I'm trying to figure out how to get the relative file path starting from the workspace folder. So for example if the workspace folder is called My App and the file path is to the current folder is
file:///c%3A/Users/Bob/Documents/My%20App/backend/connections/server.js
Then I want to be able to store in a variable only backend/connections/server.js so it would exclude the workspace folder with the %20 in it. Currently
vscode.window.activeTextEditor.document.uri;
outputs
file:///c%3A/Users/Bob/Documents/My%20App/backend/connections/server.js.
Is there a method I can call that I'm not using or an easy way to do this without having to do some regex expression?
The short answer: no
Visual Studio Code will does not pass any awareness of itself into the code it hosts. While you can access the $workspaceFolder variable inside Code's launch.json and tasks.json, your code itself doesn't have access to those variables.
And even if it could, as soon as you would move your code outside of your development environment, it would break, as that $workspaceFolder would no longer be there.
If you want to make provisions for doing special stuff while in a dev environment, you'll have to define what you consider the workspace folder to be - there is no way for you to get that dynamically.
Related
I am looking to build a VSCode Extension around a CLI tool which we have been working on. An example command would be
myCLI retrieve SourceName
This would be run from a specific directory (for example c:/workspace/myproject) which has been setup and contains a settings.json file for some config arguments.
This CLI has been designed that the methods which are called (for example 'retrieve') are exposed directly so the CLI itself is a wrapper also.
When trying to call these methods directly from a VS Code Extension, it is always checking in the C:/Program Files/Microsoft VS Code directory, which I understand is where the Extension is excuting from.
Now, the question: Is there any way for me to force that any time we call the method (for example 'retrieve') that this would look into the current workspace folder (C:/workspace/myProject) , and not the VS Code one (C:/Program Files/Microsoft VS Code)?
Notes which may change answers
CommonJS (not yet ESM)
We currently cannot pass in a full qualified path (for example C:/workspace/myProject), it is only looking for ./settings.json since it depends on where the CLI has bene run from
I want to avoid calling the CLI directly, as I would like to bring many of the CLI features into the VS Code Extension directly to improve user friendliness.
Is it possible to have vscode automatically open the newest folder within a specific path?
For example, with this configuration:
{
"folders": [
{
"path": "\\\\FromABC\\Archive",
"name":"From ABC"
},
{
"path": "\\\\FromXYZ\\Archive",
"name":"From XYZ"
}
]
}
I would expect these folders in the workspace to be pointing to \07\07 because those were created today:
\\\\FromABC\\Archive\\2021\\07\\07
\\\\FromXYZ\\Archive\\2021\\07\\07
Is it possible to create a workspace where the folders are opened to the latest folder within each configured path?
There's not enough information in the original question to fully answer it, however, I can suggest a few avenues of attack
Custom Command (error-prone and picky)
Modify Upstream Process (likely the best overall)
Combining Both (perhaps the best for your immediate case)
Creating a Custom Command
Create a new command per https://code.visualstudio.com/api/extension-guides/command#creating-new-commands
VSCode Commands Listing: https://code.visualstudio.com/api/references/commands
new command
detect latest folder through whatever logic you like
call vscode.openFolder to navigate to it
call your custom command through Activation Events (activationEvents) at either onStartupFinished or * (Start Up; less-preferable, but may be required to avoid confusing the editor)
https://code.visualstudio.com/api/references/activation-events#Start-up
Check out Start app when opening project in VS Code? for a few answers related to this
Modifying the Upstream Process
Cutting the gordian knot, it's likely some process (perhaps a human) is creating the directories for you
Change the upstream process so when it creates the directories, it also creates/updates a link to the directory labeled something like latest
/FromABC/Archive/2021/06/03
/FromABC/Archive/2021/07/05
/FromABC/Archive/2021/07/07
/FromABC/Archive/latest --> /FromABC/Archive/2021/07/07
/FromXYZ/Archive/2020/04/12
/FromXYZ/Archive/2021/08/18
/FromXYZ/Archive/latest --> /FromXYZ/Archive/2021/08/18
Then you can always refer to the latest directory and it will always be correct
This is quite common when something can change frequently, but another process is expecting a static path and/or has no way to know the schedule of change
I don't have any Windows systems to try this out with and you may be able to create a regular shortcut for this .. however, you may need a Junction (soft-link) or Hard Link to convince VSCode that the directory is a real directory
https://learn.microsoft.com/en-us/windows/win32/fileio/hard-links-and-junctions
This also provides an opportunity to include more files, such a beta versions of some software, which it's desirable to package into the same directory structure, but not truly the latest stable!
Combining Both
If your upstream process is either not modifiable (or some manual process it's annoying or error-prone to add extras steps to) you can likely combine both solutions to get what you really want
Use the * Action Event to call a script to detect and create the new directory - create a binary or PowerShell script to make your link
In this and with the upstream change, just point VS Code to the latest directory and it shouldn't mind
Not sure on which platform you are, I assume windows, but essentially similar.
Instead of trying to get VSCode to open the latest folder, I would create a script that updates a softlink for each folder to the latest subfolder in it. Then you can point VSCode to the softlink, which can be updated whenever needed to the latest subfolders.
I'm programming a client side applications using SharePoint Designer 2013.
I want to change to VSCODE since it supports a lot of extensions for some Javascript library like angular, jQuery. And because of the Chrome/Node.js debugger extension.
But when I try to start any Debugger, I got the error:
Unable to create 'launch.json' file inside the '.vscode' folder (Error: UNKNOWN: unknown error, mkdir '\\servername\DavWWWRoot\sitename\Style Library\.vscode').
I get this error because it's impossible to create a folder in SharePoint where the name starts with dot.
So there's a possibility to change the name of this folder or the file location to any directory in my local computer?
No, it's not possible to move/rename that folder. VS code is a tool that bases project management on folder content. So it is essential that the project settings reside in the folder being managed.
You can move the "extensions" folder, but unfortunately not the argv.json (so the ".vscode" will, at least be recreated on vscode launch)
https://github.com/microsoft/vscode/issues/17691#issuecomment-559234574
I hope that'll finally change sometime .
https://github.com/microsoft/vscode/issues/3884
https://github.com/OmniSharp/omnisharp-roslyn/issues/953
I'm automating my build process, but I wasn't able to change the model_target_rtw folder to something different.
I'm not talking about CodegenFolder, but about the folder that's created inside it during compilation.
I'm currently working this around by renaming the folder after compilation, but it would be grate to remove that step.
The folder you are referring to is the RTW (Real Time Workshop) BuildDirectory.
You can get the value of BuildDirectory by running the command:
RTW.getBuildDir('MyModel')
See:
https://se.mathworks.com/matlabcentral/answers/274082-how-can-i-change-the-build-folder-of-a-model
Also look at this question:
Save generated code in a special folder in "rtwbuild"
If you run this command in MATLAB:
set_param(0, 'CodeGenFolder', 'C:\MyBuildDir')
and then run the RTW.getBuildDir command again you will see that the BuildDirectory has changed.
I want to run my working pydev project python code by double clicking the main module (outside of eclipse): xxx.py
The problem is that due to my imports being in different packages:
from src.apackage.amodule import obj
when xxx.py is double clicked it complains it doesn't know where the imports are (even though when I run xxx.py in pydev it magically knows what I'm importing).
A simple workaround is to remove all of the packages and move all of the modules into one directory (that obviously works but is very inconvenient)
How can I run my code in the file system without doing that work around?
This page answers my question excellently:
http://blog.habnab.it/blog/2013/07/21/python-packages-and-you/
Bottom line is always execute your code from the top, highest level, root directory (e.g. using a minimal main.py file that executes the main script of your program). Then, use absolute imports always and you never have a missing module issue since you start the program from the top directory and all imports are based off that 'home' path.
The problem you encountered is the natural behavior of most languages. A programm only knows about its working path (the path it is started in), the paths which are registered in the environment variables and at least relative paths.
The "magic" of the executable you created is therefore: It collects all scripts/modules needed, and copies/combines them next to/in the executable. The Executable then runs within the directory where all other scripts also reside and voila ...
If you are not happy with your workaround of creating an executable every time you want to run your project without PyDev there are two alternatives.
First but not the one I would suggest is registering the working path into in the environment variables.
Second and the one I think is much better: Create a link to the python executable and alter the calling string of the textfield "Target:". Append the path to your script you would like to run. Then alter the textfield "Start in:" and enter the project directory. After you did this you should be able to start your project with a simple double click.
(If you rely on external libraries which are neither on the path nor in you project you could search for appending paths temporarily to the pythonpath via the sys module.)
I hope I could help a bit.