【发布时间】:2022-01-14 19:15:37
【问题描述】:
我正在查看以下documentation。
以下将文档插入集合classes。
db.classes.insertMany( [
{ _id: 1, title: "Reading is ...", enrollmentlist: [ "giraffe2", "pandabear", "artie" ], days: ["M", "W", "F"] },
{ _id: 2, title: "But Writing ...", enrollmentlist: [ "giraffe1", "artie" ], days: ["T", "F"] }
] )
还有members 集合:
db.members.insertMany( [
{ _id: 1, name: "artie", joined: new Date("2016-05-01"), status: "A" },
{ _id: 2, name: "giraffe", joined: new Date("2017-05-01"), status: "D" },
{ _id: 3, name: "giraffe1", joined: new Date("2017-10-01"), status: "A" },
{ _id: 4, name: "panda", joined: new Date("2018-10-11"), status: "A" },
{ _id: 5, name: "pandabear", joined: new Date("2018-12-01"), status: "A" },
{ _id: 6, name: "giraffe2", joined: new Date("2018-12-01"), status: "D" }
] )
他们使用以下聚合来连接数组字段enrollmentlist上的两个集合。
db.classes.aggregate( [
{
$lookup:
{
from: "members",
localField: "enrollmentlist",
foreignField: "name",
as: "enrollee_info"
}
}
] )
返回以下内容:
{
"_id" : 1,
"title" : "Reading is ...",
"enrollmentlist" : [ "giraffe2", "pandabear", "artie" ],
"days" : [ "M", "W", "F" ],
"enrollee_info" : [
{ "_id" : 1, "name" : "artie", "joined" : ISODate("2016-05-01T00:00:00Z"), "status" : "A" },
{ "_id" : 5, "name" : "pandabear", "joined" : ISODate("2018-12-01T00:00:00Z"), "status" : "A" },
{ "_id" : 6, "name" : "giraffe2", "joined" : ISODate("2018-12-01T00:00:00Z"), "status" : "D" }
]
}
{
"_id" : 2,
"title" : "But Writing ...",
"enrollmentlist" : [ "giraffe1", "artie" ],
"days" : [ "T", "F" ],
"enrollee_info" : [
{ "_id" : 1, "name" : "artie", "joined" : ISODate("2016-05-01T00:00:00Z"), "status" : "A" },
{ "_id" : 3, "name" : "giraffe1", "joined" : ISODate("2017-10-01T00:00:00Z"), "status" : "A" }
]
}
如何将enrolle_info 缩减为包含所有names 的字符串数组?
这是我追求的结果:
{
"_id" : 1,
"title" : "Reading is ...",
"enrollmentlist" : [ "giraffe2", "pandabear", "artie" ],
"days" : [ "M", "W", "F" ],
"enrollee_info" : [
"artie",
"pandabear"
"giraffe2"
]
}
{
"_id" : 2,
"title" : "But Writing ...",
"enrollmentlist" : [ "giraffe1", "artie" ],
"days" : [ "T", "F" ],
"enrollee_info" : [
"artie",
"giraffe1"
]
}
我还通过在$lookup 操作中引入pipeline 字段来研究使用多个连接。我可以使用$project 来获取仅使用{"name": "example"} 的数组,但我不确定如何删除"name"。我尝试过使用{"$unwind": "$enrollee_info.name"},但这并没有给我想要的东西。我是否需要在我进行连接之后向我的聚合管道引入另一个阶段?
【问题讨论】:
-
您可以在
$lookup之后引入$addFields阶段 - 并使用$map运算符按您的预期转换"enrollee_info"。
标签: mongodb aggregation-framework aggregation