In my linux python app for Fedora I want to save user's work to /home/user/Documents/MyCoolApp/TheirGreatWork.txt
But I am not sure how to find the "Documents" folder if the user is not using English as their default language.
What is the right way to determine the right path so that files go in their "Documents" folder.
EDIT
Here is a which comes up if you change locales... showing how paths can get easily changed.
I'd use the subprocess module to get the output of the command xdg-user-dir DOCUMENTS. For example:
import subprocess
documents_dir = subprocess.check_output(["xdg-user-dir", "DOCUMENTS"])
print documents_dir # This is what you're looking for.
There is no right way as the user may have changed their locale, which (fortunately) does not rename the directory. If you want a fixed place for files managed by your app, use ~user/.MyCoolApp or let the user specify the directory.
Related
I'm using python to download an application from a content distribution network. The application downloads as self extract file cabinet. When executed it creates a directory using a version naming format. For example app-version2/../app.exe, Thus I cannot rely on the folder name as it may change in the future. I'm trying to find the best way to work with the content inside the folder without depending on the actual folder name.
My idea was to rename the folder using os.listdir() and then os.rename(app-version2, myapp) This would work but is not automated. What would be the best automated method to find a folder name that contains version numbers and change that to something more static?
Assuming you want to find the path of the directory which begins with app, you can accomplish this using pathlib.Path:
from pathlib import Path
app_path = next(Path().glob('app*'))
This will give you the path to the first file or directory in your current directory whose name begins with "app".
TLDR
I need to get paths to system directories like "Screenshots":
On an English system. I can just use this one:
C:/Users/User/Pictures/Screenshots
How do I get the path to "Screenshots" directory on a non-English system?
C:/Users/User/Pictures/[NAME]
Description
I have a file manager app, it displays system directories and loads them on click.
The app can run system commands via Powershell and use Node.js (preferred)
Problem
The problem is, it only works if the system has English system language.
Currently, to resolve the "Screenshots" directory path, the app simply joins the User directory with the word "Screenshots"
const pictures = electronRemote.app.getPath('pictures')
const screenshots = PATH.join(pictures, 'Screenshots')
link to the line in code
Expectedly, the C:/Users/User/Screenshots path only exists on English systems.
One way to solve this is to use short names, at least on Windows, I know that system directories have short names like SCREEN~1 and WALLPA~1 for Screenshots and Wallpapers directories, but if I use these names the paths will look like this:
C:/Users/User/SCREEN~1 instead of C:/Users/User/Screenshots throughout the app.
And even if I were to run these paths through a function to convert it to readable name, how would I know which word to replace it with? I need to get the name in the system's language.
Are these translations stored somewhere on the system? Can I just retrieve the translated directory name and use that in the code above?
Question
How do I make it to get / resolve the actual path of system directories like Screenshots and Wallpapers, independently of system locale?
If you know how to do it, could you please suggest the solution for all platforms (Win, Mac, Linux)?
Should I just use the short names like SCREEN~1 and then automatically replace all the occurrences in UI and also filter all paths through a function that replaces this short name with the actual path throughout the whole app? Seems like a lot of work, this approach
I've been developing an nw.js project and use node.js file system functions in it as normal. In my application there is a file manager and I list folders and files according to user navigation. In Windows, for example, if I scan drive C: I get the Turkish named folder "Kullanıcılar" as "Users". I know it's real name in operating system is "Users" and just seen on the screen according to Languages. I can replace names of such folders when dispaying in my file manager but I'm searching for better solution if exists. Thanks in advance.
There's an SO answer here that reads the localized name of a folder in C# using the SHGetFileInfo function which might help you along.
Now I know you didn't ask, but in case you want to know where the information is stored... It's within the directory, in the Desktop.ini file.
For instance, my Windows 10 installation has this in it for "Users":
[.ShellClassInfo]
LocalizedResourceName=#%SystemRoot%\system32\shell32.dll,-21813
And this for the Images folder within my user folder (bringing this up to show you the additional keys):
[.ShellClassInfo]
LocalizedResourceName=#%SystemRoot%\system32\shell32.dll,-21779
InfoTip=#%SystemRoot%\system32\shell32.dll,-12688
IconResource=%SystemRoot%\system32\imageres.dll,-113
IconFile=%SystemRoot%\system32\shell32.dll
IconIndex=-236
The #%SystemRoot%\system32\shell32.dll,-21813 points to having to read the MUI (multilingual user interface) resources, key 21813 for the given file (presumably the # means that it's in this file, not this literal value, but don't quote me on that). %SystemRoot% is an environment variable that points to the Windows directory.
The actual MUI files and their locations are handled by Windows (see the MSDN link above), but we'll just happen to handily know that the MUI file for the US English localization of shell32.dll is system32\en-US\shell32.dll.mui.
Opening up that file with Resource Hacker, we can search for 21813 -- and voila! We can find STRINGTABLE resource #1364 that contains:
[...snip...]
21812, "Extras and Upgrades"
21813, "Users"
21814, "Saved Games"
[...snip...]
I unfortunately don't have tr-TR/shell32.dll.mui available, so you'll just have to trust me that you'd find the Kullanıcılar string there.
I have created a package and am now creating my tests within the package. For one test my inputs are a set of files and my outputs will be a different set a files created within the test.
I am saving the input files in the test directory of my package and would like to save the output files there too. Since others may run this test, I do not want to specify the input/output file location using my own path eg /home/myname/.julia/v4.0/MyPackage/test/MyInputFile.txt
How do I specify that the input location is within the package's test folder?
So basically how do I tell Julia to look in the packages's folder under the test directory and not have to worry about specifying the entire path including user name etc?
For example currently I have to say
readtable(/home/myname/.julia/v4.0/MyPackage/test/MyInputFile.txt, separator = '\t', header = false)
But I'd like to just be able to say
readtable(/MyPackage/test/MyInputFile.txt, separator = '\t', header = false)
so that no matter who the user of the package is and where they may store the package, they can still run the test?
I know that LOAD_PATH gives the path Julia looks for packages but I can't find any information on where it looks when importing files.
joinpath(Pkg.dir("MyPackage"), "test") is what you need.
As #GnimucK mentioned in a comment, a better solution is
dirname(#__FILE__)
Why is this better? A package could be installed and used from somewhere else (not the standard package directory). Pkg.dir is "stupid" and does not know better. This is rare, of course, and in most cases it won't matter.
I don't want to specify the full directory of a folder or object within my program. I do not want to do this because if a user decides to change the installation folder, it will not function properly. I've seen in HTML you can do something like: ./folder/directory/name and it would work perfectly fine. Is there a way to do something like that within Python?
From https://docs.python.org/3/reference/datamodel.html
__file__ is the pathname of the file from which the module was loaded
You may find it helpful to apply os.path.abspath() to '.' or __file__.