This might seem a little confusing, but I have a server I want to SSH into but the password is kinda long and complex, I understand for security I shouldn't save my password in plain text but for my convenience, I'm not too worried. Regardless, I'm trying to make it so I can just start up a Linux distro in WSL that automatically connects me to my SSH server and logs in but I'm having trouble. My settings.json block looks kinda like:
{
"guid": "{46ca431a-3a87-5fb3-83cd-11ececc031d2}",
"hidden": false,
"name": "SSH",
"source": "Windows.Terminal.Wsl",
"commandline": "/usr/bin/sshpass -p 'password' ssh -o StrictHostKeyChecking=no user@host"
},
then when I start that distro I get:
[error 0x80070002 when launching `/usr/bin/sshpass -p 'password' ssh -o StrictHostKeyChecking=no user@host']
Is there another way to do this? And yes I have sshpass installed on literally all my distros just to be 100% sure. I googled around for an sshpass Windows version but can't find it. I've also tried just using sshpass
instead of /usr/bin/sshpass
but it doesn't work either.
As far as I know error 0x80070002
is a file not found error, but I don't know where the command isn't being found. Does the "commandline"
setting get loaded before even the Linux Kernel? Is there a way to launch the command AFTER Linux initializes?
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