def get_queryset(self, **kwargs): serialnumber = self.kwargs.get('serialnumber') if serialnumber: serial_number = int(serialnumber) result = ModelName.objects.fiter(serialnumber=serialnumber) return result

Where can i use this i am using class based views ? Do i need to do this before writing queryset ?

Ok let me try this

How to do that in class based view i am doing Def get_queryset(self,**kwargs): Return model.objects.filter(serialnumber= self.kwargs['serialnumber'])

I don't have to use return ?

Hi guys, Im Mimi and I'm new here. I am currently developing system that used django-viewflow pro as the engine for workflow. Does anybody here familiar with django viewflow? I need some help 😄

You should)

Hoe can i show media is not allowed in this group

dpaste.org

Do you want me to paste my whole class view ?

https://dpaste.org/Kpdz#L6

One question but i am getting the url link - 127.0.0.1:8001/testResults/192920083 Why do i still get error . Also after doing changes as how you mention i am still getting same error - got an unexpected keyword argument serial number

try to print serialnumber after get it. what the type of field Part_SerialNumber? you can share all your cbv code, and tbl_TestResults model

try to print serialnumber after get it. what the type of field Part_SerialNumber? you can share all your cbv code, and tbl_TestResults model

Hi the comple code is here - dpaste.org/od43

https://dpaste.org/GeJg#L10 try this

https://dpaste.org/GeJg#L10 try this

May be i will paste error for you

or this https://dpaste.org/feuO#L10,5

yes) and could you share urls.py

yes) and could you share urls.py

Please have a look i have share the error as well as url.py

I haven't use old django versions with urls, I'm usthing path, so I can miss something, but you can try this: url(r'testResults2/(?P<Part_SerialNumber>\w+)/$', login_required(views_class.TestResult2TableView.as_view()), name='testResults2'), and in the view: serialnumber = self.kwargs.get('Part_SerialNumber')

Changing to part_serialNumber is also not working i have used the url in the same way i have sent you before, the way i did for other URL link is working let me share with you the other url

Changing to part_serialNumber is also not working i have used the url in the same way i have sent you before, the way i did for other URL link is working let me share with you the other url

Same error

Same error

Wht type of error it is

could you for testing replace url with this: url(r'^testResults2/test/$', views_class.TestResult2TableView.as_view(), name='testResults2'), and try to print something in view (first string in get_queryset), to know where the problem is

could you for testing replace url with this: url(r'^testResults2/test/$', views_class.TestResult2TableView.as_view(), name='testResults2'), and try to print something in view (first string in get_queryset), to know where the problem is

def get_queryset(self, **kwargs): print("I'm inside the view!!!")

now try to step by step return old version of path to see where it'll break, start from login required

Yes it print this

now try to step by step return old version of path to see where it'll break, start from login required

django version doesn't support path, url only?)

Didn't get you what exactly i have to try?

url(r'^testResults2/test/$', login_required(views_class.TestResult2TableView.as_view()), name='testResults2'),

I think there is mistake of case in view mode..

What mistake?

Check view.py and check spelling in function which have been created queryset....inside it check the naming and parameters are valid or not ...i am 90% sure uh made error there

It works after removing / after test

Ok

remove it =) url(r'^testResults2/(?P<test>\w+)$', login_required(views_class.TestResult2TableView.as_view()), name='testResults2'),

Does it work?

def get_queryset(self, **kwargs): print("I'm inside the view!!!")

Same error unexpected keyword argument

did it print "I'm inside the view!!!"?

Do i need to define slug in model?

from django.urls import path path('testResults2/<slug:test>', login_required(views_class.TestResult2TableView.as_view()), name='testResults2'),

Something is wrong with this url , it is searching for url

first string before urlpatterns

did it print "I'm inside the view!!!"?

Same error

Hi, everyone ✌️. I want to create a website on Django, where can I start to learn it?

I give up :(