Mongoose:find() - 匹配两个值并且 return 仅匹配的文档
Mongoose: find() - match two values and return only the documents matched
我环顾四周,找不到解决这个问题的确切方法。我查看了 Mongoose 文档并试图找到一种方法。我是 MongoDB 的新手,在学习了很多教程后,我的目标是制作自己的 API。
目标:
GET 请求 http://localhost:3000/seasons/19/1
return 剧集详情 19
和所选剧集 1
类似于:
{
"_id": "5b893aff76c12c166c4e75ae",
"season_number": 19,
"name": "Season 19",
"number_of_episodes": 24,
"episode": [
{
"_id": "5b895e56089a9a152c16af1f",
"episode_number": 1,
"name": "19.01",
"length": "1:23:33",
"author": "Aaron Wright",
"release": "2018-01-13T00:00:00.000Z",
"description": "We kick off our 2018 season with the worst of CES and the future of smart toilets before Aaron unravels the controversial “Dear David” saga that has sparked controversy. We then look at state sponsored mind control programs and the hypnotic experiments performed on unwitting Princeton students in the 1960s."
},
}
问题:
我似乎无法弄清楚如何只用选定的一集 return 上面的对象。由于所有数据都在 episodes
中,但如果我手动使用 episodes[0]
等,我可以 return 它
这是回购文件,所以不要让这个 post 太长:
https://github.com/DannnB/mysterious-universe-api/blob/master/api/controllers/seasons.js
- 行:
63
- 姓名:
seasons_get_episode_number
感谢您的帮助!
数据库 - 托管在 MongoDB Atlas - 免费层所以不能使用 $where
并且不想使用 $where
作为一个沉重的命令。
"data": [
{
"_id": "5b893aff76c12c166c4e75ae",
"season_number": 19,
"name": "Season 19",
"number_of_episodes": 24,
"episodes": [
{
"_id": "5b895e56089a9a152c16af1f",
"episode_number": 1,
"name": "19.01",
"length": "1:23:33",
"author": "Aaron Wright",
"release": "2018-01-13T00:00:00.000Z",
"description": "We kick off our 2018 season with the worst of CES and the future of smart toilets before Aaron unravels the controversial “Dear David” saga that has sparked controversy. We then look at state sponsored mind control programs and the hypnotic experiments performed on unwitting Princeton students in the 1960s."
},
{
"_id": "5b895fee089a9a152c16af20",
"episode_number": 2,
"name": "19.02",
"length": "1:22:11",
"author": "Benjamin Grundy",
"release": "2018-01-20T00:00:00.000Z",
"description": "This week we unravel the cosmic serpent to find the origins of life and the link between DNA and Ayahuasca visions. Could the building blocks of all life also be a conscious force that is capable of direct communication with our altered states of mind?"
}
]
},
{
"_id": "5b893b2276c12c166c4e75b0",
"season_number": 20,
"name": "Season 20",
"number_of_episodes": 9
}, *and so on...*
]
编辑:使用 $filter
最初找到解决方案
db.dumb.aggregate(
{
$match:{a:'ok', eps:{$elemMatch:{id:1}}}
},
{
$project:{
a:1,
eps:{
$filter:{
input:'$eps',
as:'ep',
cond:{$eq:['$$ep.id',1]}
}
}
}
}
)
输出
{ "_id" : ObjectId("5b8e3eb8e5ab5ef9a9b9b673"), "a" : "ok", "eps" : [ { "id" : 1, "comment" : "ok" } ] }
{ "_id" : ObjectId("5b8e3ec3e5ab5ef9a9b9b676"), "a" : "ok", "eps" : [ { "id" : 1, "comment" : "ok too" } ] }
也许使用 unwind?
构建数据集
> db.dumb.insert({a:'ok', eps:[{id:1, comment('ok')}]})
WriteResult({ "nInserted" : 1 })
> db.dumb.insert({a:'ko', eps:[{id:1, comment:'ok'}]})
WriteResult({ "nInserted" : 1 })
> db.dumb.insert({a:'ok', eps:[{id:2, comment:'ko'}]})
WriteResult({ "nInserted" : 1 })
> db.dumb.insert({a:'ok', eps:[{id:2, comment:'ko'}, {id:1, comment:'ok too'}]})
WriteResult({ "nInserted" : 1 })
匹配系列(此处命名为'ok')
> db.dumb.find({a:'ok', eps:{$elemMatch:{id:1}}})
{ "_id" : ObjectId("5b8d447b21edbcdd7f3b3d24"), "a" : "ok", "eps" : [ { "id" : 1, "comment" : "ok" } ] }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : [ { "id" : 2, "comment" : "ko" }, { "id" : 1, "comment" : "ok too" } ] }
匹配名为 'ok' 的系列的 id 1 的电影
使用 unwind,数组的每一项都是 "spread"。请注意 "dupplicate objectIds"(您正在跟踪展开的文档)
> db.dumb.aggregate({$match:{a:'ok'}},{$unwind:'$eps'})
{ "_id" : ObjectId("5b8d447b21edbcdd7f3b3d24"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok" } }
{ "_id" : ObjectId("5b8d449121edbcdd7f3b3d26"), "a" : "ok", "eps" : { "id" : 2, "comment" : "ko" } }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : { "id" : 2, "comment" : "ko" } }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok too" } }
之后的标准匹配
> db.dumb.aggregate({$match:{a:'ok'}},{$unwind:'$eps'},{$match:{'eps.id':1}})
{ "_id" : ObjectId("5b8d447b21edbcdd7f3b3d24"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok" } }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok too" } }
您可以在 Aggregration 上使用 $filter 运算符。这是符合您要求的查询:
db.seasons.aggregate([
{ "$match": { "season_number": 19 } },
{ "$project": {
"season_number": 1,
"name": 1,
"number_of_episodes": 1,
"episodes": {
"$filter": {
"input": "$episodes",
"as": "episode",
"cond": { "$eq": [ "$$episode.episode_number", 1 ] }
}
}
}
}])
$filter 从版本 3.2
开始可用
db.users.aggregate(
// Pipeline
[
// Stage 1
{
$unwind: {
path : "$episodes",
preserveNullAndEmptyArrays : false // optional
}
},
// Stage 2
{
$match: {
"season_number" : 19.0,
"episodes.episode_number" : 1
}
},
// Stage 3
{
$group: {
"_id":"$_id",
"season_number" :{$last:"season_number"},
"name" :{$last:"name"},
"number_of_episodes" :{$last:"number_of_episodes"},
"episodes":{$push:"$episodes"}
}
},
]);
我环顾四周,找不到解决这个问题的确切方法。我查看了 Mongoose 文档并试图找到一种方法。我是 MongoDB 的新手,在学习了很多教程后,我的目标是制作自己的 API。
目标:
GET 请求 http://localhost:3000/seasons/19/1
return 剧集详情 19
和所选剧集 1
类似于:
{
"_id": "5b893aff76c12c166c4e75ae",
"season_number": 19,
"name": "Season 19",
"number_of_episodes": 24,
"episode": [
{
"_id": "5b895e56089a9a152c16af1f",
"episode_number": 1,
"name": "19.01",
"length": "1:23:33",
"author": "Aaron Wright",
"release": "2018-01-13T00:00:00.000Z",
"description": "We kick off our 2018 season with the worst of CES and the future of smart toilets before Aaron unravels the controversial “Dear David” saga that has sparked controversy. We then look at state sponsored mind control programs and the hypnotic experiments performed on unwitting Princeton students in the 1960s."
},
}
问题:
我似乎无法弄清楚如何只用选定的一集 return 上面的对象。由于所有数据都在 episodes
中,但如果我手动使用 episodes[0]
等,我可以 return 它
这是回购文件,所以不要让这个 post 太长: https://github.com/DannnB/mysterious-universe-api/blob/master/api/controllers/seasons.js
- 行:
63
- 姓名:
seasons_get_episode_number
感谢您的帮助!
数据库 - 托管在 MongoDB Atlas - 免费层所以不能使用 $where
并且不想使用 $where
作为一个沉重的命令。
"data": [
{
"_id": "5b893aff76c12c166c4e75ae",
"season_number": 19,
"name": "Season 19",
"number_of_episodes": 24,
"episodes": [
{
"_id": "5b895e56089a9a152c16af1f",
"episode_number": 1,
"name": "19.01",
"length": "1:23:33",
"author": "Aaron Wright",
"release": "2018-01-13T00:00:00.000Z",
"description": "We kick off our 2018 season with the worst of CES and the future of smart toilets before Aaron unravels the controversial “Dear David” saga that has sparked controversy. We then look at state sponsored mind control programs and the hypnotic experiments performed on unwitting Princeton students in the 1960s."
},
{
"_id": "5b895fee089a9a152c16af20",
"episode_number": 2,
"name": "19.02",
"length": "1:22:11",
"author": "Benjamin Grundy",
"release": "2018-01-20T00:00:00.000Z",
"description": "This week we unravel the cosmic serpent to find the origins of life and the link between DNA and Ayahuasca visions. Could the building blocks of all life also be a conscious force that is capable of direct communication with our altered states of mind?"
}
]
},
{
"_id": "5b893b2276c12c166c4e75b0",
"season_number": 20,
"name": "Season 20",
"number_of_episodes": 9
}, *and so on...*
]
编辑:使用 $filter
最初找到解决方案
db.dumb.aggregate(
{
$match:{a:'ok', eps:{$elemMatch:{id:1}}}
},
{
$project:{
a:1,
eps:{
$filter:{
input:'$eps',
as:'ep',
cond:{$eq:['$$ep.id',1]}
}
}
}
}
)
输出
{ "_id" : ObjectId("5b8e3eb8e5ab5ef9a9b9b673"), "a" : "ok", "eps" : [ { "id" : 1, "comment" : "ok" } ] }
{ "_id" : ObjectId("5b8e3ec3e5ab5ef9a9b9b676"), "a" : "ok", "eps" : [ { "id" : 1, "comment" : "ok too" } ] }
也许使用 unwind?
构建数据集
> db.dumb.insert({a:'ok', eps:[{id:1, comment('ok')}]})
WriteResult({ "nInserted" : 1 })
> db.dumb.insert({a:'ko', eps:[{id:1, comment:'ok'}]})
WriteResult({ "nInserted" : 1 })
> db.dumb.insert({a:'ok', eps:[{id:2, comment:'ko'}]})
WriteResult({ "nInserted" : 1 })
> db.dumb.insert({a:'ok', eps:[{id:2, comment:'ko'}, {id:1, comment:'ok too'}]})
WriteResult({ "nInserted" : 1 })
匹配系列(此处命名为'ok')
> db.dumb.find({a:'ok', eps:{$elemMatch:{id:1}}})
{ "_id" : ObjectId("5b8d447b21edbcdd7f3b3d24"), "a" : "ok", "eps" : [ { "id" : 1, "comment" : "ok" } ] }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : [ { "id" : 2, "comment" : "ko" }, { "id" : 1, "comment" : "ok too" } ] }
匹配名为 'ok' 的系列的 id 1 的电影 使用 unwind,数组的每一项都是 "spread"。请注意 "dupplicate objectIds"(您正在跟踪展开的文档)
> db.dumb.aggregate({$match:{a:'ok'}},{$unwind:'$eps'})
{ "_id" : ObjectId("5b8d447b21edbcdd7f3b3d24"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok" } }
{ "_id" : ObjectId("5b8d449121edbcdd7f3b3d26"), "a" : "ok", "eps" : { "id" : 2, "comment" : "ko" } }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : { "id" : 2, "comment" : "ko" } }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok too" } }
之后的标准匹配
> db.dumb.aggregate({$match:{a:'ok'}},{$unwind:'$eps'},{$match:{'eps.id':1}})
{ "_id" : ObjectId("5b8d447b21edbcdd7f3b3d24"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok" } }
{ "_id" : ObjectId("5b8d449d21edbcdd7f3b3d27"), "a" : "ok", "eps" : { "id" : 1, "comment" : "ok too" } }
您可以在 Aggregration 上使用 $filter 运算符。这是符合您要求的查询:
db.seasons.aggregate([
{ "$match": { "season_number": 19 } },
{ "$project": {
"season_number": 1,
"name": 1,
"number_of_episodes": 1,
"episodes": {
"$filter": {
"input": "$episodes",
"as": "episode",
"cond": { "$eq": [ "$$episode.episode_number", 1 ] }
}
}
}
}])
$filter 从版本 3.2
开始可用db.users.aggregate(
// Pipeline
[
// Stage 1
{
$unwind: {
path : "$episodes",
preserveNullAndEmptyArrays : false // optional
}
},
// Stage 2
{
$match: {
"season_number" : 19.0,
"episodes.episode_number" : 1
}
},
// Stage 3
{
$group: {
"_id":"$_id",
"season_number" :{$last:"season_number"},
"name" :{$last:"name"},
"number_of_episodes" :{$last:"number_of_episodes"},
"episodes":{$push:"$episodes"}
}
},
]);