Function to Open Arbitrary Program from WSL terminal

Question

I'm having trouble writing a program that takes the /path/to/file in an WSL Ubuntu environment and opens that file in its default Windows program. The biggest issue here is converting the /path/to/file/for/WSL into a path\to\file\for\windows. This is what I wrote and put in my .bashrc file:

# Function to get open-wsl to work  
  function open-from-wsl() {  
          echo "opening"  
          cmd_directory = echo "$1" | sed 's/^\///' | sed 's/\//\\/g' | sed 's/^...../\0C:/' | sed 's/^\mnt//'  
          cmd_directory = echo "$cmd_directory" | sed 's/^..//'  
          cmd.exe /C start $cmd_directory  
  }

Ideally what this is supposed to accomplish is if I type :

open-from-wsl /mnt/c/Users/DavidG/Google\ Drive/folder/file.PDF

I'll have my file.PDF open in my default PDF viewer. As of right now however, I get the error cmd_directory: command not found and then my command prompt window opens up. This of course is being written so that I can open any file from WSL and have it open in it's default program, it doesn't just pertain to PDFs.

EDIT: I've adjusted the code like so thanks to Socowi's input:

# Function to get open-wsl to work  
  function open-from-wsl() {  
          echo "opening"
          cmd_directory=$(echo "$1" | sed 's#\##')  
          cmd_directory=$(echo "$cmd_directory" | sed 's/^\///' | sed 's/\//\\/g' | sed 's/^...../\0C:/' | sed 's/^\mnt//')  
          cmd_directory=$(echo "$cmd_directory" | sed 's/^..//')  
          cmd.exe /C start $cmd_directory  
  }

My issue now seems to be that I can't remove the \ from /mnt/c/Users/DavidG/Google\ Drive/folder/file.PDF, which was my goal by adding the line

cmd_directory=$(echo "$1" | sed 's#\##')

Remove the spaces around = and encapsulate echo ... | sed ... | ... in $(). — Socowi
– Socowi, Commented Mar 14, 2019 at 17:37
I was under the impression that wslpath only took Windows paths and converted them to a Ubuntu path — David G.
– David G., Commented Mar 14, 2019 at 21:31

Damo · Accepted Answer · 2019-12-13 12:17:00Z

1

A much simpler solution is to create a file in /usr/bin/ called v (v for view, i like short commands, but you can call it anything you like) containing the following:

cmd.exe /C start $(wslpath -w "$(pwd)/$1")

Now assuming /usr/bin is in the wsl path just call v foo.txt to open a file in its default windows editor.

wslpath is used to convert the wsl path to a dos path. This is now included in the wsl install (link)

answered Dec 13, 2019 at 12:17

Damo

6,6235 gold badges46 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

David G. · Accepted Answer · 2019-03-15 02:12:33Z

After searching among various questions pertaining to adjacent-esque issues. I finally found the solution. This is the function that I ended up putting into my .bashrc file:

# Function to get open an arbitrary file in its default Windows program
function open-from-wsl() {
     echo "opening"
     cmd_directory=$(echo "$1" | sed 's#[\]##')
     cmd_directory=$(echo "$cmd_directory" | sed 's/^\///' | sed 's/\//\\/g' | sed 's/^...../\0C:/' | sed 's/^\mnt//')
     cmd_directory=$(echo "$cmd_directory" | sed 's/^..//')
     echo "$cmd_directory"
     cmd.exe /C start ""  "$cmd_directory"
}

This will allow you to open any file in your computer (assuming you know the WSL path to it) in it's default Windows program.

Collectives™ on Stack Overflow

Function to Open Arbitrary Program from WSL terminal

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related