Question

我疲惫的Scala技能令我感到困惑的是正确的做事方式。以下代码仅在我包含t match {...行时编译（并且有效）。如果我消除了那一行，当然还有上一行的诊断println，我得到了编译时错误，如图所示。显然，编译器将折叠的返回视为单位（对我来说很惊讶）。这可能是合理的，但我不明白。有人请告诉我一个更好的编码方法，也许会给我更多的见解吗？

[error] /home/bill/activator/cherry/app/controllers/Application.scala:34: type mismatch;
[error]  found   : Unit
[error]  required: play.api.mvc.Result
[error]   }
[error]   ^

来源：

  def ptweets = Action { implicit request =>
    import play.api.data._
    val rqForm = Form(Forms.mapping(
      "k" -> Forms.number,
      "who" -> Forms.text,
      "what" -> Forms.text)(TweetInquiry.apply)(TweetInquiry.unapply))
    val t = rqForm.bindFromRequest.fold(
      formWithErrors => BadRequest("That's not good"),
      rq =>  Ok((views.html.properForm("POST tweets TBD.")(Html("<em>Blah</em>"))))
    ) // expect a play.api.mvc.Result
    println(t.getClass.getName) // this confirms it in both run-time cases
    t match { case v:Result => v } // yet this is required for compile
  }

Answer 1

正如m-z在评论中所说，改变

val t = rqForm.bindFromRequest.fold(
      formWithErrors => BadRequest("That's not good"),
      rq =>  Ok((views.html.properForm("POST tweets TBD.")(Html("<em>Blah</em>"))))
    ) // expect a play.api.mvc.Result
    println(t.getClass.getName) // this confirms it in both run-time    cases
    t match { case v:Result => v } // yet this is required for compile

只是：

rqForm.bindFromRequest.fold(
      formWithErrors => BadRequest("That's not good"),
      rq =>  Ok((views.html.properForm("POST tweets TBD.")(Html("<em>Blah</em>"))))
   )

折叠正在评估结果，但在您发布的代码中，您将该结果分配给值t。因此，它不是评估折叠结果的Action块，而是评估一个赋值（单位，见here）。

返回bindFromRequest.fold的类型

1 个答案: