CODEWITHSUNDEEP
rubysymlink
grautur
grautur

Reputation: 30495

After symlinking a file, how do I get the path of the original file in Ruby?

I have a Ruby script with path /foo/bar/gazook/script.rb. I also created a symlink to it in $HOME/bin.

Now, I want my Ruby script to access some other file in directory /foo, and to keep paths relative, I have a variable FOO_DIRECTORY = File.expand_path(File.dirname(__FILE__) + "/../../") in my script.

The problem is that if I run my script from its symlink, this relative directory is wrong (since I guess its expanding from a different location).

How do I fix this? Is there a way besides using an absolute path?

Upvotes: 15

Views: 4390

Answers (4)

mu is too short
mu is too short

Reputation: 434765

You can use File.readlink to resolve a symlink but you'll want to check File.symlink? first.

path = File.symlink?(__FILE__) ? File.readlink(__FILE__) : __FILE__

Then you can work with path instead of __FILE__. You might want to use $0 instead of __FILE__ as well, __FILE__ is the current filename whereas $0 is the name of the current script.

Upvotes: 15

hagello
hagello

Reputation: 3255

To get any path relative to the location of your script, always use __dir__.

__dir__ is a concise way of saying File.dirname(File.realpath(__FILE__)). It's available in Ruby >= 2.0. On __dir__.

File.realpath(__FILE__) (or Pathname#realpath) has three advantages compared to File.readlink:

  • It expands symlinks anywhere in the path. readlink only expands paths that are the last part of the argument.
  • It (recursively) resolves symlinks to symlinks to... readlink resolves only the first level.
  • You do not have to check whether path is a symlink at all. Thus you can drop the if File.symlink?.

Consequently it would be good to use FOO_DIRECTORY = File.join(__dir__, '..', '..') or FOO_DIRECTORY = File.dirname(File.dirname(__dir__))

Upvotes: 7

neoneye
neoneye

Reputation: 52221

Try this

require 'pathname'
p File.dirname(Pathname.new(__FILE__).realpath)

Upvotes: 2

Marek Příhoda
Marek Příhoda

Reputation: 11198

Try this:

FOO_DIRECTORY = File.expand_path("../../../", __FILE__)

I's say the problem is that your symlink file is interpreting File.dirname(__FILE__)

Upvotes: -2

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