playframework - Convert result set to Seq[(String,String)] using Scala and anorm in play -

i have tried result set mysql database table using anorm. here code.

package models import play.api.db._ import play.api.play.current import scala.collection.mutable._ import anorm._ import anorm.sqlparser._      case class brand(id: int, name: string)           object brand {               /**             * parse brand resultset             */             val simple = {                 get[int]("m_brand.idbrand") ~                 get[string]("m_brand.brandname") map {                 case id~name => brand(id, name)                 }             }              /**             * construct map[string,string] needed fill select options set.             */             def options: seq[(string,string)] = db.withconnection { implicit connection =>                  sql("select * m_brand order brandname").as(brand.simple *).                     foldleft[seq[(string, string)]](nil) { (cs, c) =>                       c.id.fold(cs) { id => cs :+ (id.tostring -> c.name) }                 }             }  }  

i tried change code experiments not worked.

but got error

read stdout: d:\projects\test\project_vendorm8\app\models\brand.scala:69: type mismatch; read stdout: found : scala.collection.immutable.nil.type read stdout: required: scala.collection.mutable.seq[(string, string)] d:\projects\test\project_vendorm8\app\models\brand.scala:69: type mismatch; found : scala.collection.immutable.nil.type required: scala.collection.mutable.seq[(string, string)] read stdout: foldleftseq[(string, string)] { (cs, c) => foldleftseq[(string, string)] { (cs, c) => read stdout: ^

as asked in comments, simpler solution not use map , write:

def options: seq[(string,string)] = db.withconnection { implicit connection =>   sql("select * m_brand order brandname").as(simple *)     .map( b => (b.id.tostring, b.name))     .toseq }