Scala Salat和嵌入式MongoDb文档

我有一个case类，它由两个嵌入的文档组成，其中一个是列表。我在检索列表中的项目时遇到一些问题

请参阅下面我的代码：

package models

import play.api.Play.current
import com.novus.salat._
import com.novus.salat.dao._
import com.mongodb.casbah.Imports._
import se.radley.plugin.salat._
import com.novus.salat.global._

case class Category(
  _id: ObjectId = new ObjectId,
  category: Categories,
  subcategory: List[SubCategories]
)

case class Categories(
  category_id: String,
  category_name: String
  )

case class SubCategories(
  subcategory_id: String,
  subcategory_name: String
)

object Category extends ModelCompanion[Category, ObjectId] {
  val collection = mongoCollection("category")
  val dao = new SalatDAO[Category, ObjectId](collection = collection) {}
  val CategoryDAO = dao

  def options: Seq[(String,String)] = {
    find(MongoDBObject.empty).map(it => (it.category.category_id, it.category.category_name)).toSeq
  }

  def suboptions: Seq[(String,String,String)] = {
    find(MongoDBObject.empty).map(it => (it.category.category_id, it.subcategory.subcategory_id, it.subcategory.subcategory_name)).toSeq
  }
}

我发现错误：

值子类别\u id不是列表[models.SubCategories]

的成员，这对我来说没有任何意义。

您正在执行以下操作：

def suboptions: Seq[(String,String,String)] = {
  find(MongoDBObject.empty).map(category => {
    val categories: Categories = category.category
    val categoryId: String = categories.category._id
    val subcategory: List[Subcategory] = category.subcategory
    val subcategoryId: String = subcategory.subcategory_id //here you are trying to 
      //get id from list of subcategories, not one of them
    val subcategoryName: String = subcategory.subcategory_name //same here
    (categoryId, subcategoryId, subcategoryName)).toSeq  
  }
}

case class Category(
  _id: ObjectId = new ObjectId(), 
  id: String,
  name: String,
  subcategories: List[Subcategory]
)
case class Subcategory(
  id: String,
  name: String
)
//not tested
def suboptions: Seq[(String, String, String)] = {
  val categories = find(MongoDBObject.empty)
  for{
    category <- categories;
    subcategory <- category.subcategories
  } yield (category.id, subcategory.id, subcategory.name)
}

顺便说一句，在Scala中使用snake_大小写是很少见的，val/var名称应该是camelCase，请参阅

编辑：您可以通过执行以下操作使其变得简单：

def suboptions: Seq[(String,String,String)] = {
  find(MongoDBObject.empty).map(category => {
    val categories: Categories = category.category
    val categoryId: String = categories.category._id
    val subcategory: List[Subcategory] = category.subcategory
    val subcategoryId: String = subcategory.subcategory_id //here you are trying to 
      //get id from list of subcategories, not one of them
    val subcategoryName: String = subcategory.subcategory_name //same here
    (categoryId, subcategoryId, subcategoryName)).toSeq  
  }
}

case class Category(
  _id: ObjectId = new ObjectId(), 
  id: String,
  name: String,
  subcategories: List[Subcategory]
)
case class Subcategory(
  id: String,
  name: String
)
//not tested
def suboptions: Seq[(String, String, String)] = {
  val categories = find(MongoDBObject.empty)
  for{
    category <- categories;
    subcategory <- category.subcategories
  } yield (category.id, subcategory.id, subcategory.name)
}

案例类别(
_id:ObjectId=新ObjectId（），
id:String，
名称：String，
子类别：列表[子类别]
)
案例类别子类别(
id:String，
名称：String
)
//未经测试
def子选项：Seq[（字符串、字符串、字符串）]={
val categories=find（MongoDBObject.empty）
为了{
类别