No output when using 'service service-name status' on a stopped service about spur.py

[mwilliamson]

What happens if you try using shell_type=spur.ssh.ShellTypes.minimal in the call to SshShell()?

[kandarp1691]

I used shell_type=spur.ssh.ShellTypes.minimal and added the following line in my code:
except spur.RunProcessError as e: print e.stderr_output
On debugging, it shows
e: return code: 3\noutput: b'apache-tomcat is stopped\n'\nstderr output: b''

However, I am not sure how to get it as an output in the terminal.

[mwilliamson]

Ah, it's raising an exception because of the return code -- was another part of your code swallowing the exception?

In any case, if you want to always return a result (rather than raising) regardless of the return code, use allow_error=True in the call to run().

[kandarp1691]

Runs perfectly now. Thanks Mike !

P.S. What do you mean by swallowing an exception. I'm 2 months old to this and would definitely like to learn more about it. Can you point me to a link/page I could refer to get in depth and avoid such mistakes ?

Thanks again !

[mwilliamson]

Glad you got it working.

The code you originally was raising an exception. Normally, unless you have an except block, Python should exit with the exception printed out. For instance:

$ python -c 'raise Exception()'
Traceback (most recent call last):
  File "<string>", line 1, in <module>
Exception

I was curious whether you saw such a problem, and if not, why not.

[mwilliamson]

In fact, there's a try in your code snippet, so being able to see the rest of the actual code would probably help.

[kandarp1691]

This is my code:

def heathcheck(username, password, server_address, service_name):
    try:
        shell = spur.SshShell(hostname=server_address,
                              username=username,
                              password=password, missing_host_key=spur.ssh.MissingHostKey.accept)
        with shell:
            result = shell.run(['service', service_name, 'status'], allow_error=True)
            if result.output.__contains__('stopped'):
                print result.output
                result_new = shell.run(['service', service_name, 'start'])
                return result_new.output + 'on' + server_address
            else:
                return result.output + 'on' + server_address

    except spur.RunProcessError as e:
        print e.stderr_output
    except IOError as e:
        print e.message
    except Exception as e:
        failure_list.append(e.message + ' ' + server_address)
        print e.message

[mwilliamson]

Presumably that's the new code though, given that allow_error=True is in there? I would guess the problem you had before was that the output is in .output rather than stderror_output when you were catching the RunProcessError, which is why you weren't seeing any output, but I could be mistaken.

By "swallowing an exception", I mean behaviour where an exception is caught, but not handled, for instance:

try:
    something_that_might_raise_an_exception()
except Exception as error:
    pass

I've no idea what your code is meant to do, so can't comment on whether just printing the message is appropriate, but I generally prefer to either allow the exception (wrapped, if appropriate), or return an object that contains the error, so the caller can decide how best to handle the problem.

On an unrelated note: you shouldn't normally need to call magic methods such as __contains__ directly. In this case, "stopped" in result.output would be idiomatic. Similarly, I think the convention is to call str(error) rather than error.message for exceptions (I think message is deprecated in Python 2 and removed in Python 3).