I am using django-tables2 and trying to hide url and rename it on field. for example, url link is www.youtube.com but on actual field, I want it to show as 'link' instead of revealing entire url link. how do I achieve this?
tables.py
class MyVideoTable(tables.Table):
class Meta:
model = PPVideo
fields = ('title', 'url')
models.py
class PPVideo
title = models.CharField('Title', max_length=100, null=True)
url = models.URLField('URL', max_length=150, null=True)
You can define a .render_url(…) method to specify how to render this column:
from django.utils.html import format_html
class MyVideoTable(tables.Table):
def render_url(self, value, record):
return format_html('link', value)
class Meta:
model = PPVideo
fields = ('title', 'url')
Related
models.py
from django.db import models
from django.contrib.auth.models import User
STATUS = (
(0,"Draft"),
(1,"Publish")
)
class BlogModel(models.Model):
id = models.AutoField(primary_key=True)
blog_title = models.CharField(max_length=200)
blog = models.TextField()
status = models.IntegerField(choices=STATUS, default=0)
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
class Meta:
ordering = ['-created_at']
def __str__(self):
return f"Blog: {self.blog_title}"
class CommentModel(models.Model):
your_name = models.CharField(max_length=20)
comment_text = models.TextField()
created_at = models.DateTimeField(auto_now_add=True)
blog = models.ForeignKey('BlogModel', on_delete=models.CASCADE)
class Meta:
ordering = ['-created_at']
def __str__(self):
return f"Comment by Name: {self.your_name}"
admin.py
from django.contrib import admin
from blog.models import BlogModel,CommentModel
class PostAdmin(admin.ModelAdmin):
list_display = ('blog_title', 'status','created_at','updated_at')
list_filter = ('status',)
search_fields = ('blog_title', 'content',)
admin.site.register(BlogModel, PostAdmin)
admin.site.register(CommentModel)
I created a simple blog post website with comments and I want to create reports and on the admin panel I have to see how to achieve this.
Like how many posts are created and how many have comments and how many post are draft and published
I checked this module but I don't understand how to implement it https://pypi.org/project/django-reports-admin/
You already have most of this, by using PostAdmin. The list_display already shows you how many posts are published/draft, and the change list has filters for that as well.
To show the comment count, simply add that to list_display:
class PostAdmin(admin.ModelAdmin):
list_display = ('blog_title', 'status', 'comment_count', 'created_at', 'updated_at')
def comment_count(self, obj):
return obj.commentmodel_set.count()
comment_count.short_description = 'Comment count'
This thus defines a custom method on the PostAdmin, that displays the comment count as a column, and gives it a user-friendly name as column header.
You can expand this with more statistics if you like. The Django admin is highly customizable.
Note: model names should be in CamelCase, so BlogModel and CommentModel should be Blog and Comment respectively.
I want to be able to sort a table column defined using a custom method in the Django admin.
I narrowed down the problem to this simple example in Django:
models.py:
from django.db import models
class MyObject(models.Model):
name = models.CharField(_("name"), max_length=255)
layers = models.URLField(_("Layers"), blank=True, max_length=1024)
choices = models.TextField(
verbose_name=_("Choice values"),
blank=True,
help_text=_("Enter your choice"),
)
class Meta:
verbose_name = _("Object config")
verbose_name_plural = _("Objects config")
def __str__(self): # my custom method
return self.name
and admin.py:
from django import forms
from django.contrib import admin
class MyObjectAdminForm(forms.ModelForm):
"""Form"""
class Meta:
model = models.MyObject
fields = "__all__"
help_texts = {
"layers": "URL for the layers",
}
class MyObjectAdmin(admin.ModelAdmin):
form = MyObjectAdminForm
list_filter = ["name",]
search_fields = ["name",]
# I want the first column (__str__) to be sortable in the admin interface:
list_display = ["__str__", ...] # the ... represent some other DB fields
but for the moment I cannot sort that first column (it is grayed out, I cannot click on its title):
So how could I sort the first column in this admin table as defined by the __str__() method of the MyObject model? (please note that I cannot change the model itself. I'm also brand new to Django, so don't hesitate to detail your answer as if you were speaking to a kid.)
I am trying to import data from an csv file into a django db using django-import-export. My problem is trying to upload data with a ForeignKey as an object. I have migrated, followed docs, and still no solution. You can see my error below in the django admin:
Here is my csv data with a blank 'Id' column:
models.py
from django.db import models
from django.shortcuts import reverse
from urllib.parse import urlparse
class States(models.Model):
name = models.CharField(max_length=96, blank=False, unique=True)
abbrv = models.CharField(max_length=2, null=True, blank=True)
class Meta:
ordering = ['name']
verbose_name = 'State'
verbose_name_plural = 'States'
def __str__(self):
return f'{self.name}'
class Person(models.Model):
last_name = models.CharField(
max_length=255, help_text="Enter your last name.")
first_name = models.CharField(
max_length=255, help_text="Enter your first name or first initial.")
address = models.CharField(
max_length=255, blank=True, help_text="Enter your street address.")
city = models.CharField(
max_length=255, blank=True, help_text="Enter your city.")
state = models.ForeignKey('States', to_field='name', on_delete=models.SET_NULL, null=True)
zipcode = models.CharField(max_length=50)
website = models.URLField(
max_length=255, blank=True)
profession = models.CharField(max_length=50, blank=True)
# META CLASS
class Meta:
verbose_name = 'Person'
verbose_name_plural = 'Persons'
ordering = ['last_name', 'first_name']
# TO STRING METHOD
def __str__(self):
"""String for representing the Model object."""
return f'{self.last_name}, {self.first_name}'
admin.py:
from django.contrib import admin
from .models import Person, States
from import_export.admin import ImportExportModelAdmin
from import_export.widgets import ForeignKeyWidget
from import_export import fields, resources
class PersonResource(resources.ModelResource):
state = fields.Field(
column_name='state',
attribute='state',
widget=ForeignKeyWidget(States, 'name'))
class Meta:
model = Person
class PersonAdmin(ImportExportModelAdmin):
list_display = ('last_name', 'first_name', 'state')
search_fields = ('first_name', 'last_name' )
resources_class = PersonResource
admin.site.register(Person, PersonAdmin)
admin.site.register(States)
I think you need to specify both in your question here, as well as to Django how you want the id field treated.
Do you want it propagated with the Django id or pk (sometimes the same sometimes not)? Then you would have id=self.id or id=self.pk somewhere in your view for the datatable.
Do you want your database to create a unique key?
You would need to add some functionality someplace to tell Django how to fill in that field.
Also, if you want it to create an id different from the Django id or pk then you would need to add the field to your model.
https://docs.djangoproject.com/en/3.1/ref/forms/validation/
https://docs.djangoproject.com/en/3.1/ref/validators/
https://docs.djangoproject.com/en/3.1/ref/forms/api/
Or, perhaps after Validation of the form, when you create the object. Add something to the effect of id=[database function to create unique id].
Another solution might be a templateTag or templateFilter to create a value on the form side if you want to create the id based on info contained in the form. Like combining last 4 of name with time of submission.
https://docs.djangoproject.com/en/3.1/ref/templates/builtins/
https://docs.djangoproject.com/en/3.1/howto/custom-template-tags/
Having just re-read your question, also, I'm not sure but you might be asking if the database can support an embedded reference to another object. Is ID a reference to another model's key? That's a whole different question. And it is database specific.
Last Suggestion: Perhaps a re-read of:
https://docs.djangoproject.com/en/3.1/ref/forms/fields/#fields-which-handle-relationships
This error is occur because your id did not received an id or int value it received a str type of value Wyoming try to pass int value in id
Update
just update your PersonResource Meta class like this
class PersonResource(resources.ModelResource):
state = fields.Field(
column_name='state',
attribute='state',
widget=ForeignKeyWidget(States, 'name'))
class Meta:
model = Person
import_id_fields = ['id']
The default field for object identification is id, you can optionally
set which fields are used as the id when importing
check official doc. for more information.
i tried to make search field to search by author in the admin panel but i got an error
Related Field got invalid lookup: icontains
i follow the documentation and other stackoverflow question but it doesn't work
#model.py
from django.contrib.auth import get_user_model
User = get_user_model()
class Author(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
def __str__(self):
return str(self.user)
# Create your models here.
class Post(models.Model):
title = models.CharField(max_length=256)
content = models.TextField(verbose_name='content')
date_published = models.DateTimeField(auto_now_add=True)
date_edited = models.DateTimeField(auto_now=True)
author = models.ForeignKey(Author, on_delete=models.CASCADE)
thumbnail = models.ImageField(blank=True)
def __str__(self):
return self.title
#admin.py
from django.contrib import admin
from .models import Post, Author,
class PostAdmin(admin.ModelAdmin):
list_display = ['title',
'date_published',
'date_edited',
'author', ]
search_fields = ['title',
'author__user',]
admin.site.register(Post, PostAdmin)
admin.site.register(Author)
it works when i changed the search_field[1] to author__id, but since it only accept id, it can't get the username. any idea how to solve it? should i make custom user model?
I have a model, configuration, in Django and wish to fill the author field with get_username
Can this be done within the model or must it be done from the form? If it must be on the form, how can I change the standard admin page to have this functionality?
At present, the model reads thus:
class Configuration(models.Model):
title = models.CharField(max_length=100,unique=True,blank=False)
author = models.CharField(max_length=50,blank=False)
created = models.DateTimeField("date created",auto_now_add=True)
modified = models.DateTimeField("date modified",auto_now=True)
description = models.CharField(max_length=512)
drawing = models.ForeignKey(Drawing)
instruments = models.ManyToManyField(Instrument)
def __unicode__(self):
return self.title
Use models.ForeignKey:
#models.py
from django.contrib.auth.models import User
class Configuration(models.Model):
author = models.ForeignKey(User)
...
#admin.py:
class Configuration_admin(admin.ModelAdmin):
fields = ('title', 'author',....)
something like that:
from django.contrib.auth.models import User
class ...
...
username = models.ForeignKey(User)
If you want to make some relationship between your model and default User model then you can extends the User model into your own custom model , like this:
models.py
from django.contrib.auth.models import User
class Configuration(models.Model):
author = models.OneToOneField(User)
..
..