When the definition is method(self, **kwargs), you ARE asking
variable named kwargs. If you want to accept one dict object,
it should be: method(self, mydict). If you want to accept variable
list of args, it should be method(self, *mylist)
-ak
On Wednesday, April 17, 2013 12:59:21 AM UTC-4, jayh
I think this might be a useful article for you
http://agiliq.com/blog/2012/06/understanding-args-and-kwargs/
On 17 April 2013 15:45, Brad Pitcher wrote:
> I see what you're saying. Maybe you're looking for something more like
> this?
>
> def method1(self, param1=sensible_default, param2=sensibl
I see what you're saying. Maybe you're looking for something more like this?
def method1(self, param1=sensible_default, param2=sensible_default,
param3=sensible_default):
print param1, param2, param3
The variable names must be specified in the method arguments like above if
you wish to refere
but if I did that then i would have a variable named kwargs.
and it would be print kwargs['param1'], etc
Why even unpack the dictionary? I could just pass the dictionary instead.
On Tuesday, April 16, 2013 11:08:42 PM UTC-4, Brad Pitcher wrote:
>
> It should be:
> model.method1(**params)
> On
It should be:
model.method1(**params)
On Apr 16, 2013 7:49 PM, "jayhalleaux" wrote:
> Not really a Django question but I'm trying to create a model method that
> does not have a fixed set of parameters.
>
> models.py
> Class Model_A(model.Model):
> ...
>
> def method1(self, **kwargs):
>
> print p
Not really a Django question but I'm trying to create a model method that
does not have a fixed set of parameters.
models.py
Class Model_A(model.Model):
...
def method1(self, **kwargs):
print param1, param2, param3
views.py
params = dict(
param1=something1,
param2=something2,
param3=somethi
6 matches
Mail list logo
