Question

我有一个案例类：

case class A(field1: String, field2: Int)

我想在某些代码中引用确切的字符串“field1”，如：

val q = Query("field1"-> "hello")
performQuery(q)

现在我必须小心，我一直正确输入“field1”。为避免这种情况，我想使用对A.field1的引用作为字符串，如下所示：

val q = Query(A.field1 -> "hello")
performQuery(q)

这将帮助我避免任何输入错误，而且如果我稍后重命名这些字段，我不需要更新其余的代码库字符串。我可以依靠IDE。

如何将案例类字段名称提取为字符串？

EDIT1

以下是用法的另一种变体：

// capture the field name
val FIELD1_NAME = A.field1 // this implementation can change

此时FIELD1_NAME应包含"field1"字符串

// Now use the string obtained above to create a query:
val q = Query(FIELD1_NAME -> "hello")
performQuery(q)

Answer 1

找到https://gist.github.com/dwickern/b47606a1ed2319afa6e2：

import scala.annotation.tailrec
import scala.language.experimental.macros
import scala.reflect.macros.blackbox

object Macros {
  def nameOfImpl(c: blackbox.Context)(x: c.Tree): c.Tree = {
    import c.universe._

    @tailrec def extract(x: c.Tree): String = x match {
      case Ident(TermName(s)) => s
      case Select(_, TermName(s)) => s
      case Function(_, body) => extract(body)
      case Block(_, expr) => extract(expr)
      case Apply(func, _) => extract(func)
    }

    val name = extract(x)
    q"$name"
  }
  def nameOfMemberImpl(c: blackbox.Context)(f: c.Tree): c.Tree = nameOfImpl(c)(f)

  def nameOf(x: Any): String = macro nameOfImpl
  def nameOf[T](f: T => Any): String = macro nameOfMemberImpl
}

//
// Sample usage:
//

val someVariable = ""
Macros.nameOf(someVariable) // "someVariable"

def someFunction(x: Int): String = ???
Macros.nameOf(someFunction _) // "someFunction"

case class SomeClass(someParam: String)
val myClass = SomeClass("")
Macros.nameOf(myClass.someParam) // "someParam"

// without having an instance of the class:
Macros.nameOf[SomeClass](_.someParam) // "someParam"

对于您的用例，Query(nameOf[A](_.field1) -> "hello")。

在Scala中提取案例类字段名称

1 个答案: