mongodb - 在 $lookup 之后获取值作为元素数组

Question

对于 MongoDB，当使用 $lookup 查询多个集合时，是否可以为 $lookup 中返回的字段获取仅值列表？

我不想要的是包含所有键/值的完整对象的列表。

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

这个查询看起来差不多了，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()

Answer 1

只需在 name 字段中使用 .dot 符号

db.foo.aggregate([
  { "$lookup": {
    "from": "bar",
    "localField": "name",
    "foreignField": "foo",
    "as": "bars"
  }},
  { "$addFields": { "bars": "$bars.name" }}
])

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