将数据类型序列String更改为String

时间:2016-04-25 08:03:44

标签: scala playframework slick

我的DAL:

  def login(username: String,pass :String) =  
        sql"""SELECT AD1SYS.PKG_STC_LOGIN.USER_LOGIN ('#$username','#$pass')FROM DUAL""".as[String]

      def getEmpName(username: String,empName :String) =  
        sql"""SELECT AD1SYS.PKG_STC_LOGIN.GET_EMPL_NAME ('#$username','#$empName') FROM DUAL""".as[(String)]

我的服务:

def login(username:String,pass:String): Future[Seq[(String)]] = {
    db.run(DalLogin.login(username, pass))
  }

  def getEmpName(username:String,empName:String): Future[Seq[(String)]] = {
    db.run(DalLogin.getEmpName(username, empName))
  }

我的控制器:

def login = Action.async(BodyParsers.parse.json) { request =>
   val username = (request.body \ "username").as[String]
   val pass = (request.body \ "pass").as[String]
   ServiceLogin.login(username, pass).map { data => 
   ServiceLogin.getEmpName(username,data).map(result => 
     Ok(Json.toJson(result.map( p =>
            ModelKodeCabang(p)
        ))
        ))}
      }

Service.login的结果是Seq(String)" data",我不能使用Seq(String)参数ServiceLogin.getEmpName(用户名,数据)..如何转换Seq(String)to String ??

0 个答案:

没有答案