muhammad haris salamat just make a form in your template, then have a
name='any name here' and then make a view in which you have to get the name
in the following way.
query = request.POST.get('any name here')
and then do this:
value = ModelName.objects.filter(Q(model_field = query)
On Thu, Jun 2
Hi all can u help me plz how I can use filter with Contain query in Django
I wanna fetch all data from my database with the help of filter query
On Thu, 24 Jun 2021, 9:17 pm Richard Dushime,
wrote:
> THANK U I HAVE DONE TO FIX IT ITS NOW WORKING
>
> On Thu, Jun 24, 2021 at 6:04 PM Julio Cojom
>
THANK U I HAVE DONE TO FIX IT ITS NOW WORKING
On Thu, Jun 24, 2021 at 6:04 PM Julio Cojom
wrote:
> Django framework uses urls to math the html files. This work with the same
> syntax of static files, you would use {% url 'app:viewname' %} to render
> any page with Django.
>
> You should check th
Django framework uses urls to math the html files. This work with the same
syntax of static files, you would use {% url 'app:viewname' %} to render
any page with Django.
You should check the docs of urls to understand how it works and get this
done.
Regards
El jue., 24 de junio de 2021 9:00 a. m
Did you test this code? I think this code has errors...
If your objective is parse html/xml/json I think you've to check if the
application doesn't provide an API, and use the correct python library to
parse. If you need to parse by force, I think you can use BeautifulSoup.
2013/6/26 lx
> hi :
On 25/06/13 12:35, lx wrote:
> hi all:
> I want to do someing after "return HttpResponse(html)".
> for example
> ""
> from django.http import HttpResponse
> import datetime
> def current_datetime(request):
> now = datetime.datetime.now()
> html = "It is now %s.
Thank you.
But I want to fix it like this, because the time of "do something" is not
long.
###
from django.http import HttpResponseimport datetimedef
current_datetime(request):
now = datetime.datetime.now()
if now == 2013:
You can use Celery for that, I've done something similar, just with a CSV
instead of a XML, check
this http://docs.celeryproject.org/en/latest/django/first-steps-with-django.html
once installed and configured, its as simple as:
from celery import task
if (the xml format is ERROR) == True:
If the "do_something" will take a long time to finish, you've 2 options:
Find a better (optimized) way to "do_something" or inform your client the
application is doing something and he will have to wait a little.
So, I can't see any advantage to say to your client there's something was
ok if the a
Thank you.
The real demand is:
1.
receive the xml file from client.
2.
if (the xml format is ERROR) == True:
return HttpResponse(ERROR FORMAT")
else:
return HttpResponse(RIGHT FORMAT")
#I must give a response first, And do other thing. Because the spend
time of do_something() may be
The question is: What you're trying to do?
1. You can do any other process before the return line, the return will be
the same... for example:
from django.http import HttpResponse
from datetime import datetime
def current_datetime(request):
now = datetime.now()
html = "It is n
You cannot do something after the "return" line. This is Python
functionality. When you return, the function exits and does not process
anything else.
What exactly are you trying to do? There is probably another way to
accomplish it.
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