Using Windows, I've experienced a slight annoyance when using __FILE__ to get the current location of a file or the absolute path of another file with
File.expand_path("lib/other", File.dirname(__FILE__))
This doesn't work though if the folder has characters like äöüè and similar. This get's especially annoying if the windows username of a client contains such a character and my script necessarily lives inside the %appdata% folder.
To demonstrate my problem, C:\äüé\test.rb contains only
puts __FILE__
Running it:
> ruby C:\äüé\test.rb
C:/"?'/test.rb
Is there a reliable way to get the current file path?
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 am trying to write some data in one Ruby file to a file in another folder but I am having trouble identifying the path to get to the file I want to write to.
My current code is:
File.write('../csv_fixtures/loans.csv', 'test worked!')
And my folder structure is as follows:
Where I am trying to run my code in 'run_spec.rb' and write to 'loans.csv'.
Additionally, this is the error I am getting:
Give the path relative to the working directory, not the file that you call File.write from. The working directory is the place you've navigated to through cd before calling the ruby code. If you ran rspec from the root of your project, then the working directory will also be the root. So, in this case, it looks like it would be ./spec/csv_fixtures/loans.csv. You can run puts Dir.pwd to see the working directory that all paths should be relative to.
If you wanted to something more like require_relative, you have to use some sort of workaround to turn it into an absolute path, such as File.dirname(__FILE__) which gives the absolute path of the folder containing the current file:
path = File.join(File.dirname(__FILE__, "../csv_fixtures/loans.csv"))
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 './'?".
So I'm making a game with Ruby/Gosu and the lines to load all the images look like this:
#image_name = Gosu::Image.new(self, 'C:\Users\Carlos\Desktop\gamefolder\assets\bg.jpg', false)
I want to refer to them based on their location relative to the referring file. The file which includes the above line is in C:\Users\Carlos\Desktop\gamefolder\, so I would think I could just change the above to '\assets\bg.jpg' or 'assets\bg.jpg', but this doesn't work.
The specific error is "Could not load image assets/bg.jpg using either GDI+ or FreeImage: Unknown Error (Runtime Error)."
If you want to get the current directory (of your execution context, not necessarily the file you're 'in'), just use Dir.pwd. Output this to console to check that your current directory is actually gamefolder.
To get the current directory of your actual ruby file (relative to Dir.pwd), use __FILE__, e.g.
File.dirname(__FILE__)
Pass that to File.expand_path to get a fully-qualified path. You can do a little sanity check by making sure File.exists?("#{File.expand_path File.dirname __FILE__}/assets/bg.jpg") returns true.
(Try File.expand_path('assets/bg.jpg')...that might be all you need here.)
Ok so with siriproxy it my lib folder along with the rb file for the plugin I have created a myconfig.yml file so I can change certain settings by writing to that file.
I have been able to write to the file but only if I include the full path all the way from the home directory down.
But is there not a way to open the file from the same directory i am in? I have tried every path combination I can think of.
There has to be one i am missing
If you use the following in your ruby file, you should get the absolute path where it is
File.expand_path(__FILE__)
From doc __FILE__
The name of the file currently being executed, including path relative to the directory where the application was started up (or the current directory, if it has been changed)
From doc File.expand_path
Converts a pathname to an absolute pathname.
As you probably want the directory, you should use File.dirname(__FILE__), so the path of your file myconfig.yml should be obtained with
File.join(File.expand_path(File.dirname(__FILE__)), 'myconfig.yml')
In more recent Ruby (>=2.0.0), you can use __dir__ (from Archonic's comment):
Returns the canonicalized absolute path of the directory of the file from which this method is called. It means symlinks in the path is resolved. If FILE is nil, it returns nil. The return value equals to File.dirname(File.realpath(FILE)).