imp module
Magnus Heino
magnus.heino at rivermen.se
Fri Sep 29 01:20:03 EDT 2000
In article <970165987.458767 at rexx.com>, "Dave Kuhlman" <dkuhlman at rexx.com>
wrote:
>
> Does the following do what you want?
>
> >> foo = 'aaa'
> >> bar = 'bbb'
> >> cmd = 'import %s.%s' % (foo, bar)
> >> exec(cmd)
Yes. However, I realized after a couple of hours that I was doing it right all
the time ;-)
m below is bar imported, ready to use, like;
>> m.methodname()
>>> However, since I dont know the names foo and bar, I cant do it that way.
>>> Therefore I had a look at the imp module, and tried something like this:
>>>
>>> import imp fp, pathname, desc = imp.find_module('bar' ['foo'])
>>>
>>> m = imp.load_module('bar', fp, pathname, desc)
>>>
>>> What should I now do with m to get foo.bar into the current namespace?
/Magnus Heino
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