Oh, I forgot to say. For while I thought I'd make hijacker a decorator: @hijacker def my_view(request): ....
But that wouldn't work, since some views need to do some processing before calling the hijacker. I think a decorator is systematically called before the view itself so no prior processing can be done by the view... On Mar 24, 11:28 pm, Julien <[EMAIL PROTECTED]> 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 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
