'understanding try except python/django
I need your help with understanding try-except python/Django.
so I have this function:
def submit_dept_head_application(request, application_id):
cv = request.FILES['cv']
letter = request.FILES['letter']
candidate_id = request.data['candidateId']
rank_id = request.data['requestedRankId']
application_state = {
'candidate_id': candidate_id,
'rank_id': rank_id,
'cv_filename': cv.name,
'letter_filename': letter.name,
}
creator = Profile.objects.get(user=request.user.id)
department = creator.department
applicant = Profile.objects.get(user=candidate_id)
applicant_profile_id = applicant.id
rank = Rank.objects.get(id=rank_id)
try:
application = Application.objects.get(id=application_id)
# TODO - update application
except Application.DoesNotExist:
application = None
if application is None:
application = Application(creator=creator, applicant=applicant, desired_rank=rank,
application_state=application_state, department=department
)
application.save()
create_application_directory(application.id)
ApplicationStep.objects.update_or_create(
application=application, step_name=Step.STEP_1,
defaults={'can_update': True, 'can_cancel': True, 'currentStep': True}
)
copy_to_application_directory(cv, application.id)
copy_to_application_directory(letter, application.id)
send_email(settings.SENDGRID_SENDER, ['[email protected]'], 'new application submitted',
'new application submitted')
return Response(application.id, status=status.HTTP_200_OK)
right now what is happening is that if there is no open application I create a new one. what I'm trying to do is to add another check, which is Application.objects.filter(applicant=applicant_profile_id) so if I have an opened application for this candidate it won't get to the application.save() but it will send an error. i don't really know how to do so and that's where i need your help please :)
Solution 1:[1]
Check first. This is what queryset.exists() is for.
if Application.objects.filter(applicant=applicant_profile_id).exists() :
# tell applicant he's being naughty
#carry on as in the code you posted
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | nigel222 |