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_______________________________________________
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