How to pass SSH flag -i for id_rsa

[chetankatke]

🤔

[gufastian]

Same for port.
Can't get it work :(

[kylejeske]

Take a look at Line31

and at Line111

sshCmdStr = fmt.Sprintf(
"ssh -tt -q -L %v %v %v 'cd %v; %v --host 127.0.0.1 --allow-http --no-auth --port=%v'", 
localPort + ":localhost:" + localPort, *sshFlags, host, dir, codeServerPath, localPort,
)

sshCmd = exec.Command("sh", "-c", sshCmdStr)

The whole thing just wraps SSH on the command line .. so based on the code above. The second %v allows for SSH parameters.

Using that .. it could look something like ..
Line 36

Usage: [-skipsync] %v HOST [DIR] [SSH ARGS...]
sshcode 10.0.1.10 ~/mydirectory '-i ~/.ssh/id_rsa'

[gufastian]

I tried passing parameters with quotes and without quotes, but the result is the same. Script tries to connect only to port 22.

MacBook:Downloads $ ./sshcode-darwin-amd64 [email protected] ~/projects '-p 1234'
2019-04-22 19:28:25 INFO ensuring code-server is updated...
ssh: connect to host xxx.xxx.com port 22: Operation timed out
2019-04-22 19:29:41 FATAL failed to update code-server: exit status 255

[SscSPs]

Currently, I don't think they are supported, but I think you can do all that by using ~/.ssh/config file to save your SSH targets

[ammario]

Sorry, the command's help is currently out-of-date with the flags. Going to fix that today in #37

sshcode --ssh-flags "-p 1234" [email protected] should work

[ammario]

Similarly, sshcode --ssh-flags '-i .ssh/id_rsa' [email protected] should work.

Feel free to re-open this issue if that's still not working for you.