scala - 在没有显式匹配的情况下转换 Doobie ConnectionIO[Option[Int]]

Question

我有一个ConnectionIO[Option[Int]]和映射Option来产生ConnectionIO[Option[String]]一个查询，Some[Int]否则保持Nones. I was able to do this with a匹配comprehension and a`：

def findWidgetByOwner(name: String): ConnectionIO[Option[String]] = for {
    opt <- sql"SELECT owner_id FROM owners WHERE name = $name".query[Int].option
    widget <- opt match {
        case None => None.pure[ConnectionIO]
        case Some(id) => sql"SELECT widget_name FROM widgets WHERE owner_id = $id".query[String].option
    }
} yield widget

我知道我被ConnectionIO容器绊倒了，但我找不到更干净的映射方法来转换ConnectionIO[Option[Int]]为ConnectionIO[Option[String]].

Answer 1

使用 SQL 而不是 scala 加入会更干净：

def findWidgetByOwner(name: String): ConnectionIO[Option[String]] = 
  sql"""
    SELECT widgets.widget_name FROM widgets WHERE owner_id = $id
    INNER JOIN owners ON widgets.owner_id = owners.owner_id
    WHERE owners.name = $name
  """.query[Int].option

但是如果你想清理原始的，一些咒语sequence可能会起作用（未经测试）：

import cats._, cats.data._, cats.implicits._

...
widget <- opt.map {id => 
  sql"SELECT widget_name FROM widgets WHERE owner_id = $id".query[String].unique
}.sequence

注意：您必须更改query[String].option为，.query[String].unique否则如果小部件查询可以为空，则它可能是可取的，但最后需要widgeta。Option[Option[String]].flatten