我想查询mongodb文件以返回特定文件
假设我有以下文档，并且只想搜索给定房间的特定详细信息。说，我想查询以返回room_num = 210的注释。我该怎么做？
我能够通过db.locations.findOne（{“ facility.room.room_num”：room_num}）进行查询；我不知道如何只投影房间。概括这个问题，我如何只投影嵌入式文档特定的条目。
我的文档看起来像这样--->
{
_id: Object("xxx"),
name: "Campus",
facilities: [
{ name : "Science",
rooms : [
{
room_num: 210,
notes: "Chemistry Lab 1",
Dimension: "x*x"
},{
room_num: 120,
notes: "Chemistry Lab 2",
Dimension: "x*x"
}
]
},
{ name : "Arts",
rooms: [
{
room_num: 90,
notes: "Drawing 1",
Dimension: "x*x"
},{
room_num: 100,
notes: "Drawing 2",
Dimension: "x*x"
}
]
}
]
由于您有一个嵌入式阵列（一个在另一个阵列中），因此您确实可以使用聚合来完成。您只需要执行“双重”展开操作：
db.collection.aggregate([
{ "$unwind": "$facilities" },
{ "$unwind": "$facilities.rooms" },
{ "$match": { "facilities.rooms.room_num": 210 } }
])
如果您真的希望它看起来像原始文档，那么请进一步：
db.collection.aggregate([
{ "$unwind": "$facilities" },
{ "$unwind": "$facilities.rooms" },
{ "$match": { "facilities.rooms.room_num": 210 } },
{ "$group": {
"_id": "$_id",
"name": { "$first": "$name" },
"facName": { "$first": "$facilities.name" },
"rooms": { "$push": "$facilities.rooms" }
}},
{ "$group": {
"_id": "$_id",
"name": { "$first": "$name" },
"facities": { "$push": {
"name": "$facName" ,
"rooms": "$rooms"
}}
}}
])
因此，从您的数据中可以得出以下结果：
{
"_id" : {
"0" : "x",
"1" : "x",
"2" : "x"
},
"name" : "Campus",
"facities" : [
{
"name" : "Science",
"rooms" : [
{
"room_num" : 210,
"notes" : "Chemistry Lab 1",
"Dimension" : "x*x"
}
]
}
]
}
或者最后，如果您需要的只是数组的内部房间部分，则只需$project在最后使用一条语句，如第一个示例所示：
db.campus.aggregate([
{ "$unwind": "$facilities" },
{ "$unwind": "$facilities.rooms" },
{ "$match": { "facilities.rooms.room_num": 210 } },
{ "$project": {
"_id": 0,
"room_num": "$facilities.rooms.room_num",
"notes": "$facilities.rooms.notes",
"Dimension": "$facilities.rooms.Dimension"
}}
])
结果如下：
{ "room_num" : 210, "notes" : "Chemistry Lab 1", "Dimension" : "x*x" }
那应该清除它。
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