Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8369289
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 9, 20262026-06-09T13:36:17+00:00 2026-06-09T13:36:17+00:00

Using Play! in Scala with Squeryl ORM. I have no idea what’s causing the

  • 0

Using Play! in Scala with Squeryl ORM. I have no idea what’s causing the problem, but when I added a new model to Play! it now won’t compile with a very strange error:

value update is not a member of models.OauthCred

The compiler points specifically to the zero-argument constructor of this() which is required by Squeryl.

The code is below:

case class OauthCred(
  val id: Long,

  @Column("vendor")
  val vendor: String,

  @Column("user_id")
  val userId: Long,

  @Column("remote_id")
  val remoteId: Option[String],

  @Column("scope")
  var scope: String,

  @Column("access_code")
  var accessCode: Option[String],

  @Column("unauthorized_token")
  var unauthorizedToken: Option[String],

  @Column("authorized_token")
  var authorizedToken: Option[String],

  @Column("refresh_token")
  var refreshToken: Option[String],

  @Column("is_active")
  var isActive: Boolean,

  @Column("is_revoked")
  var isRevoked: Boolean,

  @Column("created")
  val created: Timestamp,

  @Column("modified")
  var modified: Timestamp

) extends KeyedEntity[Long] {

  this() = this(0L, "", 0L, Some(""), "", Some(""), Some(""), Some(""), Some(""), false, false, new Timestamp(), new Timestamp())

}

object OauthCred {

  def get(id:Long) = DB.oauthcred.lookup(id)

  def save(o:OauthCred) = DB.oauthcred.update(o)

  def getByRemoteId(remoteId:String) = { from(DB.oauthcred)( o => where(o.remoteId === Some(remoteId)) select(o) ).headOption }
}

What could be causing this?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-09T13:36:18+00:00Added an answer on June 9, 2026 at 1:36 pm

    Solved the problem, it was actually a lot simpler than I thought:

    this() actually needed to be def this()

    Hope this helps future users.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I'm new at using Play Framework 2.0 (I am using Scala) and have a
I'm using Scala Play! framework with Anorm to persist the data model to the
I am using Play Framework 2.0 with Scala. So I have an action that
I'm trying to test a model method using Play Framework 2.0 and Specs2. Global.scala
Using Play! 1.2.4, Scala, and Squeryl with Postgres Performing a simple DB insert like
I am able to debug scala controller in play framework using netbeans IDE but
Using play framework 1.2.4 with scala. I have few play jobs that looks like
I'm using the Play! framework and I have a model (an Entity) that has
I created a new project using the play console now, by default I got
Since Intellij does not yet support the Play-Scala-Template-Engine I was thinking about using plain

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.