如何快速运行shell命令? [英] How do I run shell command in swift?

问题描述

我正在做一个macOS应用，试图在子进程中运行shell命令.如果我未将launchPath设置为/usr/bin/env，则会收到错误Couldn't posix_spawn: error 13，为什么会这样?如何在其他路径中运行shell命令?

I'm doing a macOS app, trying to run shell command in child process. I'll get an error Couldn't posix_spawn: error 13 if I don't set launchPath to /usr/bin/env, why is it like this? How can I run shell command in other path?

class Helper {
    static func shell(launchPath path: String, arguments args: [String]) -> String {
        let task = Process()
        task.launchPath = path
        task.arguments = args

        let pipe = Pipe()
        task.standardOutput = pipe
        task.standardError = pipe
        task.launch()

        let data = pipe.fileHandleForReading.readDataToEndOfFile()
        let output = String(data: data, encoding: .utf8)
        task.waitUntilExit()

        return(output!)
    }
}

let res = Helper.shell(launchPath: "/Users/myUserName", arguments: ["ls"]) //error

推荐答案

可以访问/Users/myName/myWorkspace之类的路径， 但首先，您需要禁用应用程序沙箱:

It's possible to access to a path like /Users/myName/myWorkspace, but you have first, to disable the app sandbox doing so:

如果要在自定义目录中运行ls，则可以尝试以下示例:

If you want to run ls in a custom directory, you might try this example:

let res = Helper.shell(launchPath: "/bin/ls", arguments: ["/Users/myUserName/myworkspace"])
print("*** ls ***:\n\(res)")

在我的情况下，我有以下输出:

in my case I have the following output:

*** ls ***:
file1.txt
file2.txt
file3.txt

这篇关于如何快速运行shell命令?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！