ให้เราสร้างคอลเลกชันที่มีเอกสาร -
> db.demo136.insertOne(
... {
...
... "Name":"Chris",
... "Details":[
... {
... "Id":"101",
... "EmployeeName":"Mike",
... "EmployeeDetails":[
... {
... "EmpId":1001,
... "Salary":1000
... },
... {
... "EmpId":1002,
... "Salary":2000
... }
... ]
... },
... {
... "Id":"102",
... "EmployeeName":"David",
... "EmployeeDetails":[
... {
... "EmpId":1004,
... "Salary":4000
... },
... {
... "EmpId":1005,
... "Salary":5000
... }
... ]
... }
... ]
... }
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5e31b15cfdf09dd6d085399f")
} แสดงเอกสารทั้งหมดจากคอลเล็กชันโดยใช้วิธี find() -
> db.demo136.find();
สิ่งนี้จะสร้างผลลัพธ์ต่อไปนี้ -
{
"_id" : ObjectId("5e31b15cfdf09dd6d085399f"), "Name" : "Chris", "Details" : [
{ "Id" : "101", "EmployeeName" : "Mike", "EmployeeDetails" : [ { "EmpId" : 1001, "Salary" : 1000 },
{ "EmpId" : 1002, "Salary" : 2000 } ] },
{ "Id" : "102", "EmployeeName" : "David", "EmployeeDetails" : [ { "EmpId" : 1004, "Salary" : 4000 },
{ "EmpId" : 1005, "Salary" : 5000 } ] }
]
} ต่อไปนี้เป็นแบบสอบถามเพื่อจัดเรียงเอกสารย่อย -
> db.demo136.aggregate([
... { $unwind: "$Details" },
... {
... $project: {
... Name: 1,
... Details: 1,
... SalaryTotal: {
... $sum: "$Details.EmployeeDetails.Salary"
... }
... }
... },
... { $sort: { SalaryTotal: -1 } },
... {
... $group: {
... _id: "$_id",
... Name: { $first: "$Name" },
... Details: { $push: "$Details" }
... }
... }
... ]).pretty() สิ่งนี้จะสร้างผลลัพธ์ต่อไปนี้ -
{
"_id" : ObjectId("5e31b15cfdf09dd6d085399f"),
"Name" : "Chris",
"Details" : [
{
"Id" : "102",
"EmployeeName" : "David",
"EmployeeDetails" : [
{
"EmpId" : 1004,
"Salary" : 4000
},
{
"EmpId" : 1005,
"Salary" : 5000
}
]
},
{
"Id" : "101",
"EmployeeName" : "Mike",
"EmployeeDetails" : [
{
"EmpId" : 1001,
"Salary" : 1000
},
{
"EmpId" : 1002,
"Salary" : 2000
}
]
}
]
} 