I have a problem that I just can't figure out. After creating a work order I want to redirect to the detail page of that work order. Here is my models.py class
In my frontend i'm logging into another app's api in the browser, I'm then redirected back to my app, that hits a View in my backend which gets a code from the
models.py from django.db import models class Images(models.Model): def upload_path(instance, filename): return '/images/'.join([filename]) im
I am using react components and to test my code I wrote a quick custom authentication class and defined in the settings for the rest_framework as follow: DEFAUL
<table id="tabledata" class="tablecenter" style=" position: static; top:50%;"> <tr> <th>Username</th> <th>Email</th>
I've setup a relationship using django's ForeignKey against 2 unmanaged tables like so: class Product(BaseModel): publish_name = models.CharField(unique=Tru
I have seemingly redundant code in Django Views which evaluates the same query-set (Model.objects.all()) but with different field names: def overview_view(reque
I have a Django app that requests data from an external API and my goal is to convert that data which is returned as list/dictionary format into a new REST API
I'm writing tests for my djangocms app, first time using selenium to replicate the browser/user. I'm trying to test the login page, user simply fills out the us
I have a simple code like: subprocess.run( ["python3", "action.py", "--options"] ) In action.py I import some packages, perform a small script It
After experimenting with setting up CustomUser and many-to-many relationships, I decided to make some (what turned out to be major) changes to the models. I had
Is there any way to refresh another page by clicking a button on the current page? (any way for the server to send a packet that makes every browser on a certai
I want to separate the files in my models based on extension type. Right now I am able to print all the files but now I want to separate them based on extension
I am doing a simple form site with Django. This is what my sites url is looks like: mysite.com/register/12345678 I want to print the part after the register (12
I have a use case of using two celery instances in a same django application. One celery instance is used for incoming events to the app and the other is to pub
i want to display a form to create posts. i use crispy-form and it currently displays: with html template: {% extends 'blog_app/base.html' %} {% load crispy_fo
I have a model class Entry(models.Model): maca = models.CharField(max_length=100, blank=True, null=True) This field will accept only numbers (cannot set ch
In my django project, I have made a view class by using TemplateView class. Again, I am using django channels and have made a consumer class too. Now, I am tryi
class ObjectSEO(models.Model): object= models.ForeignKey(Objects, related_name='seo', on_delete=models.CASCADE) meta_description = models.CharField(max_
my code is like this and I send data through httpClient and it returns error on upload my angular version is 13 and django version is 4 and I'm using django res