僅顯示MongoDB文件中特定薪資的員工姓名,且文件包含員工記錄?
若要僅顯示特定薪資的員工姓名,請在MongoDB $in 中設定薪資並提取姓名。讓我們建立一個包含以下文件的集合
> db.demo666.insertOne({"EmployeeName":"John","EmployeeSalary":25000});
{
"acknowledged" : true,
"insertedId" : ObjectId("5ea1c04824113ea5458c7d0d")
}
> db.demo666.insertOne({"EmployeeName":"Chris","EmployeeSalary":35000});
{
"acknowledged" : true,
"insertedId" : ObjectId("5ea1c05524113ea5458c7d0e")
}
> db.demo666.insertOne({"EmployeeName":"David","EmployeeSalary":65000});
{
"acknowledged" : true,
"insertedId" : ObjectId("5ea1c06024113ea5458c7d0f")
}
> db.demo666.insertOne({"EmployeeName":"Carol","EmployeeSalary":40000});
{
"acknowledged" : true,
"insertedId" : ObjectId("5ea1c06f24113ea5458c7d10")
}使用find()方法顯示某個集合中的所有文件−
> db.demo666.find();
這將產生以下輸出 −
{ "_id" : ObjectId("5ea1c04824113ea5458c7d0d"), "EmployeeName" : "John", "EmployeeSalary" : 25000 }
{ "_id" : ObjectId("5ea1c05524113ea5458c7d0e"), "EmployeeName" : "Chris", "EmployeeSalary" : 35000 }
{ "_id" : ObjectId("5ea1c06024113ea5458c7d0f"), "EmployeeName" : "David", "EmployeeSalary" : 65000 }
{ "_id" : ObjectId("5ea1c06f24113ea5458c7d10"), "EmployeeName" : "Carol", "EmployeeSalary" : 40000 }以下是使用$in並提取特定薪資員工姓名的查詢−
> db.demo666.find({"EmployeeSalary":{$in:[35000,40000]}},{_id:0,"EmployeeSalary":0});這將產生以下輸出 −
{ "EmployeeName" : "Chris" }
{ "EmployeeName" : "Carol" }
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