Cast a squeryl Query to a case class scala

Question

I am new to Squeryl and trying to figure out how to cast the results I get back from a query into case class. I have something like this

def getUsers = {
    val data = getUserData
    ...
}

def getUserData = {
    transaction {
        from(users)(s =>
            select(s.id, s.firstName, s.lastName, s.userName, s.email, s.lastLoginDate, s.dateJoined)
        )
    }
}

case class UserData(userId: Long, firstName: String, lastName: String, userName: String, email: String, lastLoginDate: Timestamp, dateJoined: Timestamp)

case class UserDataRecords(users: List[UserData])

Ideally, I would like to get the data back in the from of UserDataRecords. Right now it is returned as Query[Tuple7]. Such as

... 
(5,Suzie,Queue,squeue,SQueue@example.com,2014-01-15 22:02:12.0,2014-01-15 22:02:12.0)
...

What I cant figure out is how to cast this data. Any help on this would be great!

score 2 · Accepted Answer · answered Apr 29 '14 at 04:20

The simplest way would be to use map to transform the result list. Something like the example below should be a starting point:

def getUserData = {
    transaction {
        from(users)(s =>
            select(s.id, s.firstName, s.lastName, s.userName, s.email, s.lastLoginDate, s.dateJoined)
        ).toList.map { row =>
            UserData(row._1, row._2, row._3, row._4, row._5, row._6, row._7)
        }
    }
}

This will iterate through each row and create a UserData object out of the tuple, leaving you with List[UserData].

That said, if your schema users is of type Table[UserData], then you could simply do:

from(users)(s => select(s))

Cast a squeryl Query to a case class scala

1 Answers1