everyone!
Implementing the edit/add functions to the app, I found that Django ignores my clean_<field_name> method in some classes. When I tried to debug it I found out that the methods are not even called. Here are the relevant code snippets:
models.py
class StoredItem(models.Model):
item = models.ForeignKey(Item)
quantity = models.IntegerField(default=0)
class Meta:
ordering = ["item"]
def __str__(self):
return "{0} - {1}, {2} pcs.".format(self.item.serial, self.item.name, self.quantity)
def get_absolute_url(self):
return reverse('almacen:storeditem-detail', args=[str(self.id)])
forms.py
# Stored items related forms
class NewOrEditStoredItemForm(forms.Form):
item = forms.ModelChoiceField(queryset=StoredItem.objects.all())
quantity = forms.IntegerField()
def clean_item(self):
item = self.cleaned_data['item']
storeds = map(lambda x: x.item, StoredItem.objects.all())
if item in storeds:
raise ValidationError(_("This item has already been stored - you only can change its quantity in the stock!"))
def clean_quantity(self):
quantity = self.cleaned_data['quantity']
if quantity < 0:
raise ValidationError(_("Items quantity can not be negative!"))
return quantity
views.py
# Stored items related ---------------------------------------------------------------------
class CreateStoredItem(CreateView):
model = StoredItem
fields = '__all__'
class UpdateStoredItem(UpdateView):
model = StoredItem
fields = '__all__'
storeditem_form.html
{% extends 'base_template.html' %}
{% block header %}
<h2 id="center">Stocked item add/delete </h2>
{% endblock %}
{% block content %}
<form action="" method="post">
{% csrf_token %}
<table>
{{ form.as_table }}
</table>
<input type="submit" value="Submit"/>
</form>
{% endblock %}
EDIT
Resolved by rewriting the classes into functions. For classes I was recommended to use validators, but anyway, functions worked as I need.
Resolved by rewriting the classes into functions. For classes I was recommended to use validators, but anyway, functions worked as I needed
Related
Here i am using Django 3 and Python 3.7 where i want to display the the signature as image in my html page
Here is my views.py
def jobsignature_view(request, pk):
job = CreateJob.objects.filter(id=pk)
job_data={}
for val in job_id:
job_data['job_id'] = pk
job_data['signature'] = val.signature
......
....
Here is my models.py
class Cre(models.Model):
signature = models.TextField(default='', blank=True, null=True)
here in my database it is saving as
id signature
1 b'b3V0cHV0'
Here is my template.html
I confirm that this work has been completed:
<div class="span12">
{{ form.media }}
{% if job_data.signature != None and job_data.signature != '' %}
<img src="data:image/png;base64,{{job_data.signature}}"/>
{% endif %}
</div>
here is how its is displaying
How can i make my signature display here please help me
In your life I think for a search on a primary key it is more appropriate to use get(), your primary key is unique, we expect only one object. Filter() is used to return a queryset of otherwise multiple items :
def jobsignature_view(request, pk):
job = CreateJob.objects.get(id=pk)
return render(request, 'myapp/yourpage.html', {'job': job})
In the html template :
{% if job %}
<img src="{{job.signature}}"/>
{% endif %}
For the model I think you should use ImageField :
from django.core.files.storage import FileSystemStorage
from django.db import models
fs = FileSystemStorage(location='/media/photos')
class CreateJob(models.Model):
...
signature = models.ImageField(storage=fs)
trying to display all the ingredients that are associated to a recipe, however:
my models:
class IngredientList(models.Model):
name = models.CharField(max_length=100)
class RecipeList(models.Model):
name = models.CharField(max_length=100)
ingredients = models.ManyToManyField('IngredientList')
instructions = models.TextField(max_length=400)
amount = models.IntegerField()
my views:
from django.shortcuts import render
from .models import IngredientList, RecipeList
def index(request):
ing = RecipeList.objects.all()
context = {'ing': ing}
return render(request, 'myrecipes/home.html', context)
my template:
<div class="card-body">
<h4>{{ details.name}} <span class="badge badge-info">{{details.cuisine}}</span></h4>
<p class="card-text">Ingredients: {{details.ingredients}}</p>
<p class="card-text">Instructions: {{details.instructions}}</p>
<p class="card-text">This makes {{ details.amount}} meals</p>
</div>
my output is "myrecipes.IngredientList.None"
You need to access the queryset of the ingredients, so details.ingredients.all, furthermore you need to iterate over these ingredients, so you work with:
{% for ingredient in details.ingredients.all %}
{{ ingredient }}
{% endfor %}
In the view you can load all the related Ingredients with one extra query with:
def index(request):
ing = RecipeList.objects.prefetch_related('ingredients')
return render(request, 'myrecipes/home.html', {'ing': ing})
This will avoid making an extra query per RecipeList.
I have a long list of elements (50 elements) and I wanted to use that as a list_filter in django admin panel. But when I use this list, on very right side where FILTER panel appears I can see only 20 items and it is not scrollable so rest of 30 is hidden. How can I get those 30 items in FILTER section or How scrolling will be possible so that all 50 items can be viewed.
You can create a custom filter for your field and use a dropdown as selector.
I strongly recommend using https://github.com/mrts/django-admin-list-filter-dropdown which already provide a couple of filters.
If you want to create your own:
admin.py
from django.contrib.admin.filters import RelatedOnlyFieldListFilter
class RelatedOnlyDropdownFilter(RelatedOnlyFieldListFilter):
template = 'filters/related_filter.html'
#admin.register(ModelName)
class YourModelAdmin(admin.ModelAdmin):
list_filter = (
('field', RelatedOnlyDropdownFilter),
)
related_filter.html
{% load i18n %}
<script type="text/javascript">var filter_by_select = function(opt) { window.location = window.location.pathname + opt };</script>
<h3>{% blocktrans with title as filter_title %} By {{ filter_title }} {% endblocktrans %}</h3>
<ul class="admin-filter-{{ title|cut:' ' }}">
{% if choices|slice:"4:" %}
<li>
<select class="form-control"
onchange="filter_by_select(this.options[this.selectedIndex].value)">
{% for choice in choices %}
<option{% if choice.selected %} selected="selected"{% endif %}
value="{{ choice.query_string|iriencode }}">{{ choice.display }}</option>
{% endfor %}
</select>
</li>
{% else %}
{% for choice in choices %}
<li{% if choice.selected %} class="selected"{% endif %}>
{{ choice.display }}</li>
{% endfor %}
{% endif %}
</ul>
I suggest you to use this cool third-party.
django-admin-autocomplete-filter
It's easy to use and user-friendly in case of your problem.
For example look at this. models.py
from django.db import models
class Artist(models.Model):
name = models.CharField(max_length=128)
class Album(models.Model):
name = models.CharField(max_length=64)
artist = models.ForeignKey(Artist, on_delete=models.CASCADE)
cover = models.CharField(max_length=256, null=True, default=None)
admin.py
from django.contrib import admin
from admin_auto_filters.filters import AutocompleteFilter
class ArtistFilter(AutocompleteFilter):
title = 'Artist' # display title
field_name = 'artist' # name of the foreign key field
class ArtistAdmin(admin.ModelAdmin):
search_fields = ['name'] # this is required for django's autocomplete functionality
# ...
class AlbumAdmin(admin.ModelAdmin):
list_filter = [ArtistFilter]
"""
defining this class is required for AutocompleteFilter
it's a bug and I am working on it.
"""
class Media:
pass
# ...
I want to update the User and Lab model. I am able to see the form but it is not pre-populated with existing database information even after setting the instance parameter. If I submit a blank form then all fields are reset to blank values in the database. I have tried several solutions available online but nothing works.
My queries -
How do I pre-populate my form with existing data?
If the user doesnt fill out a particular field, I want the previous information to be stored as it is and not as a blank value. How do I achieve this?
I have the following models.py
class Lab(models.Model):
uid = models.OneToOneField(User, on_delete=models.CASCADE)
company=models.CharField(max_length=200,blank=True)
#receiver(post_save, sender=User)
def create_lab_profile(sender, instance, created, **kwargs):
if created:
Lab.objects.create(uid=instance)
#receiver(post_save, sender=User)
def save_lab_profile(sender, instance, **kwargs):
instance.lab.save()
Forms.py
class UserForm(forms.ModelForm):
email=forms.EmailField(max_length=300)
class Meta:
model = User
fields = ('first_name', 'last_name', 'email',)
class LabForm(forms.ModelForm):
class Meta:
model = Lab
fields = ('company',)
views.py
#login_required
def update_user_details(request,pk):
if request.method == 'POST':
user_form = UserForm(request.POST,instance=request.user)
lab_form = LabForm(request.POST,instance=request.user.lab)
if user_form.is_valid() and lab_form.is_valid():
user_form.save()
lab_form.save()
messages.success(request,'Your profile was successfully updated!')
return redirect('user_details')
else:
messages.error(request,('Please correct the error below.'))
else:
user_form = UserForm(instance=request.user)
lab_form = LabForm(instance=request.user.lab)
return render(request, 'update_user_details.html', {'user_form': user_form,'lab_form': lab_form})
template -
{% extends 'base.html' %}
{% block content %}
{% csrf_token %}
<H3> Update Personal information - </H3>
<br>
<form method="post" enctype="multipart/form-data">
{% csrf_token %}
{{ user_form.as_p }}
{{ lab_form.as_p }}
<button type="submit" class="btn btn-primary">Save changes</button>
</form>
{% endblock %}
Any help/suggestions will be appreciated!
I am trying to show the related field of the user on the webpage using Django.
I have models:
Models.py
class Companies(models.Model):
company_name = models.TextField()
company_email = models.EmailField()
company_owner = models.ForeignKey(User, on_delete=models.CASCADE)
def __str__(self):
return self.company_name
class Cars(models.Model):
company = models.ForeignKey('Companies', on_delete=models.CASCADE)
car_model = models.TextField()
views.py:
class UserCompanyCars(ListView):
model = Cars
template_name = 'home/company_cars.html'
context_object_name = 'cars'
slug_field ="username"
paginate_by = 100
def get_queryset(self):
company_n = get_object_or_404(Companies, company_owner=self.request.user)
return Cars.objects.filter(company=company_n)
and my html is:
{% extends 'home/base.html' %}
{% block content %}
<h1 class="mb-3"> Cars of {{user.Companies.company_name}}</h1>
{% for car in cars %}
<div class="media-body">
<div class = "article-metadata">
<p class="article-content">{{car.company}}</p>
{{car.car_model}}
<p class="article-content">{{car.car_carry }}</p>
</div>
</div>
{% endfor %}
{% endblock content %}
What I am trying to achieve is to write out "Cars of TestCompany (26)" on the webpage, but I cannot figure out how to get the Company_Name which is owned by the user. I have been trying all those Companies.objects... variations but none of them seem to work.
Try this (sample code taken from my project):
category = get_object_or_404(Category, pk=self.kwargs['pk'])
context['allcategories'] = MyPosts.objects.filter(category=category)
return context
This will filter all posts in the database based on certain categories. You can apply the same solution to your code.
If this solution does not work, please try to give more details about the problem.