假设这是我的数据的外观:
{
"_id" : ObjectId("545dad3562fa028fb48832f0"),
"number" : "123456",
"persons" : [
{
"name" : "A",
"country" : "US"
},
{
"name" : "N",
"country" : "Australia"
},
{
"name" : "Z",
"country" : "US"
}
]
}
{
"_id" : ObjectId("545dad3562fa028fb48832f0"),
"number" : "123457",
"persons" : [
{
"name" : "Q",
"country" : "India"
},
{
"name" : "B",
"country" : "Brazil"
},
{
"name" : "U",
"country" : "UK"
}
]
}
我想在C#中返回:(所有仅包含国家为美国的子文档的文档)
{
"_id" : ObjectId("545dad3562fa028fb48832f0"),
"number" : "123456",
"persons" : [
{
"name" : "A",
"country" : "US"
},
{
"name" : "Z",
"country" : "US"
}
]
}
目前,我可以获取包含匹配的子文档的文档,但也可以获取不相关的子文档。 查询我尝试过:
filter = Builders<BsonDocument>.Filter.Eq(c => c.number, "123456") & Builders<BsonDocument>.Filter.ElemMatch(c => c.persons, x => x.country == "US");
var result = client.GetDatabase(MyMongoDB).GetCollection<MyCollection>(CollectionName).Find<MyCollection>(filter);
答案 0 :(得分:0)
你可以用这样的聚合来做到这一点:
db.collection.aggregate([
{
$match:{
"persons.country":"US"
}
},
{
$redact:{
$cond:{
if:{
$or:[
{
$eq:[
"$country",
"US"
]
},
{
$or:"$number"
}
]
},
then:"$$DESCEND",
else:"$$PRUNE"
}
}
}
])
这将输出:
{
"_id":ObjectId("545dad3562fa028fb48832f0"),
"number":"123456",
"persons":[
{
"name":"A",
"country":"US"
},
{
"name":"Z",
"country":"US"
}
]
}