I have created an .erlang file to specify my working directory and to load certain modules at startup. I used the answer from here and it worked well when all the modules to be loaded resided in the current working directory:
shell_default:cd("C:/Users/Some/Default/Working Directory/ebin").
LP = fun() ->
[code:ensure_loaded(list_to_atom(filename:rootname(filename:basename(F))))
|| P <- code:get_path(), F <- filelib:wildcard(P ++ "/*.beam")]
end.
spawn(LP).
When I try to load modules from other directories, I ran into some trouble. I changed the .erlang file:
shell_default:cd("C:/Users/Some/Default/Working Directory/ebin").
code:add_patha("C:/Users/Some/Other/File Location/ebin").
LP = fun() ->
[code:ensure_loaded(list_to_atom(filename:rootname(filename:basename(F))))
|| P <- code:get_path(), F <- filelib:wildcard(P ++ "/*.beam")]
end.
spawn(LP).
For some reason, the modules from the added code path are not loaded (I checked that the path is indeed added to the code path). When I tested filelib:wildcard(code:get_path() ++ "/*.beam") separately with the desired code path, I saw that it returns an empty string (i.e. it does not seem to find the .beam files in the added code path).
Any ideas of how to solve this problem? Thanks!
Related
I have a directory and inside of the directory are subdirectories and a python file trying to run other files called Main.py. In each of the subdirectories there will be one file called runner.py with its only function I need to execute is “func”. If I want to use importlib.importmodule how would I go about doing that. As of right now I’m iterating over the directory and using:
filepath = sys.argv\[1\] + "/runner.py"
student_module = importlib.import_module("func", filepath)
But I’m afraid I don’t know how to use it properly as I’m new to python.
In case my words weren’t clear the directory architecture is as follows
Directory
->Main.py
->Subdirectory(1)
->runner.py
->Subdirectory(2)
->runner.py
.
.
.
->Subdirectory(n)
->runner.py
I had tried using the snippet of code above and I was expecting to be able to run the func from the different runner.py files. Unfortunately I’m getting a:
ModuleNotFoundError: No module named ‘func’
I'm trying to create a python program that allows you to dynamically choose the folder to save the download.
'outtmpl' lets you choose an output as a simple path such that
'e:/python/downloadedsongs/%(title)s.%(ext)s'
is a valid path and will let you save to the downloadedsongs folder
But With the function that defines the choose folder button
def openLocation():
global Folder_Name
Folder_Name = filedialog.askdirectory()
print(Folder_Name)
why does
f'{Folder_name}/%(title)s.%(ext)s'
not return a valid path, the program outright ignores the variable. I thought at first i was having a fstring issue because %s was pythons old format for f'strings but I have also tried these variances with no success
'outtmpl': Folder_Name + '/' + '%(title)s.%(ext)s', (no errors but cannot find output)
'outtmpl': '%(Folder_Name)s%(title)s.%(ext)s',
%()s is the old version of fstrings I was conflating the two.
answer ended up looking like 'outtmpl': Folder + '/%(title)s.%(ext)s'
I tried to import System.Directory in my Frege program (In Eclipse) in order to use functions as getDirectoryContent, etc., and it writes me this error :
Could not import module frege.system.Directory (java.lang.ClassNotFoundException: frege.system.Directory)
What do I have to do ?
It is because the module frege.system.Directory doesn't exist in Frege. A good way to find out about a module is to use Hoogle for Frege at this URL: http://hoogle.haskell.org:8081. If we search for that module there, we can see that it doesn't list any module as opposed to, say, if you search for frege.data.List, we would see the module in the result.
Now for the functions you need like getDirectoryContent, if you look at the search result for frege.system.Directory, the first result is about processes and the third and fourth results are about jars and zip files. If you click on the second result, it would open the module frege.java.IO and you can see some relevant functions that might be useful for you (list for example). However the Haskell module you are trying to find is not yet ported to Frege but it should, of course, be possible to port that module backed by native Java implementations.
Update for OP's comment
Here is a simple snippet to return the files under a given directory:
ls :: String -> IO [String]
ls dir = do
contents <- File.new dir >>= _.list
maybe (return []) (JArray.fold (flip (:)) []) contents
Regarding createTempFile, the following works for me:
frege> File.createTempFile "test.txt"
String -> STMutable RealWorld File
How can one use code in a LiveScript file from another LS file? For example:
# In script-one.ls
foo = 5
# In script-two.ls
bar = -> foo + 3
Simply including both files in the HTML via script tags does not seem to work. Changing the first script to export foo = 5 and using require! './script-one' (or variants) in the second script doesn't work either.
And what about circular dependencies?
LiveScript simply compiles to javascript. The module format is your decision just like in JS.
The export keyword simply compiles to a commonjs exports.foo = right now and will not work in browsers without using something like browserify (http://browserify.org/) to bundle your modules (ES6 compat is planned in the future).
Say I have a project, whose folder structure looks like below:
| main.lua
|
|---<model> // this is a folder
| |a.lua
| |b.lua
|
|---<view>
|a.lua
|b.lua
model/a.lua requries model/b.lua: require "b"
view/a.lua requries view/b.lua: require "b"
main.lua requries files in model and view.
Now I have problem to get these modules loaded correctly. I know I can fix it by changing the require calls to:
model/a.lua: require "model.b"
view/a.lua: require "view.b"
But if I do that, I have to modify these files every time when I change the folder structure.
So my questions are:
How to fix the module path issue without hard code paths in module files?
Why Lua doesn't use the module search rule of Node.js, which looks easier?
When you require a module, the string parameter from require gets passed into the module which you can access using the variable-argument syntax .... You can use this to include other dependent modules which reside in the same path as the current module being requireed without making it dependent on a fixed hard-coded module name.
For your example, instead of doing:
-- model/a.lua
require "model.b"
and
-- view/a.lua
require "view.b"
You can do:
-- model/a.lua
local thispath = select('1', ...):match(".+%.") or ""
require(thispath.."b")
and
-- view/a.lua
local thispath = select('1', ...):match(".+%.") or ""
require(thispath.."b")
Now if you change directory structure, eg. move view to something like control/subcontrol/foobar, then control/subcontrol/foobar/a.lua (formerly view/a.lua) will now try to require control/subcontrol/foobar/b.lua instead and "do the right thing".
Of course main.lua will still need to fully qualify the paths since you need some way to disambiguate between model/a.lua and view/a.lua.
How to fix the module path issue without hard code paths in module files?
I don't have any better cross-platform solution, maybe you should plan the folder structure early on.
Why Lua doesn't use the module search rule of Node.js, which looks easier?
Because Lua tries its best to rely only on ANSI C, which is really successful. And in ANSI C, there's no such concept of directories.
There are a couple approaches you can use.
You can add relative paths to package.path as in this SO answer. In your case you'd want to add paths in main.lua that correspond to the various ways you might access the files. This keeps all the changes required when changing your directory structure local to one file.
You can add absolute paths to package.pathusing debug.getinfo -- this may be a little easier since you don't need to account for all the relative accesses, but you still need to do this in main.lua when changing your directory structure, and you need to do string manipulation on the value returned by debug.getinfo to strip the module name and add the subdirectory names.
> lunit = require "lunit"
> info = debug.getinfo(lunit.run, "S")
> =info.source
#/usr/local/share/lua/5.2/lunit.lua
> =info.short_src
/usr/local/share/lua/5.2/lunit.lua
The solution is to add the folder of main.lua (project root) to package.path in main.lua.
A naive way to support folders of 1 level deep:
-- main.lua
package.path = package.path .. ";../?.lua"
Note for requires in (project root) will look up files outside of project root, which is not desirable.
A better way of to use some library (e.g.: paths, penlight) to resolve the absolute path and add it instead:
-- main.lua
local projectRoot = lib.abspath(".")
package.path = package.path .. ";" .. projectRoot .. "/?.lua"
Then in you source use the folder name to scope the files:
-- model/a.lua
require "model.b"
-- you can even do this
require "view.b"
and
-- view/a.lua
require "view.b"