如何为group by编写MongoDBObject和case类

我对mongo、scala和salat都是新手

我将json集合作为

{[
    {"firstName":"John", "lastName":"Doe", department="IT", skills="java"},
    {"firstName":"Anna", "lastName":"Smith", department="accounts, skills="tally"},
    {"firstName":"Peter", "lastName":"Jones", department="It" skills="java" }
]}

我对case类的定义如下

case class Employee(firstName:String,lastName:String,department:String,skills:String)

使用的沙拉酱如下所示

object EmployeeDao extends SalatDAO[Employees, ObjectId](collection =employee_collection)

还有让员工这样做的方法

  def getEmployees(IT : String):List[Employees]={

    val listOfEmployees:List[Employees] = OrderDAO3.find(MongoDBObject({"department"->IT})).toList

    return listOfEmployees
  }

  {"IT":[
    {"firstName":"John", "lastName":"Doe"},
    {"firstName":"Anna", "lastName":"Smith"},
    {"firstName":"Peter", "lastName":"Jones"}
  ]
  {"accounts":[
    {"firstName":"Johnac", "lastName":"Doeac"},
    {"firstName":"Annaac", "lastName":"Smithac"},
    {"firstName":"Peterac", "lastName":"Jonesac"}
 ]}
 }

得到这样的结果

{[ {“firstName”：“John”，“lastName”：“Doe”，department=“IT”，skills=“java”}， {“firstName”：“Peter”，“lastName”：“Jones”，department=“IT”skills=“java”} ]}

到目前为止还不错，。。。现在我想按所有部门分组，希望得到这样的结果

  def getEmployees(IT : String):List[Employees]={

    val listOfEmployees:List[Employees] = OrderDAO3.find(MongoDBObject({"department"->IT})).toList

    return listOfEmployees
  }

  {"IT":[
    {"firstName":"John", "lastName":"Doe"},
    {"firstName":"Anna", "lastName":"Smith"},
    {"firstName":"Peter", "lastName":"Jones"}
  ]
  {"accounts":[
    {"firstName":"Johnac", "lastName":"Doeac"},
    {"firstName":"Annaac", "lastName":"Smithac"},
    {"firstName":"Peterac", "lastName":"Jonesac"}
 ]}
 }

---

因此，请建议我如何编写MongoDBObject以获得此结果，并建议我保存该集合的相应案例类。

您能用这些信息自己编译您的答案吗？）

---

谢谢。）

你能用这些信息自己整理一下你的答案吗

谢谢）