Doobie无法找到或构造类型T的Read实例

Question

我正在使用doobie查询一些数据，并且一切正常，如下所示：

case class Usuario(var documento: String, var nombre: String, var contrasena: String)

  def getUsuario(doc: String) =
     sql"""SELECT documento, nombre, contrasena FROM "Usuario" WHERE "documento" = $doc"""
      .query[Usuario]
      .option
      .transact(xa)
      .unsafeRunSync()

但是如果我声明一个具有类型限制的函数，如下所示：

 def getOption[T](f: Fragment): Option[T] = {
    f.query[T]
     .option
     .transact(xa)
     .unsafeRunSync()

}

我遇到这些错误：

Error:(42, 12) Cannot find or construct a Read instance for type:
  T
This can happen for a few reasons, but the most common case is that a data
member somewhere within this type doesn't have a Get instance in scope. Here are
some debugging hints:
- For Option types, ensure that a Read instance is in scope for the non-Option
  version.
- For types you expect to map to a single column ensure that a Get instance is
  in scope.
- For case classes, HLists, and shapeless records ensure that each element
  has a Read instance in scope.
- Lather, rinse, repeat, recursively until you find the problematic bit.
You can check that an instance exists for Read in the REPL or in your code:
  scala> Read[Foo]
and similarly with Get:
  scala> Get[Foo]
And find the missing instance and construct it as needed. Refer to Chapter 12
of the book of doobie for more information.
    f.query[T].option.transact(xa).unsafeRunSync()

Error:(42, 12) not enough arguments for method query: (implicit evidence$1: doobie.util.Read[T], implicit h: doobie.LogHandler)doobie.Query0[T].
Unspecified value parameter evidence$1.
    f.query[T].option.transact(xa).unsafeRunSync()

有人知道怎么做我想要的吗？我认为这很简单，但我不知道如何解决。

Answer 1

为了使doobie