播放框架:错误类型不匹配;找到:需要输入:play.api.mvc.Result

Question

>我正在尝试将表单值保存到数据库，并播放时出现错误：

type mismatch;  found :Int required: play.api.mvc.Result 

这是我的代码：

Application.scala

import play.api._
import play.api.mvc._
import play.api.data._
import play.api.data.Forms._
import views.html.defaultpages.badRequest
import play.api.data.validation.Constraints._    
import models.User
import models.UserData    
object Application extends Controller {
   val RegisterForm = Form(
      mapping(     
         "fname" -> nonEmptyText(1, 20),
         "lname" -> nonEmptyText(1, 20),
         "email" -> email,
         "userName" -> nonEmptyText(1, 20),
         "password" -> nonEmptyText(1, 20),
         "age" -> number,
         "choice" -> text,
         "gender" -> text
      )(User.apply)(User.unapply)
       .verifying("Ag should be greater then or equal to 18", model => model.age match {
          case (age) => age >= 18
       })
  )
  def index = Action {
    Ok(views.html.index(RegisterForm))
  }
  def register = Action { implicit request => 
      RegisterForm.bindFromRequest().fold(
          hasErrors => BadRequest(views.html.index(hasErrors)), 
          success => {  
              User.save(success)
          }
      )
  }
}

User.scala

import anorm._
import play.api.db.DB
import anorm.SqlParser._
import play.api.Play.current
case class User (
  fname: String,
  lname: String,
  email: String,
  userName: String,
  password: String,
  age: Int,
  choice: String,
  gender: String
)
object User {
  val userinfo = {
    get[String]("fname") ~ 
    get[String]("lname") ~ 
    get[String]("email") ~ 
    get[String]("userName") ~ 
    get[String]("password") ~ 
    get[Int]("age") ~ 
    get[String]("choice") ~ 
    get[String]("gender") map {
        case fname~lname~email~userName~password~age~choice~gender => 
            User(fname, lname, email, userName, password, age, choice, gender)
        }
    }
  def save(ud: User) = {
     DB.withConnection { implicit c => 
         SQL ("insert into userinfo(fname,lname,email,userName,password,age,choice,gender) values ({fname},{lname},{email},{userName},{password},{age},{choice},{gender})")
         .on('fname -> ud.fname, 'lname ->ud.lname ,'email ->ud.email, 'userName->ud.userName , 'password->ud.password ,'age->ud.age, 'choice->ud.choice,'gender->ud.gender)
         .executeUpdate()  
     }
  }
}

我从 Scala 食谱中获得了此保存函数的帮助，但我遇到了问题，因为我是玩 Scala 的新手，也许这个错误很小，所以请帮助我，我想要的只是将所有输入值存储在表单中输入的数据库中

Answer 1

问题出在控制器函数register：

def register = Action { implicit request => 
    RegisterForm.bindFromRequest().fold(
       hasErrors => BadRequest(views.html.index(hasErrors)), 
       success => {
           // You need to return a `Result` here..
           // Maybe redirect to login?
           // Redirect("/login")
           User.save(success)
       }
    )
}

问题是User.save返回一个Int，而控制器函数期待一个Result，如Ok(..)、Redirect(...)等。您需要对User.save的结果执行一些操作才能返回Result。

最好使用 executeInsert() 而不是 executeUpdate() ，因为默认情况下executeInsert返回Option[Long] - 这将帮助您确定保存是否成功。

例如：

def save(ud: User): Option[Long] = {
    DB.withConnection {implicit c => 
        SQL("insert into userinfo(fname,lname,email,userName,password,age,choice,gender) values ({fname},{lname},{email},{userName},{password},{age},{choice},{gender})")
            .on('fname -> ud.fname, 'lname ->ud.lname ,'email ->ud.email, 'userName->ud.userName , 'password->ud.password ,'age->ud.age, 'choice->ud.choice,'gender->ud.gender)
            .executeInsert()  
    }
}

然后在控制器函数中像这样使用它：

def register = Action { implicit request => 
    RegisterForm.bindFromRequest().fold(
       hasErrors => BadRequest(views.html.index(hasErrors)), 
       success => {
           User.save(success) map { newId =>
               Redirect("/login") // or whatever you want this to do on success
           } getOrElse {
               InternalServerError("Something has gone terribly wrong...")
           }
       }
    )
}

您可以将我的占位符结果替换为应用程序中更有意义的内容。

Answer 2

save（ud：User） 方法返回整数值。操作应始终返回结果，因此请编写

     success => {  
       val result = User.save(success)
         Ok(result.toString)
           }