验证:隐式scalaz.Bind未找到

Question

我有scalaz验证的这个实现,似乎隐含scalaz.Bind不在范围内,所以表达式不起作用.这是代码:

import scalaz._
import Scalaz._

case class User(name: String, knowScala: Boolean, age: Int)

object PublicValidationSpec extends BasicValidation {
  def validate(user: User): Validation[String, String] = {
    for {
      b <- checkAge(user)
      c <- checkName(user)
    } yield s"Congrats, ${c.name}"
  }
}

trait BasicValidation {
  def checkName(user: User): Validation[String, User] = {
    if(user.name == "") "must have a name".fail else user.success
  }

  def checkAge(user: User): Validation[String, User] = {
    if(user.age < 3) "must be a valid age".fail else user.success
  }
}

例外是

Error:(14, 25) Implicit not found: scalaz.Unapply[scalaz.Bind, 
scalaz.Validation[String,scalaz.validation.User]]. Unable to unapply type 
`scalaz.Validation[String,scalaz.validation.User]` into a type constructor of 
kind `M[_]` that is classified by the type class `scalaz.Bind`. Check that the 
type class is defined by compiling `implicitly[scalaz.Bind[type constructor]]` and 
review the implicits in object Unapply, which only cover common type 'shapes.'
           b <- checkAge(user)

Did i miss some implicit imports here ?
                        ^

Answer 1

验证没有为其定义Bind.

在Scalaz 7.1.0-M5(M6太)Validation.flatMap已被弃用,在颠覆警告的尝试,它看起来像的优先级flatMap正在失去scalaz.syntax.bind._,这是部分Scalaz._进口.请参阅此提交https://github.com/scalaz/scalaz/commit/2727b2451eba2aa159f3fbb62bf92790ac99dc7a.尝试添加import scalaz.Validation.FlatMap._或仅导入您需要的东西,例如

import scalaz.Validation
import scalaz.syntax.validation._

我建议使用除此之外的东西Validation,因为这可能只会在将来引起更多问题.见scalaz.\/下文.

这可以使用scalaz 7.0.5进行编译.Validation.flatMap在7.0.6中定义,因此它也应该使用该版本进行编译.我不会在新代码中使用此功能(Validation在for理解中).

scalaz的权力在Validation.flatMap一段时间以来一直在贬低.flatMap是什么让它在for理解中工作.flatMap 是在工作分支虽然弃用.这有很长的背景.请参阅https://groups.google.com/forum/#!topic/scalaz/Wnkdyhebo2w.

TLDR - 验证不是monad.任何可能的实现Validation.flatMap都不会与Apply定义的行为相匹配Validation.

scalaz.\/如果你想在for理解中使用某些东西,请使用(也称为析取).如果需要累积错误,请使用Validationover \/,convert to Validation然后返回的原因\/.