On Tue, 13 Nov 2012 07:54:32 +0000, Some Developer wrote:
> On 13/11/2012 07:19, Steven D'Aprano wrote:
>> On Tue, 13 Nov 2012 06:38:31 +0000, Some Developer wrote:
>>
>>> I'm trying to find a way to get a string of the module path of a
>>> class.
>>>
>>> So for instance say I have class Foo and it is in a module called
>>> my.module. I want to be able to get a string that is equal to this:
>>> "my.module.Foo". I'm aware of the __repr__ method but it does not do
>>> what I want it to do in this case.
>>>
>>> Can anyone offer any advice at all?
>> py> from multiprocessing.pool import Pool py> repr(Pool)
>> "<class 'multiprocessing.pool.Pool'>"
>>
>> Seems pretty close to what you ask for. You can either pull that string
>> apart:
>>
>> py> s = repr(Pool)
>> py> start = s.find("'")
>> py> end = s.rfind("'")
>> py> s[start+1:end]
>> 'multiprocessing.pool.Pool'
>>
>> or you can construct it yourself:
>>
>> py> Pool.__module__ + '.' + Pool.__name__ 'multiprocessing.pool.Pool'
>>
>>
> Yeah I considered doing it this way but was wary of that method because
> of possible changes to the implementation of the __repr__ method in the
> upstream code. If the Django developers don't consider the __repr__
> method a public API then it could change in the future breaking my code.
I didn't call SomeClass.__repr__. That is an implementation detail of
SomeClass, and could change.
I called repr(SomeClass), which calls the *metaclass* __repr__. That is
less likely to change, although not impossible.
If you're worried, just use the second way:
SomeClass.__module__ + '.' + SomeClass.__name__
> Of course this might not happen but I was hoping that there was a more
> generic way of doing it that did not rely on a certain implementation
> being in existence.
SomeClass.__name__ is the official way to get the name of a class;
SomeClass.__module__ is the official way to get the name of the module or
package it comes from.
--
Steven
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