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我使用 scalikejdbc 2.0.1 和 playframework 2.3。我按照一对多http://scalikejdbc.org/documentation/one-to-x.html的说明进行操作，但还是报错：

我的数据模型是，一个帐户有许多待办事项：

待办模型：</p> 

case class Todo (id: Long, value: String, userId:Option[Long] = None, users: Option[Account] = None){}
object Todo extends SQLSyntaxSupport[Todo]{
    val todo = syntax("todo")
    override val tableName = "todo"
    private val auto = AutoSession
    def opt(m: ResultName[Todo])(rs: WrappedResultSet) = rs.longOpt(m.id).map(_ => Todo(m)(rs))
    apply(todo.resultName)(rs)
    def apply(a: ResultName[Todo])(rs: WrappedResultSet): Todo = new Todo(
        id         = rs.long(todo.id),
        userId     =rs.longOpt(todo.userId),
        value      = rs.string(todo.value)
        )

    def apply(m: ResultName[Todo], a: ResultName[Account])(rs: WrappedResultSet) =  {
        apply(m)(rs).copy(users = rs.longOpt(a.id).map(_ => Account(a)(rs)))
    }


}

模型帐户是：

case class Account(id: Int, email: String, password: String, name: String, permission: Role,todos:Seq[Todo]=Nil)

object Account extends SQLSyntaxSupport[Account] {
...
    val (a, t) = (Account.syntax, Todo.syntax)
    val accounts: List[Account] = withSQL {
    select.from(Account as a).leftJoin(Todo as t).on(a.id,t.userId)
    }.one(Account(a))
    .toMany(Todo.opt(t))
    .map { (account, todos) => account.copy( todos = todos) }
    .list.apply()
    }
}

我得到的错误是：

[error] G:\testprojects\mifun\app\models\Todo.scala:23: overloaded method apply
needs result type
[error]     apply(m)(rs).copy(users = rs.longOpt(a.id).map(_ => Account(a)(rs)))
[error]          ^
[error] G:\testprojects\mifun\app\models\Account.scala:53: type mismatch;
[error]  found   : scalikejdbc.QuerySQLSyntaxProvider[scalikejdbc.SQLSyntaxSuppo
rt[models.Todo],models.Todo]
[error]  required: scalikejdbc.ResultName[models.Todo]
[error]     (which expands to)  scalikejdbc.ResultNameSQLSyntaxProvider[scalikej
dbc.SQLSyntaxSupport[models.Todo],models.Todo]
[error]     .toMany(Todo.opt(t))
[error]                      ^
[error] two errors found
[error] (compile:compile) Compilation failed

我有两个问题：

1、为什么我不能用toMany？我想用ResultNameSQLSyntaxProvider，我写的opt函数怎么改？

2、在 Todo.scala:23 上应该给出什么 rs 类型？

4

1 回答 1

2

1、为什么我不能用toMany？我想用ResultNameSQLSyntaxProvider，我写的opt函数怎么改？

抱歉，这是由于文档中的错误。更新opt方法如下：

def opt(m: SyntaxProvider[Todo])(rs: WrappedResultSet) 
  = rs.longOpt(m.resultName.id).map(_ => Todo(m.resultName)(rs))

或者像这样添加重载方法：

def opt(m: SyntaxProvider[Todo])(rs: WrappedResultSet): Option[Todo] = opt(m.resultName)(rs)

2、在 Todo.scala:23 上应该给出什么 rs 类型？

正如错误消息所说，重载方法应该在 Scala 中使用显式返回类型来定义。

def apply(m: ResultName[Todo], a: ResultName[Account])(rs: WrappedResultSet) =  {
  apply(m)(rs).copy(users = rs.longOpt(a.id).map(_ => Account(a)(rs)))
}

上面的方法应该是这样的：

def apply(m: ResultName[Todo], a: ResultName[Account])(rs: WrappedResultSet): Todo =  {
  apply(m)(rs).copy(users = rs.longOpt(a.id).map(_ => Account(a)(rs)))
}
于 2014-06-12T09:39:51.243 回答