"),
"profile" : {
"firstname" : "user1",
"lastname" : "user1"
"avatar" : "user1.png"
}
"username" : "user1",
"friends" : [
{
"_id" : ObjectId("5b51fd7c3f4ec33546a0664f"),
"user" : ObjectId("5b51fd643f4ec33546a0664c")
},
{
"_id" : ObjectId("5b51fdb53f4ec33546a06655"),
"user" : ObjectId("5b51fd903f4ec33546a06652")
}
]
}
{
"_id" : ObjectId("5b51fd643f4ec33546a0664c"),
"profile" : {
"firstname" : "user2",
"lastname" : "user2"
"avatar" : "user2.png"
}
"posts" : [
{
"_id" : ObjectId("5b51fdcd3f4ec33546a06610"),
"_id" : ObjectId("5b51fdcd3f4ec33546a06611"),
"_id" : ObjectId("5b51fdcd3f4ec33546a06612"),
}
],
"__v" : 5,
"username" : "user2"
},
{
"_id" : ObjectId("5b51fd903f4ec33546a06652"),
"profile" : {
"firstname" : "user3",
"lastname" : "user3"
"avatar" : "user3.png"
},
"posts" : [
{
"_id" : ObjectId("5b51fdce3f4ec33546a06615"),
"_id" : ObjectId("5b51fd2a3f4ec33546a06617")
}
],
"__v" : 5,
"username" : "user3"
}
示例日期:帖子
{
"_id" : ObjectId("5b51fdcd3f4ec33546a06610"),
"user" : ObjectId("5b51fd2a3f4ec33546a06648"),
"author" : ObjectId("5b51fd2a3f4ec33546a06648"),
"date" : ISODate("2018-07-20T15:18:02.962Z"),
"desc" : "user1 gibberish",
"__v" : 0
}
{
"_id" : ObjectId("5b51fdcd3f4ec33546a06611"),
"user" : ObjectId("5b51fd643f4ec33546a0664c"),
"author" : ObjectId("5b51fd643f4ec33546a0664c"),
"date" : ISODate("2018-07-20T15:19:00.968Z"),
"desc" : "user2 gibberish",
"__v" : 0
}
{
"_id" : ObjectId("5b51fdcd3f4ec33546a06612"),
"user" : ObjectId("5b51fd903f4ec33546a06652"),
"author" : ObjectId("5b51fd903f4ec33546a06652"),
"date" : ISODate("2018-07-20T15:19:44.102Z"),
"desc" : "user3 gibberish",
"__v" : 0
}
{
"_id" : ObjectId("5b51fdce3f4ec33546a06615"),
"user" : ObjectId("5b51fd643f4ec33546a0664c"),
"author" : ObjectId("5b51fd643f4ec33546a0664c"),
"date" : ISODate("2018-07-20T15:19:00.968Z"),
"desc" : "more gibberish",
"__v" : 0
}
{
"_id" : ObjectId("5b51fdce3f4ec33546a06616"),
"user" : ObjectId("5b51fd903f4ec33546a06652"),
"author" : ObjectId("5b51fd903f4ec33546a06652"),
"date" : ISODate("2018-07-20T15:19:44.102Z"),
"desc" : "more gibberish",
"__v" : 0
}
解决方案
您可以尝试使用以下代码展开应用层中的项目
User.find({ _id: mongoose.Types.ObjectId("5b51fd2a3f4ec33546a06648") }, 'friends.user -_id')
.populate({
path: 'friends.user',
model: 'User',
select: 'posts -_id',
})
.exec(function(err, results) {
// Post.find({ _id: {$in: results
const postIDs = results.reduce((acc, user) => {
return acc.concat(
...user.friends.map(
friend => friend.user.posts
)
)
}, [])
Post.find({ _id: { $in: postIDs } })
.populate(
{
path: 'author user',
model: 'User',
select: 'profile.firstname profile.lastname profile.avatar username'
}
)
.exec(function(err, posts) {
console.log(
JSON.stringify(posts, null, 2)
);
})
});
其他选项是 map reduce,但由于您不喜欢聚合,我也不确定 map reduce 是否合适。
const mapReduceConfig = {
map: function() {
var friends = this.friends;
emit(
this._id,
friends.reduce(
(acc, friend) => {
return acc.concat(
friend.posts &&
friend.posts.reduce((pacc, p) => pacc.concat(p), [])
) || [];
},
[]
)
);
},
reduce: function(k, vals) {
return Array.sum(vals);
}
};
User.mapReduce(mapReduceConfig, function(err, results) {
console.log(
JSON.stringify(results, null, 2)
);
});
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