Question

我想编译一下：

import scala.sys.process._

val output = "scala".!!

但是告诉我这个：

java.io.IOException：无法运行程序“scala”：CreateProcess error = 2，系统找不到指定的文件

这也行不通：

val cmd = "\"C:\\Program Files (x86)\\scala\\bin\\scalac.bat\\\""

val output = "cmd".!!

我的环境变量也很好。（对于java：C：\ Program Files \ Java \ jdk1.8.0_111 \和scala： C：\ Program Files（x86）\ scala 和路径变量：％JAVA_HOME％\ bin < / strong>和％SCALA_HOME％\ bin

在cmd工作中键入“scala”。这段代码也有效：

import scala.sys.process._ val output = "java".!!

Windows 10

Answer 1

命令scala是一个带有交互式控制台的REPL。您是否尝试从交互式控制台运行交互式命令？

它应该适用于任何非交互式命令，例如ls -l（dindows中的DIR）或date，如下面的示例（在unix中），

scala> val output = "date".!!
output: String =
"Mon May 15 14:52:54 PDT 2017
"

或

scala> val output = "java -version".!!
java version "1.8.0_111"
Java(TM) SE Runtime Environment (build 1.8.0_111-b14)
Java HotSpot(TM) 64-Bit Server VM (build 25.111-b14, mixed mode)
output: String = ""

或，

scala> Seq("echo", "urayagppd") #>> new File("mylogs.log")
res2: scala.sys.process.ProcessBuilder =  ( [echo, urayagppd] #| /Users/prayagupd/myrepo/mylogs.log )

但要运行某些互动过程，例如ssh或其他内容，

scala> val scalaProcess = Process("""scala""")
scalaProcess: scala.sys.process.ProcessBuilder = [scala]

scala> val exitCode = scalaProcess.!
Welcome to Scala 2.12.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_111).
Type in expressions for evaluation. Or try :help.

scala> :quit
exitCode: Int = 1

scala> val p = scalaProcess.run
p: scala.sys.process.Process = scala.sys.process.ProcessImpl$SimpleProcess@af9dd34

scala> Welcome to Scala 2.12.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_111).
Type in expressions for evaluation. Or try :help.

scala> :quit

scala> val exitCode = p.exitValue
exitCode: Int = 1

java.io.IOException：无法运行程序“scala”：CreateProcess error = 2，

1 个答案: