Hi,
I have and existing live project running on rails. I have to expose the
API's using django for now. I have generated models and edited them to
create django model and mapping and exposed some API locally.
I have one issue which is confusing my mind.
That is how would I use the existing use
Okay, user preference aside. Is this something Django couldn't handle at
the moment? If so, it should be submitted as a bug.
On Saturday, November 11, 2017 at 10:22:03 PM UTC+8, mike wrote:
>
> i would never try to fight my database and django like that. I would add a
> separate uuid field
>
>
Hi,
Since it is static (i.e. never modified), the CSS classes dictionary (statemap)
could be extracted from the function and promoted as a module global variable,
to avoid it being reconstructed each time the filter is invoked.
This will not cut the processing time in half, but small streams
Omar (and everybody else who responded),
Thanks for that ideas and putting a method on the model to return the
appropriate CSS class based on the state. I think that would be the
cleanest solution if I am going to bite the bullet and put this
template-specific logic on the model.
It got me th
No i dont.
But i am thinking that since it is a GET http request, Then there shouldnt be a
parameter in the url pattern. So maybe it should be
url(r'callback/' call_back) instead the previous one i did.
Although i havent tried it out yet.
Yingi Kem
> On 11 Nov 2017, at 4:46 PM, Dylan Reinhold
Do you have a url entry above that is also matching it?
On Sat, Nov 11, 2017 at 3:52 AM, yingi keme wrote:
> I have a callback function(webhook) implemented below
>
> def call_back(request, param):
> # Testnet
> method='GET'
> x_signature = request.headers.get('X-Signature')
> pr
Hi treyd,
I think idea of updating the model and storing it in the db is a bit
overkill, however you could create an @propery on the model that will do
that logic for you and return the css class for you.
e.g.
@property
def css_class(self):
if self.state is SUCHANDSUCH:
return "label-soan
i would never try to fight my database and django like that. I would add a
separate uuid field
On Fri, Nov 10, 2017 at 10:54 PM, Liuyang Wan wrote:
> Let's say in my Django project I have two models A and B.
>
> class A(models.Model):
> something...
>
> class B(models.Model):
> a = model
I have a callback function(webhook) implemented below
def call_back(request, param):
# Testnet
method='GET'
x_signature = request.headers.get('X-Signature')
print(siri)
request_uri_encoded = str(request).splitlines()[0].split()[1]
request_uri = urllib.parse.unquote
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