Question

如何使用Scala进程执行需要用户交互的操作系统命令？

为了说明这一点，请考虑例如Unix / Linux OS中的passwd，如下所示，

import sys.process._
import scala.language.postfixOps

等等

scala> "passwd"!

(current) UNIX password: passwd: Authentication token manipulation error
passwd: password unchanged
Changing password for userabc
res0: Int = 10

Answer 1

您需要使用ProcessIO实例来处理读取和写入stdin / out的过程。

def execute(command: String): Int = {
  def reader(input: java.io.InputStream) = {
    //read here
  }

  def writer(output: java.io.OutputStream) = {
    //write here
  }

  val io = new ProcessIO(writer, reader, err=>err.close())
  Process(command).run(io).exitValue;
}

有些棘手的问题是，当有需要输入的内容时，您可能需要同步代码以写入流，否则您编写的字符可能会丢失。

Answer 2

使用＆＃39;！＆lt;＆＃39;而不是＆＃39;！＆＃39;

"passwd"!<

来自scala.sys.process.ProcessBuilder特质：

/** Starts the process represented by this builder, blocks until it exits, and
  * returns the exit code.  Standard output and error are sent to the console.
  */
def ! : Int

/** Starts the process represented by this builder, blocks until it exits, and
  * returns the exit code.  Standard output and error are sent to the console.
  * The newly started process reads from standard input of the current process.
  */
def !< : Int

在Scala中执行需要用户交互的OS命令

2 个答案: