macos - NSFilehandle swift

Question

我正在尝试将从命令行返回的信息作为字符串输入到我的程序中。这是我的代码，我尝试了一些其他变体但没有成功。

    let MyTask : NSTask = NSTask()
    MyTask.launchPath = "/usr/bin/cal"
    MyTask.arguments = ["04","2014"]
    MyTask.launch()
    MyTask.waitUntilExit()
    var TheHandle = MyTask.standardOutput as  NSFileHandle
    println(TheHandle)
    var Data = TheHandle.readDataToEndOfFile()
    var DataAsString = NSString(data: Data,encoding: NSUTF8StringEncoding)
    println(DataAsString)

输出

     April 2014
Su Mo Tu We Th Fr Sa
1  2  3  4  5
6  7  8  9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30

<_NSStdIOFileHandle: 0x6180000e4f80>
2014-11-14 23:22:40.014 Command Reader[6631:118648] *** -[_NSStdIOFileHandle     readDataOfLength:]: Bad file descriptor

Answer 1

任务的标准输出必须设置为进程可以写入的文件句柄。对于父进程和子进程之间的通信，通常这样做用管。子进程写入管道的写入端，而父进程从读取端读取:

let myTask  = NSTask()

myTask.launchPath = "/usr/bin/cal"
myTask.arguments = ["04","2014"]

let pipe = NSPipe()
myTask.standardOutput = pipe

myTask.launch()
myTask.waitUntilExit()

let theHandle = pipe.fileHandleForReading
let data = theHandle.readDataToEndOfFile()
if let dataAsString = NSString(data: data,encoding: NSUTF8StringEncoding) {
    println(dataAsString)
} else {
    println("could not convert output to string")
}

备注:

• myTask.standardOutput 设置为管道本身。这导致进程写入 pipe.fileHandleForWriting，完成后关闭该文件描述符。如果文件描述符未关闭，则 theHandle.readDataToEndOfFile()在父进程中会无限期地等待。
• 我使用了以小写字母开头的变量，这是 Swift 中的惯例。

如果你想将过程输出写入文件，那么你可以创建文件，将其作为文件句柄打开并将该句柄传递给 myTask.standardOutput:

let myTask  = NSTask()
myTask.launchPath = "/usr/bin/cal"
myTask.arguments = ["04","2014"]

let path = "/path/to/output.txt" // where the output should be written ...
if (!NSFileManager.defaultManager().createFileAtPath(path, contents: nil, attributes: nil)) {
    println("could not create file")
} else if let theHandle = NSFileHandle(forWritingAtPath: path) {
    myTask.standardOutput = theHandle

    myTask.launch()
    myTask.waitUntilExit()
    theHandle.closeFile()
} else {
    println("could not open file")
}