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法律研究在 MongoDB 中查詢某個欄位的所有非唯一值？
使用 aggregate() 方法獲取某個欄位的所有非唯一值。我們首先用文件建立一個集合
> db.findAllNonDistinctDemo.insertOne({"UserName":"John","UserAge":28});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c995078863d6ffd454bb647")
}
> db.findAllNonDistinctDemo.insertOne({"UserName":"Larry","UserAge":21});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c995081863d6ffd454bb648")
}
> db.findAllNonDistinctDemo.insertOne({"UserName":"Larry","UserAge":23});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c995089863d6ffd454bb649")
}
> db.findAllNonDistinctDemo.insertOne({"UserName":"David","UserAge":22});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c995093863d6ffd454bb64a")
}
> db.findAllNonDistinctDemo.insertOne({"UserName":"John","UserAge":26});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c99509d863d6ffd454bb64b")
}
> db.findAllNonDistinctDemo.insertOne({"UserName":"Robert","UserAge":24});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c9950a7863d6ffd454bb64c")
}
> db.findAllNonDistinctDemo.insertOne({"UserName":"Robert","UserAge":25});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c9950b1863d6ffd454bb64d")
}
> db.findAllNonDistinctDemo.insertOne({"UserName":"Mike","UserAge":29});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c9950bc863d6ffd454bb64e")
}以下是使用 find() 方法顯示集合中所有文件的查詢
> db.findAllNonDistinctDemo.find().pretty();
這將產生以下輸出
{
"_id" : ObjectId("5c995078863d6ffd454bb647"),
"UserName" : "John",
"UserAge" : 28
}
{
"_id" : ObjectId("5c995081863d6ffd454bb648"),
"UserName" : "Larry",
"UserAge" : 21
}
{
"_id" : ObjectId("5c995089863d6ffd454bb649"),
"UserName" : "Larry",
"UserAge" : 23
}
{
"_id" : ObjectId("5c995093863d6ffd454bb64a"),
"UserName" : "David",
"UserAge" : 22
}
{
"_id" : ObjectId("5c99509d863d6ffd454bb64b"),
"UserName" : "John",
"UserAge" : 26
}
{
"_id" : ObjectId("5c9950a7863d6ffd454bb64c"),
"UserName" : "Robert",
"UserAge" : 24
}
{
"_id" : ObjectId("5c9950b1863d6ffd454bb64d"),
"UserName" : "Robert",
"UserAge" : 25
}
{
"_id" : ObjectId("5c9950bc863d6ffd454bb64e"),
"UserName" : "Mike",
"UserAge" : 29
}以下是查詢 MongoDB 中某個欄位的所有非唯一值的查詢
> db.findAllNonDistinctDemo.aggregate([
... { "$group": {
... "_id": "$UserName",
... "Counter": { "$sum": 1 }
... }},
... { "$match": {
... "Counter": { "$gt": 1 }
... }}
... ]).pretty();這將產生以下輸出
{ "_id" : "Robert", "Counter" : 2 }
{ "_id" : "Larry", "Counter" : 2 }
{ "_id" : "John", "Counter" : 2 }
更新於: 2019 年 7 月 30 日
853 次瀏覽
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