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django model inheritance and admin application

I have a django application in which I want to use model inheritance. The application contains one super model class Article and here is its code
class Article(models.Model):
english_title = CharField(max_length=200)
arabic_title = CharField(max_length=200)
english_body = HTMLField()
arabic_body = HTMLField()
enabled = BooleanField()
def __unicode__(self):
return self.english_title
def get_body(self, locale):
if locale == "ar" :
return self.arabic_body
else:
return self.english_body
def get_title(self, locale):
if locale == "ar" :
return self.arabic_title
else:
return self.english_title
and there is a child class called History which extends this class and here is its code
class History(Article, IHasAttachments):
date = DateField(auto_now_add=True)
My problem appears in the admin application where the dateField (date) in the History model does not appear in the admin form when inserting new entry.
NOTE: I am using django-tinymce, djnago-filebrowser, and django-grappelli
What would be the problem?

I think the problem is in your History model you set auto_now_add=True, which will prevent your date field shown on admin, please check the django document about Model field reference:
As currently implemented, setting auto_now or auto_now_add to True will cause the field to have editable=False and blank=True set.
And about Field.editable:
If False, the field will not be displayed in the admin or any other ModelForm.
If you want it editable but also has a default value, try this:
class History(Article, IHasAttachments):
date = DateField(default=datetime.date.today)

Perhaps its inheritance problem, try with mixing:
http://www.linuxjournal.com/node/4540/print

If you add auto_now=True this field will be update at every modification. So it will logically normal you couldn't edit it.
If you want to add automatic date when you create an instance, I suggest you this:
from django.utils.timezone import now
class History(Article, IHasAttachments):
date = DateField(default=now())

Django multiple tag field

I'm trying to find a good tutorial for django how to create multiple tags in a model.
For example:
class Tag(models.Model):
name = models.CharField()
class Sample(models.Model):
name = models.CharField()
urlA = models.CharField()
urlB = models.CharField()
tagA = models.ManyToManyField(Tag)
tagB = models.ManyToManyField(Tag)
I would like to display the tags as an input field and separate by ',' and split in the save method. So I'd like to see 2 different input for the 2 tag field.
If you have an easy way to do or know a good tutorial, please tell me! :)
Thank you guys!

Edit: you do not have to have the actual table sets over laid. You can generate any queryset you want to inn your views. Your url conf can be set up to display the detail view from multiple url. If i am still not understanding then please refine your question.
For having multiple anything tags categories your either going m21 or m2m. So when you create your tags you can add them one by one. Are you familiar with what the Django ORM has to offer with some of its admin functionality? Please give the documentation a good look through. Your approach to this problem is anything but reasonable. Not trying to rub you the wrong way I'm no genius. You would do something like so.
class Tag(models.Model):
title = models.CharField(max_length=250, blank=True)
slug = models.SlugField(blank=True
class Meta:
verbose_name = "tag"
verbose_name_plural = "tags"
ordering = ['title']
#models.permalink
def get_absolute_url(self):
return "/tags/%s/" % self.slug
class Entry(models.Model):
title = models.CharField(max_length=250, blank=True)
body = models.TextField()
tags = models.ManyToMany('Tag')
slug = models.SlugField()
#models.permalink
def get_absolute_url(self):
return "/blog/%s/" % self.slug
There's a little more code to be done for the EntryAdmin and the TagAdmin models, Many other things that can be done as well. I am not sure what you are trying to achieve with that if you could be more clear? Thank you, the above is a rough illustration of how I would approach it.

I found a solution from here:
https://dev.to/thepylot/how-to-add-tags-to-your-models-in-django-django-packages-series-1-3704
django-taggit is very useful for tagging.
It is
a reusable application that primarily offers you a Tag model, and a manager for easily adding tags to any model.
pip install django-taggit
After that, open settings.py and edit the installed apps section:
INSTALLED_APPS = [
...
'taggit'
]
After that, edit your model and add tags like this:
tags = TaggableManager()
The TaggableManager will show up automatically as a field in a ModelForm or in the admin.
Documentation: https://django-taggit.readthedocs.io/en/latest/index.html

django forms logged in user problem

I am writing an application to help employees track projects the are working on.
Part of the form should allow logged in employees to click a drop down and see all projects they are working on. This is the part I am struggling with; specifically getting ONLY the logged in user's projects populated in a drop down. Any help or insight is much appreciated. Thanks……
models.py
class Photo(models.Model):
image = models.ImageField(upload_to='uploads/images/photo')
title = models.CharField(max_length=50)
def __unicode__(self):
return self.title
class Employee(models.Model):
user = models.ForeignKey(User, unique=True)
photo = models.ImageField(upload_to='uploads/images')
department = models.ForeignKey(Department, null=True)
phone = PhoneNumberField("Phone")
def __unicode__(self):
return self.user.get_full_name()
class Projects(models.Model):
name = models.CharField(max_length=40)
student = models.ForeignKey(Student)
photos = models.ManyToManyField(Photo, blank=True, null=True)
forms.py
class ProjectsForm(forms.ModelForm):
employee = get_object_or_404(Employee, user=user)
employee_projects = employee.projects_set.all()
name = forms.ModelChoiceField(queryset=employee_projects,
empty_label="(Select a Project)", required=True)
class Meta:
model = Projects

You need to put first two lines from ProjectsForm class definition to its initialization method and change them a bit.
class ProjectsForm(forms.ModelForm):
name = forms.ModelChoiceField(queryset=Employee.objects.all(),
empty_label="(Select a Project)", required=True)
class Meta:
model = Projects
def __init__(self, user, *args, **kwargs):
super(self, ProjectsForm).init(*args, **kwargs)
employee = get_object_or_404(Employee, user=user)
self.fields['name'].queryset = employee.projects_set.all()
Now, some explanation. Hope someone will find it useful.
In your original ProjectsForm definition, you're trying to get employee's projects when your class is defined. But this happens once your forms.py file is compiled, which takes place rarely (when you change code, for example). Also, you of course have no data that is necessary to filter projects, i.e., no user object, at that stage.
Instead, you need to do this each time the class is initialized. Initialization in Python takes place in special __init__() method. For example, when you're doing something like
form = ProjectsForm(data=request.POST)
in your view, what happens is that ProjectsForm.__init__(data=request.POST) is called to initialize ProjectsForm into an instance.
So, in our new definition, we're requiring a new argument (user) to be passed to the form when it's instantiated. So you can do something like this in your view:
form = ProjectsForm(request.user, data=request.POST)
In new initialization method, first we're calling the initialization method of parent class (which does some internal django things and should be called anyway), then use argument user to get related employee, and then assign a queryset with that employee's projects to the name field.
Sorry if I'm being too verbose.
See also:
Django, passing information when modifying queryset for ModelForm
django - dynamic select fields in forms

Why not have a many to many field in the Employee model that points to all the projects the employee can work on?
https://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.ManyToManyField
I think that would be the best way to do it.

Implementing USZipCodeField and USStateField in django

I'm looking to implement a zipcode field in django using the form objects from localflavor, but not quite getting them to work. I want to have a zipcode field in a form (or ModelForm in my case), but the fields never validate as a zipcode when calling _get_errors() on the form object. The way I'm implementing it seems right to me but is apparently wrong, does anyone know what the right way to do this might be?
I have a ModelForm that I want to use zipcode (and also USStateField) in:
from django.contrib.localflavor.us.forms import USStateField
from django.contrib.localflavor.us.forms import USZipCodeField
class FooForm(ModelForm):
class Meta:
model = Bar
fields = ('address', #This form uses a subset of fields from the model
'address_apt',
'address_city',
'address_state',
'address_zip',
'home_phone',
'mobile_phone')
widgets= {
'address_zip' : USZipCodeField(),
'address_state' : USStateField(),
}
The ModelForm 'FooForm' links to a model that looks like:
from django.contrib.localflavor.us import models as usmodels
class Bar(models.Model):
db_table = 'BAR'
address = models.CharField(max_length=255)
address_apt = models.CharField(max_length=40, blank=True)
address_city = models.CharField(max_length=90)
address_state = usmodels.USStateField()
address_zip = models.CharField(max_length=15)
home_phone = usmodels.PhoneNumberField( )
mobile_phone = usmodels.PhoneNumberField( )
#... There are more fields in the model...
But if I create an instance of the form and run it's validation, it never cares about the form level validation, only the model level validation:
foo_instance = FooForm(request.POST)
#Let's assume request.POST looks like:
#<QueryDict: {u'address_city': [u'asdf'], u'mobile_phone': [u'asdf'], u'address_state': [u'California'], u'home_phone': [u'asdf'], [u'1'], u'address': [u'123 foo'], u'address_zip': [u'asdf']}>
foo_instance._get_errors()
Yields:
<ul class="errorlist">
<li>mobile_phone<ul class="errorlist">
<li>Phone numbers must be in XXX-XXX-XXXX format.</li></ul>
</li><li>home_phone<ul class="errorlist">
<li>Phone numbers must be in XXX-XXX-XXXX format.</li></ul>
</li></ul>
I need to be able to call validation on the populated form object and have it tell me that the zipcode is formated improperly if so. Doing something wrong, just don't know what atm.

Using widgets declaratively has literally only just been added to the trunk SVN version in the last day or so. If you're using an older checkout, or a released version, it won't work - you'll need to go back to the old way of doing it, by overriding the field declarations at the top level of the form.

admin template for manytomany

I have a manytomany relationship between publication and pathology. Each publication can have many pathologies. When a publication appears in the admin template, I need to be able to see the many pathologies associated with that publication. Here is the model statement:
class Pathology(models.Model):
pathology = models.CharField(max_length=100)
def __unicode__(self):
return self.pathology
class Meta:
ordering = ["pathology"]
class Publication(models.Model):
pubtitle = models.TextField()
pathology = models.ManyToManyField(Pathology)
def __unicode__(self):
return self.pubtitle
class Meta:
ordering = ["pubtitle"]
Here is the admin.py. I have tried variations of the following, but always
get an error saying either publication or pathology doesn't have a foreign key
associated.
from myprograms.cpssite.models import Pathology
class PathologyAdmin(admin.ModelAdmin):
# ...
list_display = ('pathology', 'id')
admin.site.register(Pathology, PathologyAdmin)
class PathologyInline(admin.TabularInline):
#...
model = Pathology
extra = 3
class PublicationAdmin(admin.ModelAdmin):
# ...
ordering = ('pubtitle', 'year')
inlines = [PathologyInline]
admin.site.register(Publication,PublicationAdmin)
Thanks for any help.

Unless you are using a intermediate table as documented here http://docs.djangoproject.com/en/dev/ref/contrib/admin/#working-with-many-to-many-intermediary-models, I don't think you need to create an Inline class. Try removing the line includes=[PathologyInline] and see what happens.

I realize now that Django is great for the administration (data entry) of a website, simple searching and template inheritance, but Django and Python are not very good for complex web applications, where data is moved back and forth between a database and an html template. I have decided to combine Django and PHP, hopefully, applying the strengths of both. Thanks for you help!

That looks more like a one-to-many relationship to me, tho I'm somewhat unclear on what exactly Pathologies are. Also, so far as I understand, Inlines don't work on manytomany. That should work if you flip the order of the models, remove the manytomany and add a ForeignKey field to Publication in Pathology.
class Publication(models.Model):
pubtitle = models.TextField()
def __unicode__(self):
return self.pubtitle
class Meta:
ordering = ["pubtitle"]
class Pathology(models.Model):
pathology = models.CharField(max_length=100)
publication = models.ForeignKey(Publication)
def __unicode__(self):
return self.pathology
class Meta:
ordering = ["pathology"]