Can I declare a class within a function in Python?

Question

How can I define class Options inside of my CheckForJiraIssueRecord function?

def CheckForJiraIssueRecord(object):
    #sys.stdout = open(os.devnull)
    #sys.stderr = open(os.devnull)

    class Options:
        pass
    options = Options()

    options.user = 'user'
    options.password = 'password'

    try:
        com = jira.Commands()
        logger = jira.setupLogging()
        jira_env = {'home':os.environ['HOME']}
        command_cat= "cat"
        server = "http://jira.server.com:8080/rpc/soap/jirasoapservice-v2?wsdl"
    except Exception, e:
        sys.exit('config error')
    try:
        jira.soap = jira.Client(server)
        jira.start_login(options, jira_env, command_cat, com, logger)
        issue = com.run(command_cat, logger, jira_env, my_args)
    except Exception, e:
        print sys.exit('data error')

if __name__ == '__main__':
    commit_text_verified = verify_commit_text(os.popen('hg tip --template "{desc}"'))
    #commit_text_verified = verify_commit_text(os.popen('hg log -r $1  --template "{desc}"'))
    if (commit_text_verified):
        sys.exit(0)
    else:
        print >> sys.stderr, ('[obey the rules!]')
        sys.exit(1);

Is it possible to declare classes within functions in Python?

Yeah, this is a pretty common pattern when you need the class to be different each time for whatever reason. — Dave
– Dave, Commented Sep 21, 2011 at 21:03
Why do people keep asking "Can I do X?"? I mean, why can't people just try it? If it doesn't work, THEN maybe ask "How do I do X?" or "What can I do instead of X to accomplish my larger goal?". Especially since this is Python, for crying out loud, one of the easiest languages to just try stuff in. — John Y
– John Y, Commented Sep 21, 2011 at 21:50
maybe i should rephase the question, to what are the implictions if i declare a class within a function. @ John Y, you are right about trying, but i actually wanted to know the implications, and if this is indeed the pythonic way, since python does not seem to be like Perl, where the motto is there are many way to do the same thing, right ? — kamal
– kamal, Commented Sep 22, 2011 at 12:56

user1114 · Accepted Answer · 2011-09-21 21:26:32Z

6

Yes, just correct your indentation and that code should work. You'll be creating a new class each time the function is called.

def CheckForJiraIssueRecord(object):
    class Options:
        pass
    options = Options()

    options.user = 'user'
    options.password = 'password'

edited Sep 21, 2011 at 21:26

answered Sep 21, 2011 at 20:58

user1114

1

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Comments

Karl Knechtel · Accepted Answer · 2011-09-21 21:20:02Z

Yes, however:

1) Each time through the function, Options becomes a separate class. Not that you will really be able to write code that exploits (or is broken by) this property, because

2) You can only instantiate the class via the function, unless you explicitly expose it to the global namespace somehow.

3) Assuming that you just need an object with those attributes - i.e. Jira doesn't care about your class interface beyond being able to do .user and .password - you don't really need to create a custom class at all:

from collections import namedtuple
def CheckForJiraIssueRecord(object):
    options = namedtuple('Options', 'user password')('user', 'password')
    # 'user password' are the field names, and ('user', 'password') are the
    # initialization values. This code creates a type similar to what you had
    # before, naming it 'Options' internally, but doesn't bind it to a variable.

    # As you were...

Dietrich Epp · Accepted Answer · 2011-09-21 20:58:45Z

2

Yes, you can. Although each time the function is called, you will get a different class.

answered Sep 21, 2011 at 20:58

Dietrich Epp

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