我有两个mongo收藏：
>
defn
"_id" : ObjectId("8570bebcb7db3"),
"fields" : [ {
"control" : { "appearance" : "field-list" },
"children" : [ { "bind" : { "required" : "yes" }, ...
数据
"_id" : ObjectId("1570bf18a7db"),
"defn" : ObjectId("8570bebcb7db3"),
"data" : {
"country" : "",
"age" : 1,
"age_unit" : "years",
},
"label" : "type"
_id=data. defn
如何编写一个查询来返回Defn._id的data. label？该查询的格式为db.defn.find（{data.label where defn._id="X"}）
如果您使用的是MongoDB 3.2，$lookup阶段可以执行相当于左向外连接的操作。
留档在此提供了使用此操作的示例。
使用您的数据的示例：
db.defn.insert({
"_id" : "123456",
"some_text" : "main document"
})
db.data.insert( {
"defn" : "123456",
"label" : "data we want to access"
})
db.defn.aggregate( [
{ "$lookup" : {
"from" : "data",
"localField" : "_id",
"foreignField" : "defn",
"as" : "defns"
}
}
])
// Results:
// {
// "_id" : "123456",
// "some_text" : "main document",
// "defns" : [ {
// "_id" : ObjectId("57a2cbbbeb99ff285a1f0893"),
// "defn" : "123456",
// "label" : "data we want to access"
// } ]
// }
你可能想试试这个模块@coolgk/mongo，它允许你加入和过滤多个集合。
SELECT*from a LEFT JOIN b on a.b_id=b.id
变得
model.find({}, {
join: [ { on: 'b_id' } ]
})
结果：
[{
_id: '5a8bde4ae2ead929f89f3c42',
a_name: 'aname1',
b_id: {
_id: '5a8bde4ae2ead929f89f3c41',
b_name: 'bname1'
}
}, { ... }, ... ]
选择*from a， b WHERE a.b_id=b.id和b.b_name='bname1'
变得
model.find({}, {
join: [ { on: 'b_id', filters: { b_name: 'bname1' } } ]
})
结果：
[{
_id: '5a8bdfb05d44ea2a08fa8a4c',
a_name: 'aname2',
b_id: {
_id: '5a8bdfb05d44ea2a08fa8a4b',
b_name: 'bname2'
}
}]
SELECT*from a， b，c WHERE a.b_id=b.id和b.c_id=c.id和c.c_name='cname3'
modela.find({}, {
join: [{
on: 'b_id',
join: [{
on: 'c_id',
filters: { c_name: 'cname3' }
}]
}]
})
结果：
[{
_id: '5a8bdfc1b07af22a12cb1f0b',
a_name: 'aname3',
b_id: {
_id: '5a8bdfc1b07af22a12cb1f0a',
b_name: 'bname3',
c_id: {
_id: '5a8bdfc1b07af22a12cb1f09',
c_name: 'cname3'
}
}
}]
