Scala最佳实践:特征继承与枚举

Question

我目前正在尝试使用Scala并寻找最佳实践.我发现自己有两种相反的方法来解决单个问题.我想知道哪个更好,哪个更好,哪个更常规,如果你知道其他一些更好的方法.第二个看起来更漂亮.

1.基于枚举的解决方案

import org.squeryl.internals.DatabaseAdapter
import org.squeryl.adapters.{H2Adapter, MySQLAdapter, PostgreSqlAdapter}
import java.sql.Driver

object DBType extends Enumeration {
  val MySql, PostgreSql, H2 = Value

  def fromUrl(url: String) = {
    url match {
      case u if u.startsWith("jdbc:mysql:") => Some(MySql)
      case u if u.startsWith("jdbc:postgresql:") => Some(PostgreSql)
      case u if u.startsWith("jdbc:h2:") => Some(H2)
      case _ => None
    }
  }
}

case class DBType(typ: DBType) {
  lazy val driver: Driver = {
    val name = typ match {
      case DBType.MySql => "com.mysql.jdbc.Driver"
      case DBType.PostgreSql => "org.postgresql.Driver"
      case DBType.H2 => "org.h2.Driver"
    }
    Class.forName(name).newInstance().asInstanceOf[Driver]
  }
  lazy val adapter: DatabaseAdapter = {
    typ match {
      case DBType.MySql => new MySQLAdapter
      case DBType.PostgreSql => new PostgreSqlAdapter
      case DBType.H2 => new H2Adapter
    }
  }
}

2.基于单身的解决方案

import org.squeryl.internals.DatabaseAdapter
import org.squeryl.adapters.{H2Adapter, MySQLAdapter, PostgreSqlAdapter}
import java.sql.Driver

trait DBType {
  def driver: Driver
  def adapter: DatabaseAdapter
}

object DBType {
  object MySql extends DBType {
    lazy val driver = Class.forName("com.mysql.jdbc.Driver").newInstance().asInstanceOf[Driver]
    lazy val adapter = new MySQLAdapter
  }

  object PostgreSql extends DBType {
    lazy val driver = Class.forName("org.postgresql.Driver").newInstance().asInstanceOf[Driver]
    lazy val adapter = new PostgreSqlAdapter
  }

  object H2 extends DBType {
    lazy val driver = Class.forName("org.h2.Driver").newInstance().asInstanceOf[Driver]
    lazy val adapter = new H2Adapter
  }

  def fromUrl(url: String) = {
    url match {
      case u if u.startsWith("jdbc:mysql:") => Some(MySql)
      case u if u.startsWith("jdbc:postgresql:") => Some(PostgreSql)
      case u if u.startsWith("jdbc:h2:") => Some(H2)
      case _ => None
    }
  }
}

Answer 1

如果你声明了一个sealed trait DBType,你可以用穷举检查对它进行模式匹配(即,如果你忘记了一个案例,Scala会告诉你).

无论如何,我不喜欢斯卡拉Enumeration,而且我并不孤单.我从不使用它,如果有一些枚举实际上是最干净的解决方案,最好用Java编写它,用Java编写.

Answer 2

对于这种特殊情况,您并不需要每种数据库类型的类; 这只是数据.除非真实情况要复杂得多,否则我会使用基于地图和字符串解析的解决方案来最小化代码重复的数量:

case class DBRecord(url: String, driver: String, adapter: () => DatabaseAdapter) {}

class DBType(record: DBRecord) {
  lazy val driver = Class.forName(record.driver).newInstance().asInstanceOf[Driver]
  lazy val adapter = record.adapter()
}

object DBType {
  val knownDB = List(
    DBRecord("mysql", "com.mysql.jdbc.Driver", () => new MySQLAdapter),
    DBRecord("postgresql", "org.postgresql.Driver", () => new PostgreSqlAdapter),
    DBRecord("h2", "org.h2.Driver", () => new H2Adapter)
  )

  val urlLookup = knownDB.map(rec => rec.url -> rec).toMap

  def fromURL(url: String) = {
    val parts = url.split(':')
    if (parts.length < 3 || parts(0) != "jdbc") None
    else urlLookup.get(parts(1)).map(rec => new DBType(rec))
  }
}