Question

我正在尝试通过获取正确的linkedAccount来选择耦合用户。创建的查询是正确的，但在尝试使用属性时在dbuser例如dbuser.lastName我得到一个编译错误，因为dbuser不是类型为User但Query1 size =？
这可能是非常简单的事情，但是因为我而无法理解 scala和squeryl noob！

为什么它不返回正确的值以及我在查询中做错了什么？此外，保存工作没有任何问题。

用户：

class User(
            @Column("id") val id: Long,
            @Column("first_name") val firstName : String,
            @Column("last_name") val lastName : String,
            @Column("email") val email : String,
            @Column("email_validated") val emailValidated: Boolean = false,
            @Column("last_login") val lastLogin: Timestamp = null,
            val created: Timestamp,
            val modified: Timestamp,
            val active: Boolean = false
                 ) extends KeyedEntity[Long] {

  lazy val linkedAccounts: OneToMany[LinkedAccount] = AppDB.usersToLinkedAccounts.left(this)
}

LINKEDACCOUNT：

class LinkedAccount(
                     @Column("id") val id: Long,
                     @Column("user_id") val userId: Long,
                     @Column("provider_user_id") val providerUserId: String,
                     @Column("salt") val salt: String,
                     @Column("provider_key") val providerKey: String) extends KeyedEntity[Long] {

  lazy val user: ManyToOne[User] = AppDB.usersToLinkedAccounts.right(this)

}

AppDB：

object AppDB extends Schema {
  val users = table[User]("users")
  val linkedAccounts = table[LinkedAccount]("linked_account")
  val usersToLinkedAccounts = oneToManyRelation(users, linkedAccounts).via((u, l) => u.id === l.userId)

def userByLinkedAccount(prodivderKey: String, providerUserId: String) = {
    from(AppDB.users)(u =>
      where(u.id in
        from(AppDB.linkedAccounts)(la =>
          where(la.userId === u.id and la.providerKey === prodivderKey and la.providerUserId === providerUserId)
            select (la.userId)
        )
      )
        select (u)
    )    
  }

电话：

val dbuser = inTransaction {
      val u2 = AppDB.userByLinkedAccount(id.providerId, id.id)
      println(u2.statement)              
    }
     println(dbuser.lastName)

生成的SQL

Select
  users10.last_login as users10_last_login,
  users10.email as users10_email,
  users10.modified as users10_modified,
  users10.last_name as users10_last_name,
  users10.first_name as users10_first_name,
  users10.id as users10_id,
  users10.created as users10_created,
  users10.email_validated as users10_email_validated,
  users10.active as users10_active
From
  users users10
Where
  (users10.id in ((Select
     linked_account13.user_id as linked_account13_user_id
   From
     linked_account linked_account13
   Where
     (((linked_account13.user_id = users10.id) and (linked_account13.provider_key = 'facebook')) and (linked_account13.provider_user_id = 'XXXXXXXXXX'))
  ) ))

Answer 1

好的想通了。我需要拨打电话，在这种情况下：

.headOption

在Per

的一些提示之后修复了查询

def userByLinkedAccount(providerKey : String, providerUserId : String) = {
    inTransaction {
      from(AppDB.users, AppDB.linkedAccounts)((u,la) =>
        where (u.id === la.userId and la.providerKey === providerKey and la.providerUserId === providerUserId)
          select(u)
      ).headOption
    }
  }

Answer 2

BTW，在@Column和@ColumnBase的文档中说：

定义列元数据的首选方法不是定义它们（！）

因此，您可以将列定义为

val id: Long,

而不是

@Column("id") val id: Long

使用squeryl和scala时的奇怪结果

2 个答案: