如何将 java.util.UUID 转换为 doobie.syntax.SqlInterpolator.SingleFragment？

Question

我正在尝试使用 doobie、http4s 和 circe 设置一个带有数据库的简单 Scala 应用程序。

如何将 java.util.UUID 转换为 doobie.syntax.SqlInterpolator.SingleFragment ？

final case class User(id: UUID, details: UserDetails)

  implicit val userDecoder: Decoder[User] = deriveDecoder[User]
  implicit def userEntityDecoder[F[_]: Sync]: EntityDecoder[F, User] = jsonOf

  implicit val userEncoder: Encoder[User] = deriveEncoder[User]
  implicit def userEntityEncoder[F[_]: Applicative]: EntityEncoder[F, User] = jsonEncoderOf

  implicit val put: Put[User] =
    Put[Json].contramap(_.asJson)

  implicit val get: Get[User] =
    Get[Json].temap(_.as[User].left.map(_.show))

[info] welcome to sbt 1.3.12 (N/A Java 14.0.1)
[info] loading global plugins from /Users/ryan/.sbt/1.0/plugins
[info] loading settings for project bobbymoore-build from plugins.sbt ...
[info] loading project definition from /Users/ryan/fullStackRyan/bobbymoore/project
[info] loading settings for project root from build.sbt ...
[info] set current project to bobbymoore (in build file:/Users/ryan/fullStackRyan/bobbymoore/)
[info] sbt server started at local:///Users/ryan/.sbt/1.0/server/90f560b1e0964865fc4c/sock
sbt:bobbymoore> compile
[info] Compiling 1 Scala source to /Users/ryan/fullStackRyan/bobbymoore/target/scala-2.13/classes ...
[error] /Users/ryan/fullStackRyan/bobbymoore/src/main/scala/com/c/bobbymoore/database/UserQueries.scala:24:20: type mismatch;
[error]  found   : java.util.UUID
[error]  required: doobie.syntax.SqlInterpolator.SingleFragment
[error]          |  ${user.id},
[error]                    ^
[error] /Users/ryan/fullStackRyan/bobbymoore/src/main/scala/com/c/bobbymoore/database/UserQueries.scala:34:21: type mismatch;
[error]  found   : java.util.UUID
[error]  required: doobie.syntax.SqlInterpolator.SingleFragment
[error]          |WHERE id=$id
[error]                     ^
[error] two errors found
[error] (Compile / compileIncremental) Compilation failed
[error] Total time: 2 s, completed 13 Oct 2020, 16:31:46

Answer 1

如果您使用的是 PostgreSQL 并且 Java 中的 UUID 转换为 Postgres 表中的 UUID 列，那么您必须使用Doobie的Postgres 扩展

// other doobie imports
import doobie.postgres._
import doobie.postgres.implicits._

sql"""SELECT id, name FROM users WHERE id = ${user.id}""".query[(UUID, String)]

如果您不使用 PSQL 并且此 UUID 映射到 TEXT 字段，则您必须在查询中自己映射：

sql"""SELECT id, name FROM users WHERE id = ${user.id.toString}"""
  .query[(String, String)]
  .map { case (uuid, name) => (UUID.fromString(uuid), name) }

或定义您自己的Meta[UUID]实例，例如

implicit val uuidMeta: Meta[UUID] =
  Meta[String].imap[UUID](UUID.fromString)(_.toString)

如果您想为包含 UUID 的 case 类派生一个实例，则需要这个。 Postgres 扩展库提供了一种与 PSQL 实现相匹配的扩展库。

如果您在 DB 端使用例如Array[Byte]或其他一些数字表示来表示 UUID，则您肯定必须定义自己的 Meta 实例。