I understand I can get current directory by
$CurrentDir = Dir.pwd
How about parent directory of current directory?
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File.expand_path("..", Dir.pwd)
August
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Rob Di Marco
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15@Niklas: Or `File.expand_path('..')`, the [default `dir_string`](http://ruby-doc.org/core-1.9.3/File.html#method-c-expand_path) is `'.'`. – mu is too short Dec 28 '11 at 22:07
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@muistooshort: nice to know :) – Niklas B. Dec 28 '11 at 22:21
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6The solution suggested above did not work for me using `ruby 2.1.5`. The following did ... `File.dirname(File.expand_path('..', __FILE__))` – Som Poddar May 10 '15 at 23:23
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I think an even simpler solution is to use File.dirname:
2.3.0 :005 > Dir.pwd
=> "/Users/kbennett/temp"
2.3.0 :006 > File.dirname(Dir.pwd)
=> "/Users/kbennett"
2.3.0 :007 > File.basename(Dir.pwd)
=> "temp"
File.basename returns the component of the path that File.dirname does not.
This, of course, works only if the filespec is absolute and not relative. To be sure to make it absolute one could do this:
2.3.0 :008 > File.expand_path('.')
=> "/Users/kbennett/temp"
2.3.0 :009 > File.dirname(File.expand_path('.'))
=> "/Users/kbennett"
Keith Bennett
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My answer here works, but is not as expressive and clear as the other answers. – Keith Bennett Sep 14 '21 at 21:39
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Perhaps the simplest solution:
puts File.expand_path('../.')
Marek Příhoda
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@muistooshort indeed interesting, and definitly useful to know, thanks ;) – Marek Příhoda Dec 29 '11 at 08:43
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Your answer piqued my curiosity to check the spec for `File.expand_path` so thanks for that. – mu is too short Dec 29 '11 at 09:29
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That's pretty neat. You'd probably almost always need to call `.to_s` on that to convert it to a String (it's a Pathname). – Keith Bennett Sep 14 '21 at 21:41
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1In my experience you rarely need `to_s`, but rather you should embrace Pathname everywhere for file names, rather than String. – akim Sep 15 '21 at 14:45