Replace an image through Django admin panel - python

I want to be able to replace my homepage image from the Django admin panel. I can upload to my ../media/homepage directory just fine but I want to first delete any image named "bg.jpg" and rename my new image to "bg.jpg".
models.py
from django.db import models
from django.core.files.storage import FileSystemStorage
from datetime import datetime
class Homepage(models.Model):
homepage_image = models.ImageField(upload_to="../media/homepage",blank=True)
image_text = models.CharField(max_length=200, blank=True)
header_title = models.CharField(max_length=200, blank=True)
header_text = models.TextField(blank=True)
class Meta:
verbose_name_plural = "Homepage"
def __str__(self):
return "Homepage"
Also is there any way I can display an image preview in the panel of the current image?

Related

The model is not displayed in the django admin panel

I don't have the advertisement module displayed in the django admin panel. Here is the model code
from django.db import models
class Advertisement(models.Model):
title = models.CharField(max_length=1000, db_index=True)
description = models.CharField(max_length=1000, default='', verbose_name='description')
creates_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
price = models.FloatField(default=0, verbose_name="price")
views_count = models.IntegerField(default=1, verbose_name="views count")
status = models.ForeignKey('AdvertisementStatus', default=None, null=True, on_delete=models.CASCADE,
related_name='advertisements')
def __str__(self):
return self.title
class Meta:
db_table = 'advertisements'
ordering = ['title']
class AdvertisementStatus(models.Model):
name = models.CharField(max_length=100)
admin.py /
from django.contrib import admin
from .models import Advertisement
admin.site.register(Advertisement)
I was just taking a free course from YouTube. This was not the case in my other projects. Here I registered the application got the name in INSTALLED_APPS. Then I performed the creation of migrations and the migrations themselves. Then I tried to use the solution to the problem here , nothing helped. I didn't find a solution in Google search either.
127.0.0.1:8000/admin/
console
admins.py
The name of the file is admin.py not admins.py. Yes, that is a bit confusing since most module names in Django are plural. The rationale is probably that you define a (single) admin for the models defined.
Alternatively, you can probably force Django to import this with the AppConfig:
# app_name/apps.py
from django.apps import AppConfig
class AppConfig(AppConfig):
def ready(self):
# if admin definitions are not defined in admin.py
import app_name.admins # noqa

How can I hide current uploaded image location path in django?

I have made a feature in Django where every user can change his platform's logo. The image selected by the user will be saved in static/{user.customer.public_id}/platformLogo/image.jpg. When i save the changes, i can see the uploaded image's path which also contain unique public ID which i don't want user to see for security purpose. Can anyone help me to hide this image path in Django for user? Attaching my code part here below.
Here we can see the image path which has unique ID in path, which we need to hide
Here is the uploaded image path directory
Here is my models.py
from sre_constants import CATEGORY
from unicodedata import category
from attr import fields
from django.db import models
from datetime import date
from django.contrib.auth.models import User
import uuid
def upload_path(instance, filename):
filename = str(date.today())
name = instance.user.customer.public_id.hex
return f'{name}/platformLogo/{filename}.jpg'
class Customer(models.Model):
user = models.OneToOneField(User, null=True, blank =True, on_delete=models.CASCADE)
public_id = models.UUIDField(primary_key=True, default = uuid.uuid4, editable=False)
date_created = models.DateTimeField(auto_now_add=True, null=True)
name = models.CharField(max_length=200, null=True)
otp_code = models.CharField(max_length=6, null=True)
first_name = models.CharField(max_length=200, null=True)
last_name = models.CharField(max_length=200, null=True)
email = models.CharField(max_length=200, unique=True)
phone = models.CharField(max_length=200, null=True)
profile_pic= models.ImageField(upload_to=upload_path, default='logo.png', null=True, blank=False,)
def __str__(self):
return self.name
Here is my views.py
#login_required(login_url='login')
def accountSetting(request):
customer = request.user.customer
form = CustomerForm(instance= customer)
if request.method == 'POST':
form = CustomerForm(data=request.POST, files=request.FILES, instance=customer)
if form.is_valid():
form.save()
context = {'form': form}
if request.user.is_anonymous:
return redirect("/")
return render(request, 'account-settings.html', context)
Here is my forms.py
from django.forms import ModelForm
from django.contrib.auth.forms import UserCreationForm
from django.contrib.auth.models import User
from .models import Customer
from django import forms
class CustomerForm(ModelForm):
class Meta:
model = Customer
fields = '__all__'
exclude = ['user', 'email','name','otp_code']
class CreateUserForm(UserCreationForm):
class Meta:
model = User
fields = ['username','first_name','last_name', 'email', 'password1', 'password2']
Here is settings.py
STATIC_URL = '/static/'
STATICFILES_DIRS = [os.path.join(BASE_DIR, 'static')]
MEDIA_URL = '/platformLogo/'
MEDIA_ROOT = os.path.join(BASE_DIR, 'static/platformLogo')
Since you already made public_id UUID, why not hash the logo and image name?
In Django environments I’ve used xsendfile with Apache or nginx. You end up placing the images in a folder that is accessible by Apache and served by apache, but can only be served after a request to the Django backend. It prevents all of the logos being visible to prying eyes.

Django ImageField upload works locally but not in production

My models file:
from django.db import models
from django.utils import timezone
from django.contrib.auth.models import User
class Profile(models.Model):
user= models.OneToOneField(User,on_delete=models.CASCADE)
description = models.CharField(max_length=100,default='')
city = models.CharField(max_length=100,default='')
website = models.URLField(default='')
phone = models.IntegerField(default=0)
avatar = models.ImageField(upload_to='avatars/',blank=True,default='avatars/no.png')
genre = models.IntegerField(choices=((1, ("Homme")),
(2, ("Femme")))
)
def __str__(self):
return self.user.username
Locally when I fill the form, the image is saved in my folder media/avatars/.
But on heroku the image is not saved in this folder and therefore it can't be displayed.

How do I delete a field from my Django ModelForm?

I'm trying to hide and delete two fields from showing in a form I created in the Django administration page using ModelForm.
I looked at answers that said I should use the "exclude" meta field, but I don't know why it's not working in my case.
Here is my code:
models.py:
class Activity(models.Model):
type = models.CharField(max_length=50, default="")
title = models.CharField(max_length=200, default="")
description = models.CharField(max_length=500)
owner = models.ForeignKey(User, related_name="owner")
college = models.CharField(max_length=200)
location = models.CharField(max_length=200)
room = models.CharField(max_length=200)
startDate = models.DateTimeField(null=True, blank=True)
endDate = models.DateTimeField(null=True, blank=True)
attendee = models.ManyToManyField(Attendee, related_name="attendees",null=True, blank=True)
volunteer = models.ManyToManyField(Volunteer, related_name="volunteers",null=True, blank=True)
I'm trying to exclude the "attendee & volunteer" fields from displaying in the Django administration form.
In admin.py I have:
from django.contrib import admin
from django import forms
from KSUvity.models import Activity
class ActivityForm(forms.ModelForm):
class Meta:
model = Activity
exclude = ['attendee', 'volunteer',]
class ActivityAdmin(admin.ModelAdmin):
exclude = ['attendee', 'volunteer',]
form = ActivityForm
admin.site.register(Activity, ActivityAdmin)
You have to create an admin.py file in your app and register your models
Follow the instuctions
See the example below
from django import forms
from django.contrib import admin
from myapp.models import Person
class PersonForm(forms.ModelForm):
class Meta:
model = Person
exclude = ['name']
class PersonAdmin(admin.ModelAdmin):
exclude = ['age']
form = PersonForm
admin.site.register(Person, PersonAdmin)
You can use either fields or exclude in one class.
In your app admin field add this code.
app_name/admin.py
from django.contrib import admin
class ActivityAdmin(admin.ModelAdmin):
exclude = ('attendee', 'volunteer',)
You have to use ModelAdmin option to exclude fields from form in Django administration, either ModelAdmin.exclude or ModelAdmin.fields. Below is an example:
class ActivityAdmin(admin.ModelAdmin):
exclude = ('attendee', 'volunteer', )
To make it work, you register model like this:
admin.site.register(Activity, ActivityAdmin)
You add this code to admin.py file.

Django Admin - How can I select new data created using ForeignKey?

For example, I have this models
# models.py
from django.db import models
class Category(models.Model):
title = models.CharField(max_length=60)
class CategoryDescription(models.Model)
category = models.ForeignKey(Category)
description = models.CharField(max_length=120)
class Image(models.Model):
category = models.ForeignKey(Category)
description = models.ForeignKey(CategoryDescription)
image = models.ImageField()
And I have this admin init file
# admin.py
from django.contrib import admin
from .models import Category, CategoryDescription, Image
class CategoryDescriptionInline(admin.TabularInline):
model = CategoryDescription
class CategoryAdmin(admin.ModelAdmin):
inlines = (CategoryDescriptionInline,)
class ImageAdmin(admin.ModelAdmin):
pass
admin.site.register(Category, CategoryAdmin)
admin.site.register(Image, ImageAdmin)
I want create new Image in Admin, i create new category, using plus placed right from select input and next, i want choose Image Description, but i cant do it, because it isn't loaded.
What you may recommend for me in this situation? Thanks
Register a ModelAdmin for CategoryDescription, e.g.:
admin.site.register(CategoryDescription)

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