对于MongoDB,当使用$lookup
查询多个集合时,是否可以获得$lookup
中返回的字段的仅值列表?
我不需要的是完整对象及其所有键/值的列表。
数据:
failover_tool:PRIMARY> db.foo.find().pretty()
{
"_id" : ObjectId("5ce72e415267960532b8df09"),
"name" : "foo1",
"desc" : "first foo"
}
{
"_id" : ObjectId("5ce72e4a5267960532b8df0a"),
"name" : "foo2",
"desc" : "second foo"
}
failover_tool:PRIMARY> db.bar.find().pretty()
{
"_id" : ObjectId("5ce72e0c5267960532b8df06"),
"name" : "bar1",
"foo" : "foo1"
}
{
"_id" : ObjectId("5ce72e165267960532b8df07"),
"name" : "bar2",
"foo" : "foo1"
}
{
"_id" : ObjectId("5ce72e1d5267960532b8df08"),
"name" : "bar3",
"foo" : "foo2"
}
所需的查询输出
{
"_id" : ObjectId("5ce72e415267960532b8df09"),
"name" : "foo1",
"desc" : "first foo",
"bars" : ["bar1", "bar2"]
},
{
"_id" : ObjectId("5ce72e4a5267960532b8df0a"),
"name" : "foo2",
"desc" : "second foo",
"bars" : ["bar3"]
}
最近
该查询似乎已经存在,但是它在bars
字段中返回了太多数据:
db.foo.aggregate({
$lookup: {
from:"bar",
localField:"name",
foreignField: "foo",
as:"bars"
}
}).pretty()
答案 0 :(得分:4)
只需在.dot
字段中使用name
表示法
db.foo.aggregate([
{ "$lookup": {
"from": "bar",
"localField": "name",
"foreignField": "foo",
"as": "bars"
}},
{ "$addFields": { "bars": "$bars.name" }}
])
答案 1 :(得分:2)
希望以下查询有帮助:
db.foo.aggregate([{
$lookup: {
from:"bar",
localField:"name",
foreignField: "foo",
as:"bars"
}
},
{$unwind : '$bars'},
{
$group : {
_id : {
_id : '$_id',
name : '$name',
desc : '$desc'
},
bars : { $push : '$bars.name'}
}
},
{
$project : {
_id : '$_id._id',
name : '$_id.name',
desc : '$_id.desc',
bars : '$bars'
}
}
]).pretty()
输出:
{
"_id" : ObjectId("5ce72e4a5267960532b8df0a"),
"name" : "foo2",
"desc" : "second foo",
"bars" : [
"bar3"
]
}
{
"_id" : ObjectId("5ce72e415267960532b8df09"),
"name" : "foo1",
"desc" : "first foo",
"bars" : [
"bar1",
"bar2"
]
}