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.
Related
I'm trying to use Atom to run a Lua script. However, when I try to load files via the require() command, it always says it's unable to locate them. The files are all in the same folder. For example, to load utils.lua I have tried
require 'utils'
require 'utils.lua'
require 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua'
require 'D:\\Users\\Mike\\Dropbox\\Lua Modeling\\utils.lua'
require 'D:/Users/Mike/Dropbox/Lua Modeling/utils.lua'
I get errors like
Lua: D:\Users\Mike\Dropbox\Lua Modeling\main.lua:12: module 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' not found:
no field package.preload['D:\Users\Mike\Dropbox\Lua Modeling\utils.lua']
no file '.\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua\init.lua'
no file 'D:\Program Files (x86)\Lua\5.1\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
The messages says on the first line that 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' was not found, even though that is the full path of the file. What am I doing wrong?
Thanks.
The short answer
You should be able to load utils.lua by using the following code:
require("utils")
And by starting your program from the directory that utils.lua is in:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
The long answer
To understand what is going wrong here, it is helpful to know a little bit about how require works. The first thing that require does is to search for the module in the module path. From Programming in Lua chapter 8.1:
The path used by require is a little different from typical paths. Most programs use paths as a list of directories wherein to search for a given file. However, ANSI C (the abstract platform where Lua runs) does not have the concept of directories. Therefore, the path used by require is a list of patterns, each of them specifying an alternative way to transform a virtual file name (the argument to require) into a real file name. More specifically, each component in the path is a file name containing optional interrogation marks. For each component, require replaces each ? by the virtual file name and checks whether there is a file with that name; if not, it goes to the next component. The components in a path are separated by semicolons (a character seldom used for file names in most operating systems). For instance, if the path is
?;?.lua;c:\windows\?;/usr/local/lua/?/?.lua
then the call require"lili" will try to open the following files:
lili
lili.lua
c:\windows\lili
/usr/local/lua/lili/lili.lua
Judging from your error message, your Lua path seems to be the following:
.\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?\init.lua;D:\Program Files (x86)\Lua\5.1\?.lua
To make that easier to read, here are each the patterns separated by line breaks:
.\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?\init.lua
D:\Program Files (x86)\Lua\5.1\?.lua
From this list you can see that when calling require
Lua fills in the .lua extension for you
Lua fills in the rest of the file path for you
In other words, you should just specify the module name, like this:
require("utils")
Now, Lua also needs to know where the utils.lua file is. The easiest way is to run your program from the D:\Users\Mike\Dropbox\Lua Modeling folder. This means that when you run require("utils"), Lua will expand the first pattern .\?.lua into .\utils.lua, and when it checks that path it will find the utils.lua file in the current directory.
In other words, running your program like this should work:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
An alternative
If you can't (or don't want to) change your working directory to run the program, you can use the LUA_PATH environment variable to add new patterns to the path that require uses to search for modules.
set LUA_PATH=D:\Users\Mike\Dropbox\Lua Modeling\?.lua;%LUA_PATH%;
lua "D:\Users\Mike\Dropbox\Lua Modeling\main.lua"
There is a slight trick to this. If the LUA_PATH environment variable already exists, then this will add your project's folder to the start of it. If LUA_PATH doesn't exist, this will add ;; to the end, which Lua fills in with the default path.
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 created a project in /Projects/test that have the following files:
/Projects/test/first.rb
/Projects/test/second.rb
In first.rb, I do:
load 'second.rb'
And it gets loaded correctly. However, if I open the console and I type $:, I don't see the current directory "." in the load path. How does Ruby know where to load that 'second.rb' from?
See the documentation of Kernel#load clearly :
Loads and executes the Ruby program in the file filename. If the filename does not resolve to an absolute path, the file is searched for in the library directories listed in $:. If the optional wrap parameter is true, the loaded script will be executed under an anonymous module, protecting the calling program’s global namespace. In no circumstance will any local variables in the loaded file be propagated to the loading environment.
In case load 'second.rb' - second.rb has been internally resolved to the absolute path /Projects/test/second.rb,as your requiring file in the directory is same as required file directory. Nothing has been searched to the directories listed in$: for your case.
Just remember another way always
- The load method looks first in the current directory for files
Contrary to the currently accepted answer, the argument 'second.rb' does not resolve to an absolute path. If that were what was meant, you would also be able to require 'second.rb', since require has exactly the same wording about absolute paths.
I think what's happening here is just that the phrasing in the documentation for load is not clear at all about what the actual steps are. When it says "Loads and executes the Ruby program in the file filename," it means that literally — it treats the argument as a file name and attempts to load it as a Ruby program. If isn't an absolute path†, then Ruby goes through $LOAD_PATH and looks for it in those places. If that doesn't turn anything up, then it just goes ahead and tries to open it just as you passed it in. That's the logic that MRI actually follows.
† The actual check that Ruby does is essentially "Does the path start with '/', '~' or './'?".
I have a gem that I have written that has a number of handlers, each of which has their own ruby file in the gem. I need to add the ability to specify a file on the command line that will be loaded in the same manner as these other handlers. The file will typically not be in the default load path of the gem.
I'm not sure the best way to do this. I could take the filename, and then add the containing directory to the load path and then load the file. I could have the user specify a directory containing handlers to be read instead of specifying the file, or I'm sure there are a better way to do it that I haven't yet thought of.
This was fixed using require_relative and expanding the file path using Dir.pwd:
req_path = File.expand_path(arg, Dir.pwd)
require_relative req_path
I have two files with the same name "Foo.rb" in the directories pointed by my RUBYLIB env variable.
This causes the interpreter to fail, when 'require "Foo"' is invoked. It cannot know which to use.
How to solve this without renaming the files ?
What happens if you explicitly specify the path to the file required?
Something like:
require 'lib/the/correct/Foo_folder/Foo'