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NOT NULL constraint failed error when attempting to save using createview

This is my first post to SO, so please let me know if I've missed any important details. I am working on updates to a home-grown dJango based ticketing system.
I have two "parent" models (ParentProjects and Projects) that capture details about work we want to track. Both models have a number of columns that store information in the associated tables as well as some FK relations.
The generic class-based detail view is used to view objects in the Project table while the ParentProject table is accessed by a function based view. The function-based view accomplishes the same task of loading parent project object values as the class-based detail view does for the project objects.
The problem I am having is that I cannot add a new entry to the IDTasks model that automatically inserts the Parent Project id. I am able to add a new IDTask from within the admin site (or from the "client" site if I enable the "parent" field within the modelform) by manually selecting the parent I wish to associate the IDTask to. I can also edit and save an existing IDTask from within the Parent Project detail view without any issues. However, when I attempt to add an IDTask using the createview, dJango reports a Not NULL constraint error and the new entry is not saved.
In addition to reviewing and trying many other solutions to this problem, I have disabling the code that automatically adds the logged in user id but am still getting the same null constraint error. What's strange is that I am using the same basic createview structures for adding FK objects to the Projects model and this works perfectly. I'm still getting comfortable with Django's class-based views, so must surely be missing something obvious.
Thank you for your help!
Here are the main views related to my issue:
# Detail view for Parent Projects (function-based)
def ParentProjectDetail(request,parent_id):
parents = ParentProject.objects.get(id=parent_id)
projects = Project.objects.filter(parent_project__pk=parent_id).order_by('project_phase', '-launch_date',)
return render(request, 'otis/parent_detail.html', context={'parents':parents, 'projects':projects })
# Detail View for Projects (class-based)
class ProjectDetailView(generic.DetailView):
model = Project
context_object_name = 'project'
template_name = 'otis/project_detail.html'
# Add parent progress report
class add_idtask_view(SuccessMessageMixin, CreateView):
model = IDTasks
template_name = 'otis/id_report_form.html'
form_class = idTaskForm
success_message = "Report added"
def form_valid(self, form):
idtaskform = form.save(commit=False)
idtaskform.user = self.request.user
self.parents_id = self.kwargs.get('parent_id')
form.instance.ParentProject = get_object_or_404(ParentProject,id=self.parents_id)
return super(add_idtask_view, self).form_valid(form)
def get_success_url(self):
return reverse_lazy('otis:parent_detail', kwargs={'pk': self.parents_id})
Here is the modelform:
class idTaskForm(ModelForm):
class Meta:
model = IDTasks
fields = ('parent_phase','complete','milestones','nextsteps','concerns')
widgets = {
'milestones': Textarea(attrs={'cols': 50, 'rows': 5, 'placeholder': 'Task details...'}),
'nextsteps': Textarea(attrs={'cols': 50, 'rows': 5, 'placeholder': 'Task details...'}),
'concerns': Textarea(attrs={'cols': 50, 'rows': 5, 'placeholder': 'Task details...'}),
}
labels = {
'parent_phase': mark_safe('<span class="required">Phase</span>'),
'complete': mark_safe('<span class="required">Percentage Complete</span>'),
'milestones': ('Milestones'),
'nextsteps': ('Next steps'),
'concerns': ('Concerns'),
}
Here are the two models being accessed:
# Parent Project Model
class ParentProject(models.Model):
class Meta:
verbose_name = "parent project"
verbose_name_plural = "parent projects"
ordering = ['title']
title = models.CharField('Name', max_length=100, null=True, blank=False)
parent_group = models.ForeignKey(ProjectGroups, on_delete=models.CASCADE, blank=True, null=True)
parent_type = models.ForeignKey(ProjectTypes, on_delete=models.CASCADE, null=True, blank=False)
description = models.TextField('description', blank=True)
term_due = models.ForeignKey(Terms, on_delete=models.CASCADE, blank=True, null=True)
term_year_due = models.ForeignKey(Years, on_delete=models.CASCADE, blank=True, null=True)
launch_date = models.DateField('launch date', blank=True, null=True)
parent_phase = models.ForeignKey(SDLCPhases, on_delete=models.CASCADE, null=True, blank=False)
history = HistoricalRecords()
def __str__(self):
return str(self.title)
def get_absolute_url(self):
return reverse('otis:parent_detail', kwargs={'pk': self.pk})
# Reports Model
class IDTasks(models.Model):
class Meta:
verbose_name = "Parent task"
verbose_name_plural = "Parent tasks"
ordering = ['updated_on']
parent = models.ForeignKey(ParentProject, on_delete=models.CASCADE)
parent_phase = models.ForeignKey(SDLCPhases, on_delete=models.CASCADE, null=True, blank=False)
user = models.ForeignKey(User, on_delete=models.CASCADE, null=True, blank=True)
complete = models.IntegerField('percentage complete', blank=True, null=True, default=0)
milestones = models.TextField('milestones', blank=True)
nextsteps = models.TextField('next steps', blank=True)
concerns = models.TextField('concerns', blank=True)
updated_on = models.DateTimeField(auto_now_add=True, null=True)
history = HistoricalRecords()
def __str__(self):
return str(self.complete)
Here are the url patterns:
# url for the parent project detail view
path('parent_detail/<int:parent_id>/', views.ParentProjectDetail, name='parent_detail'),
# url for the create report accessed within the detail view
path('parent_detail/<int:parent_id>/add_id_report/', views.add_idtask_view.as_view(), name='add_id_report'),
Finally, the template link that invokes the modelform:
link_title

I was able to solve my initial problem by restructuring the view like so:
class add_idtask_view(SuccessMessageMixin, CreateView):
model = IDTasks
template_name = 'otis/id_report_form.html'
form_class = idTaskForm
success_message = "Report added"
def form_valid(self, form):
parents_pid = self.kwargs.get('parent_pid')
self.parent_id = parents_pid
form.instance.user = self.request.user
form.instance.parent_id = parents_pid
return super(add_idtask_view, self).form_valid(form)
def get_success_url(self):
return reverse_lazy('otis:parent_detail', kwargs={'parent_pid': self.parent_id})
The first thing I did was sort out what form fields and database fields I was trying to access. It seems I had confused these and not properly referenced them in my initial view.
Once these were working however, I started getting an error that stated the view was expecting an integer but was getting a string. It seems, for whatever reason, when I used the get_object_or_404 method, the view returns the title of the database object and not the primary key.

Update model each time a view is generated in django

I'm trying to update my an instance of Post model each time a view PostDetail is generated. So far I've tried multiple approaches but none of them worked. I know that there is ready solution (django-hitcounter) but I would like to write one myself so I can understand what is happening.
The goal there is to add 1 to post.views each time user accesses PostDetail view.
models.py
class Post(models.Model):
title = models.CharField(max_length=200, unique=True)
slug = models.SlugField(max_length=200, unique=True)
author = models.ForeignKey(
User, on_delete=models.CASCADE, related_name='blog_posts')
updated_on = models.DateTimeField(auto_now=True)
content = models.TextField()
created_on = models.DateTimeField(auto_now_add=True)
status = models.IntegerField(choices=STATUS, default=0)
views = models.IntegerField(default=0)
class Meta:
ordering = ['-created_on']
views.py
class PostDetail(generic.DetailView):
model = Post
template_name = 'blog/post_detail.html'
urls.py
urlpatterns = [
path('<slug:slug>/', views.PostDetail.as_view(), name='post_detail'),
]

Once you've got to the point where Django can return a response (eg: it's found the Post object successfully etc...) - you can increment your view count for the object then and proceed to returning the response, so if you change your view to be:
class PostDetail(generic.DetailView):
model = Post
template_name = 'blog/post_detail.html'
def render_to_response(self, context, **kwargs):
self.object.views += 1
self.object.save()
return super().render_to_response(context, **kwargs)

Django editing a model instance fails to find the instance

I'm in the process of making a Recipe Book. For some reason, whenever I try to pull up a recipe from the DB to edit it, I keep getting an error where it can't find the recipe I've specified. I'm using slugs, and my logic is that I'm going from a detailView where I've already pulled up the db information, to an updateView. I'm attempting to pass the recipe object I already pulled from the detailView to the updateView, but when I do, it keeps telling me that it can't find the recipe specified.
views.py:
The base views I'm calling here are only providing a default post method for handling a search so that I don't have to put it in for every view I create so I have some code reusability
class RecipeDetailView(BaseDetailView):
model = Recipe
template_name = 'RecipeBook/recipe_detail.html'
context_object_name = 'recipe_view'
queryset = None
slug_field = 'slug'
slug_url_kwarg = 'slug'
def get_context_data(self, *args, **kwargs):
context = super(RecipeDetailView, self).get_context_data()
recipe = self.object
recipe.ingredients = recipe.ingredients_list.split('\n')
context['recipe'] = recipe
return context
class RecipeEditView(BaseUpdateView):
model = Recipe
template_name = 'RecipeBook/edit_recipe.html'
context_object_name = 'recipe_edit'
queryset = None
slug_field = 'slug'
slug_url_kwarg = 'slug'
form_class = RecipeForm
def get_context_data(self, *args, **kwargs):
context = super(RecipeEditView, self).get_context_data()
recipe = self.object
print(recipe.name)
recipe.ingredients = recipe.ingredients_list.split('\n')
recipe.categories_list = ""
categories = Category.objects.filter(recipe=recipe)
for category in categories:
if category != categories[-1]:
recipe.categories_list += (category + ", ")
else:
recipe.categories_list += category
recipe_edit_form = RecipeForm(initial={'name': recipe.name, 'ingredients_list': recipe.ingredients,
'directions': recipe.directions, 'prep_time': recipe.prep_time,
'cook_time': recipe.cook_time, 'servings': recipe.servings,
'source': recipe.source, 'category_input': recipe.categories_list})
context['recipe'] = recipe
context['recipe_edit_form'] = recipe_edit_form
return context
models.py:
class Recipe(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=100, default="")
ingredients_list = models.TextField(default="")
servings = models.IntegerField(default=0, null=True, blank=True)
prep_time = models.IntegerField(default=0, null=True, blank=True)
cook_time = models.IntegerField(default=0, null=True, blank=True)
directions = models.TextField(default="")
source = models.CharField(max_length=100, default="", null=True, blank=True)
categories = models.ManyToManyField(Category, blank=True)
slug = models.CharField(max_length=200, default="")
def __str__(self):
return self.name
urls.py
# ex: /Recipes/Grilled_Chicken/
path('Recipes/<slug>/', views.RecipeDetailView.as_view(), name='view_recipe'),
path('Recipes/<path:slug>/', views.RecipeDetailView.as_view(), name='view_recipe'),
# ex: /Recipes/edit/Steak/
path('Recipes/edit/<slug>/', views.RecipeEditView.as_view(), name='edit_recipe'),
path('Recipes/edit/<path:slug>/', views.RecipeEditView.as_view(), name='edit_recipe'),
link in recipe_detail.html:
Edit Recipe
I've been going nuts trying to figure it out. By everything that I have in here, the recipe that I'm pulling up in the detailView should be able to be passed to the editView, but every time I try to open up the edit_recipe page, it keeps telling me that it can't find the recipe specified. The URL that it generates shows the proper slug and link that it should though. I don't know what I'm missing at this point...

Try this way:
Edit Recipe

I ended up having to go back through and change the view to be a DetailView. That was the only way I could get the recipe instance to be pushed through. There's something very specific about using an update view with models that isn't very clear...
Once I changed to a DetailView, the page would populate with the form initialized to the recipe values. I could then make tweaks to make sure everything worked from there.
Thanks for those who responded, it at least got my brain working in a different direction to get this figured out.

django foreignkey not working

I am still not satisfied with django foreignkey relationship.
below are two classess .
from django.db import models
from django.utils.text import slugify
from django_countries.fields import CountryField
from django.conf import settings
# Create your models here.
class PhotoGallery(models.Model):
title = models.CharField(max_length=255,unique=True)
picture_choices =
models.CharField(max_length=255,choices=PICTURE_CHOICES,default=NTR)
description = models.TextField(default='')
user = models.ForeignKey(settings.AUTH_USER_MODEL,
on_delete=models.CASCADE,related_name='image_uploader')
date_posted = models.DateTimeField(auto_now=True)
views = models.PositiveIntegerField()
likes = models.PositiveIntegerField()
country = CountryField(blank_label='(select country)')
slug = models.SlugField(allow_unicode=True,unique=True)
def __str__(self):
return self.title
from django.contrib import admin
from gallery.models import PhotoGallery
class PhotoGalleryAdmin(admin.ModelAdmin):
prepopulated_fields = {'slug': ('title',)}
And below is class Imagesmodel that has a foreignkey .Calling the PhotoGallery class from class Imagesmodel is not a problem .MY big problem is doing the reverse
class ImagesModel(models.Model):
gala_obj =
models.ForeignKey(PhotoGallery,
related_name='picture_gallery',on_delete=models.CASCADE)
post_image = models.ImageField(upload_to='gallery_pics',
verbose_name='Image')
def __str__(self):
return self.gala_obj.title
and below is my views.py file
`def detailViewPage(request,slug):
#here i wanted to get one image with the given slug
from `ImagesModel`
a = get_object_or_404(PhotoGallery, slug=slug)
selected_pic = a.picture_gallery.all()
#here i wanted to get all images excluding one
with the given slug
object_2 = ImagesModel.objects.exclude(id=selected_pic.id)
.order_by('-id')[:15]
return render(request,'gallery/photogallery_detail.html',{
'selected_pic':selected_pic,
'object_2':object_2,
})`
below is the error i get

The question is not very clear, but I assume you want to get all pictures associated with an object A. In your B model, give your foreignkey a related name:
class B(models.Model):
a = models.ForeignKey(A,related_name='post_title', related_name='pictures')
image = models.ImageField(upload_to='gallery_pics',
verbose_name='Image')
In your views.py:
a = get_object_or_404(A, slug=slug)
b = a.pictures.all()
I don't know why you named your foreignkey 'title', but the above code will get all the pictures associated with the object A.
The following will work also:
a = get_object_or_404(A, slug=slug)
pictures = a.b_set.all()
If you want to get the slug from an object B, then you can do the following:
b = get_object_or_404(B, id=1)
slug = b.a.slug

Django - Listing many different models in one page

I need to list different models in a single page/url.
#models.py
class Service(models.Model):
author = models.ForeignKey(User, related_name="services")
title = models.CharField(max_length=255)
slug = models.SlugField(max_length=255, unique=True)
objects = ServiceQuerySet.as_manager()
class Carousel(models.Model):
author = models.ForeignKey(User, related_name="carousels")
title = models.CharField(max_length=255)
content = models.TextField()
objects = CarouselQuerySet.as_manager()
This is my views, this way are listed in different pages, I tried to join the queryset, but got no success.
#views.py
class ServiceListView(generic.ListView):
model = models.Service
queryset = models.Service.objects.published()
class CarouselListView(generic.ListView):
model = models.Carousel
queryset = models.Carousel.objects.published()
This is my urls.py, this listing only those services.
urlpatterns = patterns('',
url(r'^$', views.ServiceListView.as_view(), name="service_list"),
url(r'^$', views.CarouselListView.as_view(), name="carousel_list"),
)
I need the two lists appear on the same page. How can I accomplish this task?

What about passing it through the context?
from .models import Service,Carousel
class ServiceListView(generic.ListView):
model = Service
queryset = Service.objects.published()
def get_context_data(self, **kwargs):
context = super(ServiceListView, self).get_context_data(**kwargs)
context['carousel_list'] = Carousel.objects.published()
return context