This is my views.py file and Im getting this message However is see that Valid is printing so its obvious the is_valid() method doesnt catch this duplicate key. EDIT Batting model and form How is it I catch this issue? Answer In your form you have set exclude as follows: exclude = (‘match’,). When we exclude a field from the
Tag: django
How to pass a filter from a dropdown into django-import-export view
I understand on how to pass a filter through views that have the return render(request, ‘htmlname.html, {}). I don’t know how to do it for this case of exporting data through django-import-export export option. I’d like to pass a filter from a dropdown selection for the data to be downloaded. My view My resource Answer There are a couple of
How do I get multiple rows from django database?
Hello I am trying to create an employee attendance program and I want to make it so that the employees can see the total hours they worked. This is the code that creating the problem: The error I get is and if I use it shows an error that says Answer You must use filter and iterate over results as
You must set settings.ALLOWED_HOSTS if DEBUG is False. Django-r-f
I want to prepare the server for deployment and this is the setup. settings.py is split into dev.py and prod.py in settings.py at the bottom ALLOWED_HOSTS = [] I also have tried ALLOWED_HOSTS = [‘*’] dev.py wsgy.py When I import settings in the manage.py shell, the settings.DEBUG = False .. why? How to make wsgi.py read the correct file? Answer
How to format dynamically generated HTML with django
I’m trying to create a sheet of labels to print out with a QR code a logo and phone number, the QR codes are generated separately from an xlsx file and they work. My issue is formatting them in the HTML. I’m not sure how to change the for loop so it creates a row for each time an image
Type error when trying to fetch data Python Django
I am trying to get data from the database based on request with id u_id and x_id but I am not getting any data instead getting an error TypeError: display_data() missing 2 required positional arguments: ‘u_id’ and ‘x_id’ can you help? code urls.py Answer You need to get them from the POST data instead. Also, you might want to use
I want to get all anime related to specific Genre,
model view urls Answer You should filter on the genre field of the Genre, so: Note: Models in Django are written in PascalCase, not snake_case, so you might want to rename the model from anime to Anime.
Django iterate all values from zip
first I zip all my lists just like but as you see the detail has not only one element, I can not sure the number of list of detail. And the number is variable. Than loop over it in templates like: As I can not confirm the number of detail, I got the error message like “Need 3 values to
ERROR: (gcloud.app.deploy) Error Response: [9] Cloud build c90ad64e-2c2f-4ad0-a250-160de6f315df status: FAILURE
I am trying to deploy my Django application to gcloud and I keep getting this error, any ideas what this means? File upload done. Updating service [default]…failed. ERROR: (gcloud.app.deploy) Error Response: [9] Cloud build c90ad64e-2c2f-4ad0-a250-160de6f315df status: FAILURE Error ID: c84b3231 Error type: UNKNOWN Answer never mind, I found out the error was due to a build error, I had an
How to make a “create-only” non-editable field in django admin
im looking for some solution about make a “create-only” field on django admin using models. I saw some questions before, but no one can answer the core question: the field should appear when the user are creating on admin panel, but i dont want to be able to edit. models.py admin.py the “readonly_fields” solves the problem in the matter of