Re: lambda callable in foreignkey default

For the record I figured it out but it had nothing to do with the lambda:

in my template I have a something like:

{{ my_project_lookup_class.classname.get_verbose_name }}

lookup_class.classname will the return the django model class, and the template engine apparently checks if the class is callable, then calls it which will try to instanciate a new object with no arguments. This in turns calls the default lambda.

I'm not quite sure if this is a bug or a feature to be honest, will post on django-dev to check.

On Wednesday, November 27, 2013 5:46:54 PM UTC+1, ke1g wrote:

Probably not it, but try:

  ...(RestaurantType, default = lambda: (RestaurantType.objects.all()[0]))

And note that your alternate will fail if RestaurantType object with id==1 is ever deleted.

On Tue, Nov 26, 2013 at 7:31 AM, Philippe Raoult <Philipp...@gmail.com> wrote:

I ran into a strange issue today when trying to use a lambda function for the default value of a ForeignKey.

my original code was this, and I expected it would look for RestaurantType instances whenever I would try to create a Restaurant:
type_of_restaurant = models.ForeignKey(RestaurantType, default = lambda: RestaurantType.objects.all()[0])

but to my surprise it would run the query when the models where loaded!

I changed to this and it behaved as I expected:
type_of_restaurant = models.ForeignKey(RestaurantType, default = lambda: RestaurantType.objects.get(id = 1))

Anyone could explain me why ? Is there something counter-intuitive with the evaluation order of the brackets or something ?

Regards,
Philippe

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