I have pyenv installed, however, it does not do its most basic function, namely switch Python versions. The following terminal commands demonstrate this.
the file `main.py` is equivalent to:
import sys
print (sys.version)
Admins-MacBook-Pro-4:kylefoley kylefoley$ pyenv versions
system
* 2.7.14 (set by PYENV_VERSION environment variable)
3.5.3
3.6.1
3.7.3
pypy3.6-7.1.1
Admins-MacBook-Pro-4:kylefoley kylefoley$ pyenv global 3.5.3
Admins-MacBook-Pro-4:kylefoley kylefoley$ pyenv exec python main.py
2.7.14 (default, Oct 17 2019, 00:01:43)
As you can see when I run main.py the version that comes out is 2.7. A lot of people have this problem. One common solution is putting
eval "$(pyenv init -)"
On the bash_profile which I have done and that did not help. Over here
Cannot switch Python with pyenv
it is recommended:
Put the PATH and shell environment vars into your .bash_profile (or whatever file your distro uses).
But what PATH and what shell environment vars is he talking about?
Also my .bashrc file looks like this:
export PATH="/Users/kylefoley/.pyenv/bin:$PATH"
eval "$(pyenv init -)"
eval "$(pyenv virtualenv-init -)"
Any help would be appreciated. One other things, when I run the following commands, I get the following output:
Admins-MacBook-Pro-4:kylefoley kylefoley$ python
Python 3.6.1rc1 (default, Mar 4 2017, 22:58:58)
The problem is that .bashrc is not sourced in a non-login mode.
Init files for Bash:
login mode:
/etc/profile
~/.bash_profile, ~/.bash_login, ~/.profile (only first one that exists)
interactive non-login:
/etc/bash.bashrc (some Linux; not on Mac OS X)
~/.bashrc
non-interactive:
source file in $BASH_ENV
And on macOS, the default Bash shell opened by a terminal app is a interactive login shell, but on Linux, the default shell opened by a terminal app is a interactive non-login shell.
Solution
The weird interactive, non-login loading requirement confuses people in other situations as well. The best solution is to change the loading requirement of ~/.bashrc as interactive only, which is exactly most of the Linux distros are doing.
# write content below into ~/.bash_profile
# if running bash
if [ -n "$BASH_VERSION" ]; then
# include .bashrc if it exists
if [ -f "$HOME/.bashrc" ]; then
. "$HOME/.bashrc"
fi
fi
This should be the solution you desire. And I recommend every Bash user setup this in the profile.
References
Unix shell initialization
Everything was done correctly, it's just that I thought the terminal command . ~/.bash_profile updates the bash profile without having to close the terminal or open a new one in order for changes to take effect. It turns out that . ~/.bash_profile only updates some of the bash_profile. After restarting the terminal, everything was working as planned.
Just add to your .bashrc or similar file, the line.
eval "$(pyenv init --path)"
After the "export PATH=$PYENV..." . Don't forget to reset your terminal after try again!
Worked on Fedora & Mint.
Just add this into you .bashrc file:
export PYENV_ROOT="$HOME/.pyenv"
export PATH="$PYENV_ROOT/bin:$PYENV_ROOT/shims:$PATH"
If your .bashrc is sourced from ~/.bash_profile you are done.
Official docs https://github.com/pyenv/pyenv#advanced-configuration suggests puting into .bashrc:
eval "$(pyenv init -)"
which was not working for me.
1.) configure:
pyenv global [python version]
2.)restart terminal (close all terminal windows)
Related
I'm using VSCode and the official remote-ssh extension.
I would like to be able to write code /path/to/file in an ssh terminal or in the vscode integrated terminal in a remote window in order to open a file/folder in vscode remote.
I am aware that I can use code --folder-uri=vscode-remote://ssh-remote+ADDRESS/path/to/file from the local machine's terminal, but I want to be able to run a command from within the integrated vscode terminal and any other terminal session where I've ssh'd into the remote machine)
Currently, if I run code from a remote terminal it opens up a new vscode window on the remote machine.
To achieve this goal, in the past I've used the following alias on the remote machine:
alias code="${VSCODE_GIT_ASKPASS_NODE%/*}/bin/code"
Which looks for the code executable in ~/.vscode-server/bin/<COMMIT_ID>/bin before defaulting to the local /bin/code.
I got that alias from this related stackoverflow question.
However, this doesn't seem to work right now.
Upon closer inspection, it appears that there is no code executable in the vscode-server directory.
How can I fix this?
Both machines are running MacOS and visual studio code version f80445acd5a3dadef24aa209168452a3d97cc326, if that's relevant.
I also wanted to be able to run code from the integrated terminal when running VSCode with the "remote ssh" extension. In my case, the "remote" is a Linux box (named "aorus" below), and I want to use VSCode from a laptop running macOS (named "mbp").
As for you, I used to use the VSCODE_GIT_ASKPASS_NODE trick. Recently, I had to change the alias since code (or code-insiders in my case) wasn't available in bin/ anymore. It seems it has been moved to bin/remote-cli. The correct alias (tested with vscode 1.64.2):
alias code="${VSCODE_GIT_ASKPASS_NODE%/*}/bin/remote-cli/code"
If you also want this to work from other ssh sessions (not just inside the integrated terminal), you can create a short script that I called coder (r for "remote") which I have in ~/bin on my remote ("aorus"). Note that you need to be able to reach the local machine from your remote (I do that with Tailscale). The script looks like this:
#! /bin/bash
set -ex
remotehost=$(hostname)
localhost=mbp
cmd="code"
while [ $# -gt 0 ]; do
if [ -f "$1" ]; then
cmd+=" --file-uri \"vscode-remote://ssh-remote+$remotehost$(readlink -f "$1")\""
elif [ -d "$1" ]; then
cmd+=" --folder-uri \"vscode-remote://ssh-remote+$remotehost$(readlink -f "$1")\""
else
cmd+=" $1"
fi
shift
done
exec ssh $localhost -q -t -x "exec bash -l -c '$cmd'"
On my Mac, when running VSCode connected remotely to my Linux box, I can type this in the integrated terminal to open the file main.go present on my remote Linux box:
coder main.go
The reason I have to wrap code in bash -l is due to the fact that ssh, by default, runs in a non-login shell, which means that the ~/.bashrc on my Mac isn't picked up, meaning code isn't in the PATH. The error message looks like this:
bash:1: command not found: code
Another note: there is a shorter syntax documented here:
ssh -q -t -x mbp bash -l -c "code --remote=ssh-remote+aorus main.go"
I don't use this syntax is because this method isn't able to know whether you are opening just a single file (which should be open in the most recent VSCode remote session) or a folder (which should be open as a new VSCode remote session).
Finally, if you are using VSCode Insiders, you can create a symlink so that the command code works on your local machine (in my case, on my Mac):
sudo ln -sf /usr/local/bin/code-insiders /usr/local/bin/code
As already explained by maelvls the path has been changed.
But if you use it outside integrated terminal you will got message
Command is only available in WSL or inside a Visual Studio Code terminal
To avoid this you need to export VSCODE_IPC_HOOK_CLI in your .bashrc .
Use this script in your .bashrc
export VSCODE_IPC_HOOK_CLI=`ls -t /run/user/1012/vscode-ipc-* | head -n1`
alias code="~/.vscode-server/bin/*/bin/remote-cli/code"
If you want to open your file in your current visual studio use -r option.
code -r tes.txt
Note :
I can't call VSCODE_GIT_ASKPASS_NODE so I use full path, it is working well
I don't know if VSCODE_IPC_HOOK_CLI will show in different location, just check it in your integrated terminal visual studio code
tested on remote server Centos 7
local macOS Monterey version 12.2
Visual Studio Code Version: 1.64.2 (Universal)
Commit: f80445acd5a3dadef24aa209168452a3d97cc326
extension : remote-ssh
So I have two WSLs (version 2). Kali and Ubuntu. The code command works in Kali but Ubuntu says Command not found. Since it works in WSL, CMD and PowerShell, it is there in PATH variables. Any help?
edit: Ubuntu doesn't read VS Code in PATH variables, while Kali does. I opened the Environment Variables Wizard. VS Code is there.
In a "normal" case, like you are seeing in your Kali instance, WSL's init appends the Windows path to your Linux path. There are two things that I can think of that would cause that to not be happening correctly under your Ubuntu instance:
You or a script that you ran in Ubuntu modified your shell's startup files (assuming Bash, typically ~/.bashrc or ~/.profile) to edit the PATH, possibly removing the Windows elements, or at least the VSCode element.
Try running that instance without any Bash startup files (from PowerShell or CMD) with:
PS> wsl -d Ubuntu -e bash --noprofile --norc
and then try code .. This, of course, assumes that your Ubuntu instance is named Ubuntu. If it's named something else (wsl -l) then edit the -d Ubuntu as needed. You might also try launching it as wsl -d Ubuntu -e sh -c 'code .'
If that works, examine your startup files for any modifications to PATH.
Less likely (because you should know if you did this one), is that you can disable the WSL feature that appends the Windows PATH in a particular distribution/instance. Check for the existence of a /etc/wsl.conf file -- If it exists, the line appendWindowsPath=false would cause the behavior that you are seeing as well. Simply change the offending setting to true. Exit your Ubuntu instance, run wsl --terminate Ubuntu (again, substituting the correct distro name) and then restart. Check your PATH and try code . again then.
I am writing a perl script, and I want to run a simple shell command to use a certain version of NVM:
Here is my code snippet:
print "\n*** Switching to correct nvm environment for dashboard builds\n";
system("nvm use 8.12.0") == 0 or die $?;
But I am getting the following error:
Can't exec "nvm": No such file or directory
Can someone help?
Update (June 30, 2021):
I also tried adding the command:
my $nvm_version = "8.12.0";
system ("bash", "-lic", "nvm use $nvm_version");
But nothing happens:
I'm not familiar with nwm, but I think I get the gist of what it does. And if so, the attempt is fundamentally flawed. Even if you fixed this to run the proper shell so that nvm could run, I believe all the tool does is change the shell's environment variables, a shell you immediately exit. This means it would have no effect even if if it ran successfully.
Again, it this tool does what I think it does, such tool are meant to be used in interactive shells. In other instances, you simply use the path the to correct executable instead of relying on the PATH.
With that in mind, you can use the following to run the command in bash:
# Non-interactive shell.
system("bash", "-c", "nvm use 8.12.0")
or
# Interactive shell.
# This is improper and fragile as interactive shells
# often create aliases that override basic commands.
system("bash", "-ic", "nvm use 8.12.0")
Just to reiterate, at least one of these will allow the command to run (if it normally works from bash), but I believe it's unlikely this will produce the results you expect.
The nvm command is shell function which is different from a shell command. Also the nvm command is not an exported function so it will not be seen by sub shells. For example, in Bash shell:
$ nvm ls
-> v15.0.1
$ my-test-script.sh
./my-test-script.sh: line 3: nvm: command not found
where my-test-script.sh is:
#! /bin/bash
nvm use 16.4
The error nvm: command not found is because nvm is not exported. I can source the script in the current shell context to make it work:
$ source my-test-script.sh
Now using node v16.4.0 (npm v7.18.1)
$ node --version
v16.4.0
So a Perl script cannot change the node version of the current shell, but it can calculate the version and pass it back to shell, which can set the version. For example:
$ nvm use $(perl -E'$v=15.0; print $v')
Now using node v15.0.1 (npm v7.0.3)
I have the case that I need to use VS Code installed on Windows 10 and run it with the extension Remote - SSH on a RHEL 7.x.
The default RHEL 7.x runs with git 1.8.x. I have installed a newer git version but this is not in the default $PATH environment.
I have found this instructions https://code.visualstudio.com/docs/remote/wsl#_advanced-environment-setup-script which describe how to set the environment variables specifically for VS Code when usind WSL.
If you want to run additional commands or modify the environment this can be done in a setup script ~/.vscode-server/server-env-setup
This does seem to work only if you use WSL. Why does this not work with the Remote - SSH extension?
My special case is that I only want and need the git>=2 while usind VS Code. When I am connected regularly via ssh I would like and need the OS default tools and settings.
This gives me the special request that I don't want to edit the ~/.bashrc, ~/.cshrc or any other user environment files.
I would like to be able to edit the environment for VS Code only. Some kind, maybe like:
#!/bin/bash
export PATH=/opt/rh/rh-git29/root/usr/bin\:$PATH
export LD_LIBRARY_PATH=/opt/rh/httpd24/root/usr/lib64:$LD_LIBRARY_PATH
...
#!/bin/csh
setenv PATH /opt/rh/rh-git29/root/usr/bin\:$PATH
setenv LD_LIBRARY_PATH /opt/rh/httpd24/root/usr/lib64:$LD_LIBRARY_PATH
...
Is there anything I have not found yet where I can make my requests to work or would this be some kind of request to the VS Code Team?
Regards.
I think I found the solution in this issue comment and the follow-up response:
When vscode-server initially starts, it uses a login shell, sourcing .profile in your home directory.
However, any following interactive shells started through VS Code are non-login shells and thus only source .bashrc
A complication in fiddling with this is that vscode-server apparently caches the environment during its lifetimes, so changes to these dotfiles don't become visible until the server is restarted.
I have a better solution to minimize the proxy scope
export http_proxy=<proxy here>
export no_proxy=<no proxy here>
while IFS= read -r _file; do
if ! grep -s -q "export http_proxy=" "${_file}"; then
sed -i -e "/^ROOT/i export http_proxy=${http_proxy}" -e "/^ROOT/i export https_proxy=${http_proxy}" -e "/^ROOT/i export no_proxy=${no_proxy}" "${_file}"
fi
done < <(find ~/.vscode-server/bin -type f -name "server.sh")
When I try to execute code from WSL, like this:
cmd.exe
wsl code .
I get the following error:
/bin/bash: code: command not found
However, if I execute the same command but first run cmd.exe as an Administrator, vscode opens as expected.
The question is why "code" is not recognizable when I run as non-admin, and how can I solve it?
Additional info:
WSL2 version 41959
vscode version 1.38
vscode Remote WSL extension version 0.39.5
As #Biswapriyo mentioned, this is an open bug where WSL cannot access Windows C drive.
Workaround that helped me is to restart WSL like this:
wsl --shutdown
wsl
Open the PowerShell administrator window
wsl --list --version
Show as wsl 2
Enter the Linux console
wsl
Change related files' rights
chmod u+x vscode_dir/code.exe
chmod u+x vscode_dir/bin/code
Create symbolic link
ln -s vscode_dir/bin/code code
mv code ~/.local/bin
Modify .bashrc, add a line as follows:
export PATH="$HOME/.local/bin:$PATH"
source .bashrc or restart terminal.
Then you can happily play code in wsl2-ubuntu environment or terminal of VSCode.
My problem was that I was using the root user after running sudo -i. Exiting and using my normal user on WSL solved the error.
In my case I am using Debian and echo %PATH% did not output anything.
I used the second part of suggested solution from resolved github issue
For Debian, /etc/profile contributed to this problem.
Here is the path definition in /etc/profile
if [ "`id -u`" -eq 0 ]; then
PATH="/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin"
else
PATH="/usr/local/bin:/usr/bin:/bin:/usr/local/games:/usr/games"
fi
Option 1:
You can delete above lines, then wsl --shutdown to restart Debian.
Option 2:
If you would like to keep these lines, you can also append ":$PATH" to each path like below, then wsl --shutdown
if [ "`id -u`" -eq 0 ]; then
PATH="/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:$PATH"
else
PATH="/usr/local/bin:/usr/bin:/bin:/usr/local/games:/usr/games:$PATH"
fi
Thank you to licanchua
I'd recommend checking /etc/wsl.conf to see if appendWindowsPath setting shares the Windows PATH with WSL, and also if it is specifically setting a particular user.
For the config settings, see https://learn.microsoft.com/en-us/windows/wsl/wsl-config