如何在Scala中获取当前脚本或类名？ [英] How do I get the current script or class name in Scala?

问题描述

我希望我的Scala程序能够以编程方式确定其脚本文件名或其类名，将该字符串存储在变量程序中并打印出名称。

I'd like my Scala program to be able to programmatically determine either its script filename or its class name, store the string in a variable program and print out the name.

Java有几种方法this。

Java has several methods for this.

推荐答案

// With help from huynhjl
// http://stackoverflow.com/questions/8129185#8131613

import scala.util.matching.Regex.MatchIterator

object ScriptName {
    val program = {
        val filenames = new RuntimeException("").getStackTrace.map { t => t.getFileName }
        val scala = filenames.indexOf("NativeMethodAccessorImpl.java")

        if (scala == -1)
            "<console>"
        else
            filenames(scala - 1)
    }

    def main(args: Array[String]) {
        val prog = program
        println("Program: " + prog)
    }
}

a href =http://rosettacode.org/wiki/Program_name#Scala =nofollow> Rosetta代码

Rosetta Code

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