On Fri, Mar 5, 2010 at 12:17 PM, Chris Rebert <c...@rebertia.com> wrote: > On 3/5/10, Pete Emerson <pemer...@gmail.com> wrote: >> In a module, how do I create a conditional that will do something >> based on whether or not another module has been loaded? >> >> Suppose I have the following: >> >> import foo >> import foobar >> >> print foo() >> print foobar() >> >> ########### foo.py >> def foo: >> return 'foo' >> >> ########### foobar.py >> def foobar: >> if foo.has_been_loaded(): # This is not right! >> return foo() + 'bar' # This might need to be foo.foo() ? >> else: >> return 'bar' >> >> If someone is using foo module, I want to take advantage of its >> features and use it in foobar, otherwise, I want to do something else. >> In other words, I don't want to create a dependency of foobar on foo. >> >> My failed search for solving this makes me wonder if I'm approaching >> this all wrong. > > Just try importing foo, and then catch the exception if it's not installed. > > #foobar.py > try: > import foo > except ImportError: > FOO_PRESENT = False > else: > FOO_PRESENT = True > > if FOO_PRESENT: > def foobar(): > return foo.foo() + 'bar' > else: > def foobar(): > return 'bar' > > > You could alternately do the `if FOO_PRESENT` check inside the > function body rather than defining separate versions of the function. > > Cheers, > Chris > -- > http://blog.rebertia.com >
Except I want to use the module only if the main program is using it too, not just if it's available for use. I think that I found a way in my follow-up post to my own message, but not sure it's the best way or conventional. Pete -- http://mail.python.org/mailman/listinfo/python-list
