scala - 使用 scalikeJDBC 映射结果时如何解决类型不匹配问题？

Question

我在Single / Optional Result for Query @ ScalikeJDBC Docs上输入了几行：

package org.mysample

import scalikejdbc._

// define a class to map the result
case class Emp(id: String, name: String)

// QueryDSL
object Emp extends SQLSyntaxSupport[Emp] {
  def apply(e: ResultName[Emp])(rs: WrappedResultSet): Emp =
    new Emp(id = rs.get(e.id), name = rs.get(e.name))
}

class TestClass {
  val emp: Option[Emp] = DB readOnly { implicit session =>
    sql"SELECT id, name FROM emp WHERE id = ${id}"
      .map(rs => Emp(rs.string("id"), rs.string("name"))).single.apply()
  }

  val e = Emp.syntax("e")
  val id = 123
  val emp2: Option[Emp] = DB readOnly { implicit session =>
    withSQL {
      select.from(Emp as e).where.eq(e.id, id)
    }.map(Emp(e)).single.apply()
  }
}

但是这个片段有一个编译错误：

[error] /Users/pj/git/sirloin/src/main/scala/org/mysample/emp.scala:25: type mismatch;
[error]  found   : scalikejdbc.QuerySQLSyntaxProvider[scalikejdbc.SQLSyntaxSupport[org.mysample.Emp],org.mysample.Emp]
[error]  required: scalikejdbc.ResultName[org.mysample.Emp]
[error]     (which expands to)  scalikejdbc.ResultNameSQLSyntaxProvider[scalikejdbc.SQLSyntaxSupport[org.mysample.Emp],org.mysample.Emp]
[error]     }.map(Emp(e)).single.apply()
[error]               ^

我在 sbt 中清理并重新编译，但仍然无法编译。这段代码有什么问题？

感谢您的帮助。

Answer 1

对不起，打扰你。只需指定map(Emp(e.resultName))即可。或像这样更新 Emp 伴随对象：

object Emp extends SQLSyntaxSupport[Emp] {
  def apply(e: ResultName[Emp])(rs: WrappedResultSet): Emp =
    new Emp(id = rs.get(e.id), name = rs.get(e.name))
  def apply(e: SyntaxProvider[Emp])(rs: WrappedResultSet): Emp = apply(e.resultName)(rs)
}