从 OS X 文件引用 URL 别名获取路径 (file:///.file/id=...)

Question

In PyQt4 on Apple Mac OS X, we get a file reference URL to a drag-and-dropped file in the following format:

file:///.file/id=123456.78901

This, interestingly, is not a valid file URL, or at least it doesn't transparently resolve in most applications — there is no /.file/id=... file in the filesystem.

Using just shell, how do I dereference this NSURL or file reference or alias (or whatever it's called) into a valid path to file on the filesystem?

Answer 1

In Swift you can use URL.standardized to get the valid path from a reference URL. So in Swift 4 - to get the file name from a drag-and-dropped file you'd do something like...

let fileType = NSPasteboard.PasteboardType (kUTTypeFileURL as String)

extension NSPasteboardItem {
    func fileURL ()->URL? {
        if let refURL = self.string(forType: fileType) {
            return URL (fileURLWithPath: refURL).standardized
        }
        return nil
    }
}

Answer 2

One can use osascript , an AppleScript interpreter available on default OS X installs, run with the following script:

osascript -e 'get posix path of posix file "file:///.file/id=123.456" -- kthxbai'

Prints /Users/josh/Downloads/paste-10837081.py .

Answer 3

那是一个 NSURL 对象，从中获取路径：

NSString *filePath = [fileURL path];

Answer 4

This can also be accomplished in Python using the PyObjC bridge.

from Foundation import NSURL # depends on PyObjC

# convert a URL file reference to a POSIX path
NSURL.URLWithString_('file:///.file/id=6571367.8602676692').path()

# convert a path file reference to a POSIX path
NSURL.fileURLWithPath_('/.file/id=6571367.8602676692').path()

Be sure you to select the appropriate NSURL method for your data type or else you'll get the wrong value back.

Resources

• NSURL API doc
• API discussion of file reference URLs
• PyObjC