Scala反射：如何獲取給定類的類的字符串名稱

Question

我讀了一些有關Scala反射的文章。 例如http://docs.scala-lang.org/overviews/reflection/overview.html

就我而言，我想看看給定的類名是否可以實例化。 例如org.apache.spark.sql.catalyst.SqlLexical

我不能直接使用SqlLexical（在Spark 1.6中定義），因為運行時可能是Spark 2.0

如何獲取給定String classname的類？

謝謝

Answer 1

並不是嚴格的Scala解決方案，但是您可以嘗試使用ClassLoader.loadClass加載它，然后獲取構造函數並將其調用為實例。

Answer 2

有幫助：

$ scala
Welcome to Scala 2.11.8 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_92).
Type in expressions for evaluation. Or try :help.

scala> import scala.reflect.internal.util.ScalaClassLoader
import scala.reflect.internal.util.ScalaClassLoader

scala> ScalaClassLoader(getClass.getClassLoader).tryToLoadClass("scala.Option")
res0: Option[Class[Nothing]] = Some(class scala.Option)

scala> ScalaClassLoader(getClass.getClassLoader).create("scala.UninitializedError")
res1: AnyRef = scala.UninitializedError: uninitialized value

向后兼容：

$ scala210
Welcome to Scala version 2.10.5 (OpenJDK 64-Bit Server VM, Java 1.7.0_95).
Type in expressions to have them evaluated.
Type :help for more information.

scala> import scala.tools.nsc.util.ScalaClassLoader
import scala.tools.nsc.util.ScalaClassLoader

scala> ScalaClassLoader(getClass.getClassLoader).create("scala.UninitializedError")
res0: AnyRef = scala.UninitializedError: uninitialized value

scala> 

$ scala
Welcome to Scala 2.11.8 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_92).
Type in expressions for evaluation. Or try :help.

scala> import scala.tools.nsc.util.ScalaClassLoader
import scala.tools.nsc.util.ScalaClassLoader

scala> ScalaClassLoader(getClass.getClassLoader).create("scala.UninitializedError")
warning: there was one deprecation warning; re-run with -deprecation for details
res0: AnyRef = scala.UninitializedError: uninitialized value