求 MongoDB 文件中重複列值的總得分?
要對不同文件中的值求和,請使用 MongoDB $group。讓我們建立一個包含文件的集合 −
> db.demo512.insertOne({"Name":"Chris","Score1":45,"Score2":65,"CountryName":"US"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e884d96987b6e0e9d18f588")
}
> db.demo512.insertOne({"Name":"Chris","Score1":41,"Score2":45,"CountryName":"US"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e884da2987b6e0e9d18f589")
}
> db.demo512.insertOne({"Name":"Bob","Score1":75,"Score2":55,"CountryName":"US"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e884db2987b6e0e9d18f58a")
}
> db.demo512.insertOne({"Name":"Bob","Score1":65,"Score2":90,"CountryName":"US"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e884dc2987b6e0e9d18f58b")
}藉助 find() 方法顯示集合中的所有文件 −
> db.demo512.find();
這將產生以下輸出 −
{ "_id" : ObjectId("5e884d96987b6e0e9d18f588"), "Name" : "Chris", "Score1" : 45, "Score2" :
65, "CountryName" : "US" }
{ "_id" : ObjectId("5e884da2987b6e0e9d18f589"), "Name" : "Chris", "Score1" : 41, "Score2" :
45, "CountryName" : "US" }
{ "_id" : ObjectId("5e884db2987b6e0e9d18f58a"), "Name" : "Bob", "Score1" : 75, "Score2" :
55, "CountryName" : "US" }
{ "_id" : ObjectId("5e884dc2987b6e0e9d18f58b"), "Name" : "Bob", "Score1" : 65, "Score2" :
90, "CountryName" : "US" }以下是 MongoDB 中對重複列值求和的查詢 −
> db.demo512.aggregate([
... {
... "$group": {
... "_id": "$Name",
... "Score1": { "$sum": "$Score1" },
... "Score2": { "$sum": "$Score2" },
... "CountryName": { "$first": "$CountryName" }
... }
... },
... {
... "$project": {
... "_id": 0,
... "Name": "$_id",
... "Score1": 1,
... "Score2": 1,
... "CountryName": 1
... }
... },
... { "$out": "demo513" }
... ])以下是顯示結果的查詢 −
> db.demo513.find();
這將產生以下輸出 −
{ "_id" : ObjectId("5e884f0e79442efd15c6eae0"), "Score1" : 140, "Score2" : 145, "CountryName" : "US", "Name" : "Bob" }
{ "_id" : ObjectId("5e884f0e79442efd15c6eae1"), "Score1" : 86, "Score2" : 110, "CountryName" : "US", "Name" : "Chris" }
廣告
資料結構
網路
關係型資料庫管理系統
作業系統
Java
iOS
HTML
CSS
Android
Python
C 程式設計
C++
C#
MongoDB
MySQL
Javascript
PHP