P
Pete Emerson
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.
Thanks in advance,
Pete
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.
Thanks in advance,
Pete