相关疑难解决方法(0)

如何从Scala设置环境变量?

我需要从Scala设置环境变量(PATH).

我试过这个:

val cmd = Seq("export", "PATH='bla'")
cmd.lines
Run Code Online (Sandbox Code Playgroud)

但是我得到了错误:

java.io.IOException: Cannot run program "export": error=2, No such file or directory
at java.lang.ProcessBuilder.start(ProcessBuilder.java:1041)
at scala.sys.process.ProcessBuilderImpl$Simple.run(ProcessBuilderImpl.scala:68)
at scala.sys.process.ProcessBuilderImpl$AbstractBuilder.lines(ProcessBuilderImpl.scala:140)
at scala.sys.process.ProcessBuilderImpl$AbstractBuilder.lines(ProcessBuilderImpl.scala:106)
at .<init>(<console>:12)
at .<clinit>(<console>)
at .<init>(<console>:11)
at .<clinit>(<console>)
at $print(<console>)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:606)
at scala.tools.nsc.interpreter.IMain$ReadEvalPrint.call(IMain.scala:704)
at scala.tools.nsc.interpreter.IMain$Request.loadAndRun(IMain.scala:914)
at scala.tools.nsc.interpreter.IMain.loadAndRunReq$1(IMain.scala:546)
at scala.tools.nsc.interpreter.IMain.interpret(IMain.scala:577)
at scala.tools.nsc.interpreter.IMain.interpret(IMain.scala:543)
at scala.tools.nsc.interpreter.ILoop.reallyInterpret$1(ILoop.scala:694)
at scala.tools.nsc.interpreter.ILoop.interpretStartingWith(ILoop.scala:745)
at scala.tools.nsc.interpreter.ILoop.command(ILoop.scala:651)
at scala.tools.nsc.interpreter.ILoop.processLine$1(ILoop.scala:542)
at scala.tools.nsc.interpreter.ILoop.loop(ILoop.scala:550)
at scala.tools.nsc.interpreter.ILoop.process(ILoop.scala:822)
at scala.tools.nsc.interpreter.ILoop.main(ILoop.scala:851)
at xsbt.ConsoleInterface.run(ConsoleInterface.scala:57)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) …
Run Code Online (Sandbox Code Playgroud)

scala

7
推荐指数
1
解决办法
1万
查看次数

Scala:无法设置环境变量

朋友们,

我试图设置环境变量“ASDF”在我的斯卡拉壳,如所描述这里

这些是我的命令:

scala> import scala.sys.process.Process
import scala.sys.process.Process

scala> Process(Seq("bash", "-c", "echo $asdf"), None, "asdf" -> "Hello, world!").!
Hello, world!
res18: Int = 0
Run Code Online (Sandbox Code Playgroud)

但是当我尝试读回环境变量时:

scala> sys.env.get("asdf")
res19: Option[String] = None
Run Code Online (Sandbox Code Playgroud)

输出显示“无”。如何在当前会话中正确设置环境变量?

PS - 请不要贬低这个;我真的很努力,但无法解决我的问题

scala set environment-variables

5
推荐指数
2
解决办法
8574
查看次数

标签 统计

scala ×2

environment-variables ×1

set ×1