在Swift脚本中使用管道链接shell命令

我试图在Swift脚本中将shell命令链接在一起.有问题的实际命令是gource管道输出的输出ffmpeg,但这里是一个简化的,人为的例子,我正在尝试做什么:

let echo = Process()

echo.launchPath = "/usr/bin/env"
echo.arguments = ["echo", "foo\nbar\nbaz\nbaz", "|", "uniq"]

let pipe = Pipe()
echo.standardOutput = pipe

echo.launch()
echo.waitUntilExit()

// Get the data
let data = pipe.fileHandleForReading.readDataToEndOfFile()
let output = NSString(data: data, encoding: String.Encoding.utf8.rawValue)

print(output ?? "no output")

预期产量:

foo
bar
baz

实际产量:

foo
bar
baz
baz | uniq

将|被解释为最后一个参数的一部分.如何将命令链接在一起,以便数据在Swift脚本中从一个流向另一个？我尝试了在两个es 之间分配standardIn和standardOut使用的各种组合,但要么我做错了,要么我没有将正确的部分连接在一起.Pipe()Process

我在zadr的帮助下得到了答案:

import Foundation

let pipe = Pipe()

let echo = Process()
echo.launchPath = "/usr/bin/env"
echo.arguments = ["echo", "foo\nbar\nbaz\nbaz"]
echo.standardOutput = pipe

let uniq = Process()
uniq.launchPath = "/usr/bin/env"
uniq.arguments = ["uniq"]
uniq.standardInput = pipe

let out = Pipe()
uniq.standardOutput = out

echo.launch()
uniq.launch()
uniq.waitUntilExit()

let data = out.fileHandleForReading.readDataToEndOfFile()
let output = NSString(data: data, encoding: String.Encoding.utf8.rawValue)

print(output ?? "no output")