Question

我有两个收藏集User和Post。

用户

user_id
地区
is_join

发布

post_id
user_id
身体
is_block

从rdb的角度来看，User：Post关系为1：N。

每个user可以写多个post。

例如，当前这样插入文档。

用户

> db.user.find()
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d2"), "user_id" : NumberLong(1), "region" : "US", "is_join" : true }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d5"), "user_id" : NumberLong(2), "region" : "KR", "is_join" : true }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d8"), "user_id" : NumberLong(3), "region" : "US", "is_join" : true }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28da"), "user_id" : NumberLong(4), "region" : "KR", "is_join" : false }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28dc"), "user_id" : NumberLong(5), "region" : "US", "is_join" : true }

发布

> db.post.find()
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d3"), "post_id" : NumberLong(1), "user_id" : NumberLong(1), "body" : "first", "is_block" : false }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d4"), "post_id" : NumberLong(4), "user_id" : NumberLong(1), "body" : "fourth", "is_block" : false }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d6"), "post_id" : NumberLong(2), "user_id" : NumberLong(2), "body" : "second", "is_block" : false }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d7"), "post_id" : NumberLong(3), "user_id" : NumberLong(2), "body" : "third", "is_block" : false }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28d9"), "post_id" : NumberLong(5), "user_id" : NumberLong(3), "body" : "fifth", "is_block" : true }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28db"), "post_id" : NumberLong(6), "user_id" : NumberLong(4), "body" : "sixth", "is_block" : false }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28dd"), "post_id" : NumberLong(7), "user_id" : NumberLong(5), "body" : "seven", "is_block" : true }
{ "_id" : ObjectId("5d15e41a1b48d9417ebc28de"), "post_id" : NumberLong(8), "user_id" : NumberLong(5), "body" : "eight", "is_block" : false }

要通过聚集进行联接，还需要满足更多条件。

用户region='US'，is_join=true
用户创建的帖子，结果为1
发布is_block=false
按post_id排序
（可选）如果用户未写任何帖子，除非它。我知道它可以通过preserveNullAndEmptyArrays执行，但是我认为这会导致性能问题。

所需结果

{
    "posts" : [
        {
            "post_id" : NumberLong(1),
            "body" : "first",
            "is_block" : false
        },
        {
            "post_id" : NumberLong(2),
            "body" : "second",
            "is_block" : false
        },
        {
            "post_id" : NumberLong(3),
            "body" : "third",
            "is_block" : false
        },
        {
            "post_id" : NumberLong(4),
            "body" : "fourth",
            "is_block" : false
        },
        {
            "post_id" : NumberLong(8),
            "body" : "eight",
            "is_block" : false
        }
    ]
}

post_id = 5被is_block=true排除

post_id = 6被is_join=false排除

post_id = 7被is_block=true排除

整个结果按post_id排序。

我是mongodb的新手，所以，也许我以关系数据库的形式想得太多。

我不知道它可以在NoSQL上执行。

有什么办法吗？

任何建议，非常感谢。

谢谢。

Answer 1

您可以使用查找管道运算符获得这种结果。

const region = 'US';
const is_join = true;
const is_block = false;
const query = [
    {
        $match: {
            region: region,
            is_join: is_join
        }
    },
    {
        $lookup: {
            let: { user_id: "$user_id" },
            from: 'posts',
            pipeline: [
                {
                    $match: {
                        $expr: {
                            $and: [
                                { $eq: ["$$user_id", "$user_id"], },
                                { $eq: ["$is_block", is_block] }
                            ]

                        }
                    }
                },
                {
                    $sort: {
                        post_id: 1
                    }
                }
            ],
            as: "posts"
        }
    },
    {
        $unwind: "$posts"
    },
    {
        $group:{
            _id: "mygroup",
            posts: { 
                $push: {
                    post_id: "$posts.post_id",
                    body: "$posts.body",
                    is_block: "$posts.is_block",
                }
            }
        }
    },
    {
        $project:{
            _id: false
        }
    }
]
db.users.aggregate(query)

输出

{
    "posts" : [
        {
            "post_id" : 1,
            "body" : "first",
            "is_block" : false
        },
        {
            "post_id" : 4,
            "body" : "fourth",
            "is_block" : false
        },
        {
            "post_id" : 8,
            "body" : "eight",
            "is_block" : false
        }
    ]
}

Answer 2

尝试一下-

要加入两个集合，可以将聚合用作-

  User.aggregate([{
    '$match': { 'region':'US', 'is_join': true }  // match from users collection
  }, {
    $lookup: {             // it will aggregate the result from both collection
      from: 'posts',
      localField: '_id',
      foreignField: 'user_id',
      as: 'posts'
    }
  },
   {"$unwind":"$posts"},
   {$match : { "posts.is_block" : false } }, // check inside post collection
   {$sort : { "posts.post_id" : 1}},  // sort the data
 ], (err, users) => {
    if (err) return callback(err, null);

        console.log('users :', users);

  });

MongoDB使用聚合

2 个答案: