MongoDB查询仅基于角色获取“名称”字段？
aggregate()。在这里，我们考虑了3个角色-管理员，来宾和用户。让我们创建一个包含文档的集合-> db.demo532.insertOne({"Name":"Chris","Type":"Admin"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e8b4a9def4dcbee04fbbbf9")
}
> db.demo532.insertOne({"Name":"David","Type":"Guest"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e8b4aa3ef4dcbee04fbbbfa")
}
> db.demo532.insertOne({"Name":"Bob","Type":"User"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e8b4ab0ef4dcbee04fbbbfb")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo532.find();
这将产生以下输出-
{ "_id" : ObjectId("5e8b4a9def4dcbee04fbbbf9"), "Name" : "Chris", "Type" : "Admin" }
{ "_id" : ObjectId("5e8b4aa3ef4dcbee04fbbbfa"), "Name" : "David", "Type" : "Guest" }
{ "_id" : ObjectId("5e8b4ab0ef4dcbee04fbbbfb"), "Name" : "Bob", "Type" : "User" }以下是获取唯一文档的查询，该文档需要与同一集合中的其他文档进行比较-
> db.demo532.aggregate([
... {$group: {_id:"$Name", "types":{$addToSet:"$Type"}}},
... {$project: {_id:1, "types": 1, "isSubset": { $setIsSubset: [ ["Admin"], "$types" ] }}},
... {$match: {"isSubset": false}},
... {$group: {_id:"$isSubset", "Name": {$push : "$_id"}}},
... {$project: {"_id": 0, "Name": 1}},
... {$unwind: "$Name"}
... ])这将产生以下输出-
{ "Name" : "David" }
{ "Name" : "Bob" }