mongodb将排序应用于查找结果 [英] mongodb apply sort to lookup results

问题描述

如果我有user和post集合

{"_id": 1, "name": "User 1"}
{"_id": 2, "name": "User 2"}

{"_id": 1, "title": "Post 1", "userId": 1, "createdAt": ISODate("2017-07-24T04:12:54.255Z")}
{"_id": 2, "title": "Post 2", "userId": 1, "createdAt": ISODate("2017-07-25T04:12:54.255Z")}
{"_id": 3, "title": "Post 1", "userId": 2, "createdAt": ISODate("2017-07-24T04:12:54.255Z")}

如何列出所有用户的最新帖子?就像

How can I list all users with their latest post? Would be something like

{
  "_id": 1,
  "name": "User 1",
  "post": {
    "_id": 2,
    "title": "Post 2",
    "userId": 1,
    "createdAt": ISODate("2017-07-25T04:12:54.255Z")
  }
}

我知道我可以轻松地使用$ lookup，$ unwind帖子，然后使用post.createdAt进行$ sort，但这使我拥有了冗余用户(用户1将在帖子1和帖子2中列出两次).

I know I can easily use $lookup, $unwind post, then $sort by post.createdAt, but that leave me with redundant user (User 1 will be listed twice for Post 1 and Post 2).

我不知道如何使用$ group删除重复项，同时保持其他字段(名称，post.title等)

I don't know how can I use $group to remove duplicates while keep maintaining other fields (name, post.title, etc)

推荐答案

我使用$ group和$ first

I solved the duplicates using $group and $first

db.getCollection('user').aggregate([
    {$lookup: {from: "post", localField: "_id", foreignField: "userId", as: "post"}},
    {$unwind: { path: "$post", preserveNullAndEmptyArrays: true }},
    {$sort: {"post.createdAt": -1}},
    {$group: {"_id": "$_id", "name": {$first: "$name"}, "post": {$first: "$post"}},
    {$project: {"_id": 1, "name": 1, post": 1}}
])

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