Question

如何将这两个Coffee个查询合并为一个，并指出它是coffee还是otherCoffee？

代码：

protected lazy val coffeeQuery = for {
    id <- Parameters[Long]
    coffee <- CoffeesTable if coffee.userId === id
} yield (coffee)

protected lazy val otherCoffeeQuery = for {
    id <- Parameters[Long]
    employment <- EmployeesTable if employment.employedId === id
    sup <- SuppliersTable if employment.supId === sup.id
    coffee <- CoffeesTable if (coffee.userId === id || coffee.userId === sup.userId)
} yield (otherCoffee)

现在我想获得一份咖啡和咖啡清单。区分是otherCoffee还是<{1}}：

def coffeeList(userId: Long): JValue =  withSession { implicit db: Session =>
    val coffee = coffeeQuery(userId)
    val otherCoffee = otherCoffeeQuery
    val coffeeUnion = coffee union otherCoffee
    // How would I efficiently distinguish which item is "coffee" and which is "otherCoffee"?
    ("coffee" -> 
        coffeeUnion.map { c =>
           ("name" -> c.name) ~
           ("coffee_or_other_coffee" -> // is this coffee or otherCoffee?
        }
    )
}

我愿意采取其他方式解决这个问题（不一定是工会）

Answer 1

您可以将查询结果映射到包含以下内容的元组： 'coffee / 'otherCoffee（或true / false等）

val coffee = Set(coffeeQuery(userId):_*)
val otherCoffee = Set(otherCoffeeQuery:_*)
val coffeeUnion = coffee.map(_ -> 'coffee) ++ otherCoffee.filterNot(coffee.contains(_)).map(_ -> 'otherCoffee)
// coffeeUnion is a Set[Tuple2[Coffee, Symbol]]; you can make this e.g. a Set[Tuple2[Coffee, String]] instead via
// coffee.map(_ -> "coffee") ++ otherCoffee.filterNot(coffee.contains(_)).map(_ -> "otherCoffee")
("coffee" ->
  coffeeUnion.map { tuple =>
    ("name" -> tuple._1.name) ~
    ("coffee_or_other_coffee" -> tuple._2)

您不需要使用Sets，但与使用filterNot

相比，它们会使Seqs更快

您也可以使用Sets无元组

来执行此操作

val coffee = Set(coffeeQuery(userId):_*)
val otherCoffee = Set(otherCoffeeQuery:_*)
val coffeeUnion = coffee union otherCoffee
("coffee" ->
  coffeeUnion.map { c =>
    ("name" -> c.name) ~
    ("coffee_or_other_coffee" -> if(coffee.contains(c)) 'coffee else 'otherCoffee)

Scala / Slick：区分联盟内的项目

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