[Fab-user] out: bash: sudo: command not found

[Ruslan Valiyev]

Hey guys,

I'm having some hard time with several Solaris hosts. For some reason, my function is not able to find the sudo binary.

This is the function:

def task():

  set_users('ldap')

  run('echo $PATH')

  run('type sudo')

  sudo('id')

This is my output:

$ fab --show=debug task -H 10.200.8.160

Using fabfile '/root/fab/configs/fabfile.py'

Commands to run: task

Parallel tasks now using pool size of 1

[10.200.8.160] Executing task 'task'

[10.200.8.160] run: /bin/bash -l -c "echo \$PATH"

[10.200.8.160] out: /usr/bin:/bin:/usr/sbin:/usr/sfw/bin:/opt/sfw/bin:/usr/sfw/sbin:/sbin:/bin

[10.200.8.160] out: 

[10.200.8.160] run: /bin/bash -l -c "type sudo"

[10.200.8.160] out: sudo is /opt/sfw/bin/sudo

[10.200.8.160] out: 

[10.200.8.160] sudo: sudo -S -p 'sudo password:'  /bin/bash -l -c "id"

[10.200.8.160] out: bash: sudo: command not found

[10.200.8.160] out: 

Fatal error: sudo() received nonzero return code 127 while executing!

Requested: id

Executed: sudo -S -p 'sudo password:'  /bin/bash -l -c "id"

None

Aborting.

Disconnecting from 10.200.8.160... done.

I'm guessing there's an issue with $PATH somewhere, but I just ran two commands before invoking sudo and "type" found it.

Regards,

rv