How would I, say, determine if the file ~/.my_proj_config exists on any OS in Ruby?
A call to Dir.home is a OS independent way to get to the home directory for the user. You can then use it like
File.exists?(File.join(Dir.home, ".my_proj_config"))
This works in Ruby 1.9, but note that the call to expand_path is required on some systems (e.g. Windows):
File.exists?( File.expand_path "~/.my_proj_config" )
Use the class File and its method exist?.
Take a look at the Pathname class, specifically the realpath function - This will get you the full (expanded) path to your file.
http://www.ruby-doc.org/stdlib/libdoc/pathname/rdoc/classes/Pathname.html#M001991
You then use the File class along with exists? method to find out if that exists. You shouldn't need to use realpath if you use this method, however.
http://www.ruby-doc.org/core/classes/File.html#M000045
Related
I'm using Lua in interactive mode on a Mac (thanks to rudix.org).
When I want to load a file I do:
dofile("/my/long/path/to/my/directory/file.lua")
I want to do a different thing, that is:
put all my files in a desktop directory myDirectory;
then call the file from the terminal this way dofile("file.lua");
Is this possible? How?
If the path is fixed, you can just redefine dofile:
local _dofile=dofile
local path=("/my/long/path/to/my/directory/")
function dofile(x)
return _dofile(path..x)
end
You may put this (and other initializations) in a file and set the environment variable LUA_INIT to its location. After this, every invocation of lua will see the version of dofile redefined above and the users will be able to say simply dofile("foo.lua").
Alternatively, you can use require, which looks for modules in a list of paths in package.path or LUA_PATH.
I have a JAR with a bunch of configs. I'd like to send them to the correct directory without cd'ing there.
Something like jar xf config.jar --MAGIC-PARAM PATH/TO/DIRECTORY
Is there such a thing? If it helps, this will be called by a Buildr extension (Ruby).
From the API documentation: http://buildr.apache.org/rdoc/classes/Buildr/Unzip.html
unzip(dir => zip_file).target.invoke
Alex's answer is good. If there's some special magic that jar xf does that makes you prefer it to unzipping (I'm not aware of any), here's another option:
FileUtils.cd('PATH/TO/DIRECTORY') do
system("jar xf '#{_('config.jar')'")
end
It does involve cd'ing, but when you use cd with a block, the original directory is restored after the block. You will need to use either an absolute path or a path relative to the directory you changed to; I'm using buildr's _ method to get an absolute path for a project-relative file.
So currently I have included the following in my .bashrc file.
export RUBYLIB=/home/git/project/app/helpers
I am trying to run rspec with a spec that has
require 'output_helper'
This file is in the helpers directory. My question is that when I change the export line to:
export RUBYLIB=/home/git/project/
It no longer finds the helper file. I thought that ruby should search the entire path I supply, and not just the outermost directory supplied? Is this the correct way to think about it? And if not, how can I make it so RUBY will search through all subdirectories and their subdirectories, etc?
Thanks,
Robin
Similar to PATH, you need to explicitly name the directory under which to look for libraries. However, this will not include any child directories within, so you will need to list any child sub-directories as well, delimiting them with a colon.
For example:
export RUBYLIB=/home/git/project:/home/git/project/app/helpers
As buruzaemon mentions, Ruby does not search subdirectories, so you need to include all the directories you want in your search path. However, what you probably want to do is:
require 'app/helpers/output_helper'
This way you aren't depending on the RUBYLIB environment variable being set a certain way. When you're deploying code to production, or collaborating with others, these little dependencies can make for annoying debugging sessions.
Also as a side note, you can specify . as a search path, rather than using machine-specific absolute paths.
I was trying to get path current path in PHP. I tried looking though phpinfo();, but I haven't found any interesting values which could be used to get path to my script. There is no nice values which I used on Linux, like $_SERVER["PWD"].
Now I'm wondering how I'm supposed to find current path. Maybe some function will work... I really have no idea. Because I don't want to hardcode path to script.
getcwd() is what you are looking for.
It's not entirely clear whether you mean the current working directory, or the path to the current script. For the working directory, see #Taze's answer.
For the current script, the __FILE__ magic constant will give you the full filesystem path to the current file.
Note that these constants take "current file" literally: If you include a file and call __FILE__ there, it will show the included file's path.
The getcwd() method will return the current working directory on success.
<?php
echo getcwd() . "\n";
?>
I got the filename like this:
_TCHAR filename[_MAX_PATH];
GetModuleFileName(NULL,filename,sizeof(filename));
How do I remove the filename from this full path? Should I use regex?
You can use the Windows shell API function PathRemoveFileSpec to do this. Example usage is listed on the linked page.
Since you use VS++, you can use:
_splitpath and _wsplitpath functions to break apart path