Question

使用scala 2.9 process API时，我可以执行

之类的操作

"ls -l"!

将把进程stdout和stderr发送到我自己的进程中。或者：

val output = "ls -1"!!

将把发送到stdout的内容返回到val输出。

我怎样才能同样抓住stderr？

Answer 1

您可以创建自己的ProcessLogger：

import sys.process._

val logger = ProcessLogger(
    (o: String) => println("out " + o),
    (e: String) => println("err " + e))

scala> "ls" ! logger
out bin
out doc
out lib
out meta
out misc
out src
res15: Int = 0

scala> "ls -e" ! logger
err ls: invalid option -- e
err Try `ls --help' for more information.
res16: Int = 2

编辑：上一个示例只是打印，但它可以轻松地将输出存储在某个结构中：

val out = new StringBuilder
val err = new StringBuilder

val logger = ProcessLogger(
    (o: String) => out.append(o),
    (e: String) => err.append(e))

scala> "ls" ! logger
res22: Int = 0

scala> out
res23: StringBuilder = bindoclibmetamiscsrc

scala> "ls -e" ! logger
res27: Int = 2

scala> out
res28: StringBuilder =

scala> err
res29: StringBuilder = ls: invalid option -- eTry `ls --help' for more information.

在scala中抓取进程stderr

1 个答案: