swift - 从 MacOS App Swift 启动 sudo 命令

Question

我需要向 xcode 启动终端命令。这是命令:

sudo xattr -d -r com.test.exemple /Desktop/file.extension

我试过了

   let task = Process()
        task.launchPath = "/usr/sbin/xattr"
        task.arguments = ["-d","-r", "com.test.exemple"," /Desktop/file.extension"]
        let pipe = Pipe()
        task.standardOutput = pipe
        task.standardError = pipe
        task.launch()
        task.waitUntilExit()
        let data = pipe.fileHandleForReading.readDataToEndOfFile()
        let output : String = NSString(data: data, encoding: String.Encoding.utf8.rawValue) as! String
        print(output)

Answer 1

这是一种在命令之间使用管道的方法。我验证了当我在注释掉的行中使用参数时，该文件是由 super 用户创建的。

它的作用是这样的:

echo 'password' | sudo -S /usr/bin/xattr -d -r com.test.exemple /Desktop/file.extension

func doTask(_ password:String) {
    let taskOne = Process()
    taskOne.launchPath = "/bin/echo"
    taskOne.arguments = [password]

    let taskTwo = Process()
    taskTwo.launchPath = "/usr/bin/sudo"
    taskTwo.arguments = ["-S", "/usr/bin/xattr", "-d", "-r", "com.test.exemple", " /Desktop/file.extension"]
    //taskTwo.arguments = ["-S", "/usr/bin/touch", "/tmp/foo.bar.baz"]

    let pipeBetween:Pipe = Pipe()
    taskOne.standardOutput = pipeBetween
    taskTwo.standardInput = pipeBetween

    let pipeToMe = Pipe()
    taskTwo.standardOutput = pipeToMe
    taskTwo.standardError = pipeToMe

    taskOne.launch()
    taskTwo.launch()

    let data = pipeToMe.fileHandleForReading.readDataToEndOfFile()
    let output : String = NSString(data: data, encoding: String.Encoding.utf8.rawValue) as! String
    print(output)
}