Since you are modifying my_view to add the call to hijacker anyway, why
not do this?
def hijacker(request):
if I_feel_its_necessary:
return HttpResponse(some_json_content)
def my_view(request):
hret = hijacker(request)
if hret: return hret
... # Continue as normal
return some_html_content
It's one more line than you are looking for, but is simple, and everyone
will understand it, and you don't have to find something complicated: it
works.
--Ned.
http://nedbatchelder.com/blog
Julien wrote:
> In fact, what I want is to hijack a view:
>
> def hijacker(request):
> if I_feel_its_necessary:
> hijack_caller_view_and_return(some_json_content)
>
> def my_view(request):
> hijacker(request)
> ... # Continue as normal
> return some_html_content
>
> I have some long views which I'd like to modify as little as possible
> and insert some extra behaviour handled by 'hijacker'. Adding just a
> call to 'hijacker' to all of these views would be ideal for me because
> that would be quick to modify them and also easier to maintain.
> I don't want the views to care about what 'hijacker' will do. If
> 'hijacker' thinks it's necessary, it would force returning a certain
> value (some json content in this case), otherwise it would let the
> caller view do its normal business.
>
> Any idea on how to do that, if that's even possible?
>
> Thanks!
>
> Julien
>
>
> On Mar 24, 10:35 pm, Ned Batchelder <[EMAIL PROTECTED]> wrote:
>
>> I don't know how to do this in Python, if you even can. Inspect may be
>> just the dark scary corner to explore to find this capability if it
>> exists. But why would you want to do something like this? Even if
>> inspect could do this, it looks like called would have to have some very
>> specific knowledge of caller's structure to make it work.
>>
>> Much much better would be to get caller and called to cooperate, and
>> have called simply return a value that caller would check. Or for the
>> two to be based on a template pattern (template as in the Gang of Four
>> book, not as in a Django template).
>>
>> Maybe if you told us a bit more about the problem, we could help you
>> find a solution.
>>
>> --Ned.http://nedbatchelder.com/blog
>>
>>
>>
>> Julien wrote:
>>
>>> Hi all,
>>>
>>> This is more a Python issue than a Django one, but I thought the
>>> Python-aware people that you are might be able to help me :)
>>>
>>> I would like to do something like:
>>>
>>> def called(arg)
>>> if arg==True:
>>> !!magic!!caller.return 1
>>>
>>> def caller(arg)
>>> called(arg)
>>> return 2
>>>
>>> Here, the fake !!!magic!!! represents a statement (which I ignore)
>>> that would make the caller function return a value different from what
>>> it'd return normally.
>>>
>>> For example, caller(True) would return 1, and caller(False) would
>>> return 2. The reason I want that is because I don't want the caller
>>> function to know what's going on in the called function, and be
>>> shortcut if the called function think it's necessary.
>>>
>>> Would you know if that's possible, and if so, how?
>>>
>>> I've done a bit of research and I think I've found some good pointers,
>>> in particular using the 'inspect' library:
>>>
>>> import inspect
>>>
>>> def called(arg)
>>> if arg==True:
>>> caller_frame = inspect.stack()[1]
>>> ...
>>>
>>> Here 'caller_frame' contains the frame of the caller function. Now,
>>> how can I make that frame return a particular value?
>>>
>>> Hope that was clear... :/
>>>
>>> Thanks!
>>>
>>> Julien
>>>
>> --
>> Ned Batchelder,http://nedbatchelder.com
>>
> >
>
>
--
Ned Batchelder, http://nedbatchelder.com
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