how to make django comment model field Site not required - python

I'm using django-threadedcomments from ericflo on github. This app simply extends the native django comments framework. I am running into the same issue with both frameworks. I continue to get an error relating to mysql that site_id cannot be null. I have no use for the Site field in my comments. I tried to extend the Comment model with my own making site both blank and null but I am still getting the same error. What is the proper way to override that requirement? Thanks
I tried:
class Comment(Comment):
site=models.ForeignKey(Site,null=True,blank=True)

I found it easier to just define one Site object. django-threadedcomments is not the only extension which requires that.

You will not be able to change this without monkey-patching the current model, but it shouldn't be a big deal setting the site field to Site.objects.get_current() in the view/form when saving a comment!

Related

Creating migrations for auth app in django

I need to add a couple fields to Group model in django contrib.auth app using:
field_name = models.CharField(...)
field_name.contribute_to_class(Group, 'field_name')
My issue is while creating the migrations with South, since they are created in the "migrations" directory inside the auth app and, since the system is already in production, I'm not allowed to alter the current django installation in the server in order to migrate auth.
Does anyone know how to create and load such migrations?
Thanks in advance for your help.
Django doesn't make it particularly easy to modify the standard models. I wouldn't recommend that you sublass Group, because it's quite annoying to get built-in functionality to reference the new model instead.
The usual thing to do here is to create a GroupProfile model that has a Group as a unique foreign key. It might not be elegant, but it won't have the huge maintenance overhead that comes with forking Django's source code.
Also: if you can't modify the Django code on the server, you aren't going to be able to do this with raw SQL hackery or a clever migration. South isn't going to be the problem -- the problem is that the Django ORM is going to notice that there are fields present in the SQL table that aren't specified in the code, which will cause it to throw an exception.
Since you use a hack to patch up the model, I think you should write a migration manually. Try copying another migration and changing add_column and models first, if it fails - there is always an option named "raw sql" :)

How do you add a new entry into the django admin index?

I'm working on a Django project, and I've created some custom admin views using the get_urls override method described in the documentation. It works perfectly. There is just one problem. There is no way to get to this custom admin view unless you already know the URL.
There are some ways I already know of to add a link to this view somewhere in the admin, but none of them are satisfactory. I want a link to the custom view to appear in the model listings right with all the model admins. I just don't want it to have +add or +change links next to it because it isn't a model.
I could just override the admin_site or the template, but this is no good. It puts the customization on the project level instead of the app level. It also will only put the link on the /admin/ page and not the /admin/myapp/ page.
I could also just easily add the link in a different location by overriding the app_index.html template, but that is not exactly a convenient or intuitive place to look for it.
Another solution I came up with is to create a blank model and register a blank admin for it. Then steal the url patterns for that model so clicking on its entry goes to my custom view instead of to a blank add/change view. That works, but it's an incredibly ugly hack.
Here is a picture of what I'm trying to achieve.
I still think the correct way of doing this is overwriting some parts of django admin templates. There is no easy way of adding these links.
https://docs.djangoproject.com/en/dev/ref/contrib/admin/#overriding-vs-replacing-an-admin-template
I also found this article http://coffeeonthekeyboard.com/o-hai-django-adminplus-568/ which also suggests that django-adminplus is a good tool for doing this. Personally I prefer to keep clear of any extra dependancies and would still use templates - but thats up to you.
Have you tried this app: https://github.com/jsocol/django-adminplus? Even if it does not work for the exact purpose you are trying to achieve, at least it can give you some enlightement by checking out the code
You need to override the template admin/index.html. Thenput a new pair of tags after the {% endfor %} on line 40.
You might also be able to solve it using jQuery.

Linking google-appengine native model to Django model

I've ported my GAE project to django-nonrel, and now I would like to have link from my object to Django User object:
class Opinion(google.appengine.ext.db.Model):
...
author = db.ReferenceProperty(django.contrib.auth.User)
Unfortunately, that's not possible, since you only can link GAE models this way.
Question - what's the best way to solve this? Is it possible or should I work it around somehow?
I don't want to migrate my old GAE model since I already have a bunch of data there.
My initial googling and searching brought me nothing good:
On django-nonrel list I read:
You cannot use appengine model into django-nonrel. You need alter all
your models to django proper. That is, use original
django-registration app not the appengine-patch provided registration.
Another question, exactly like mine, was left unanswered.
... and I've found another place where people recommend switching to Django models.
I guess that's that then, there is no way of doing what I need, one needs to migrate.

Storing/Retrieving/Editing project specific data with Django

I'm new to Django and I'm working on the public website for a small company.
I'm facing an issue that I guess has already been encountered by lots a django noobs,
but I can't manage to find a good solution.
My problem is that there some informations (contact address, office phone number, company description...) that I use in nearly all of my views and are by nature unique (undertand: a database table with only 1 row). I currently store these informations has a model in my databse, but I find it a bit weird issue an additional database request each time (each view)
I need to access them. However, I need my client to be able to edit these informations (by the admin interface).
So, please, is there a django idiom to handle such an use case ?
Thx in advance.
If you look into caching solutions, they will probably do what you need.
The general queryset caching solution I use in johnny-cache, but for what you need, you can probably just load it up from the db and store it in the cache.
What you want to do is use select_related('contact_profile','office_data') etc when you query the items in your view, and in the admin, instead of registering all the data separately just use the InlineAdmin class for the Admin site and you will be able to edit all the information as if it was a single entity.
Check out the django docs for more information.

Form validation in django

I just started to use django. I came across forms and I need to know which one is the better way to validate a forms. Would it be using django forms or should we use javascript or some client side scripting language to do it?
You should ALWAYS validate your form on the server side, client side validation is but a convenience for the user only.
That being said, Django forms has a variable form.errors which shows if certain form fields were incorrect.
{{ form.name_of_field.errors }} can give you each individual error of each field that's incorrectly filled out. See more here:
http://docs.djangoproject.com/en/dev/topics/forms/
There's a pluggable Django app (django-ajax-forms) that helps validate forms on the client side through JavaScript. But as AlbertoPL says, use client side validation only as a usability measure (e.g. telling a user that his desired username is already taken without reloading the registration page). There are all kind of ways to sidestep client side validation, in most cases as simple as deactivating JavaScript.
Generally speaking: presume all data coming from the outside as faulty until validated.
Just came accross django-floppyforms, which seems to do clientside validation by default. They use HTML5 which supports clientside validation by default. Not sure if they also use javascript though if the browser doesn't support HTML5. Haven't tried it myself yet.
Link to django-floppyforms: Documentation and Github
If you are using bootstrap then you can simply add required attribute in forms field. For example if there is programe field then you can validate it like:
In forms.py:
programme = forms.ChoiceField(course_choices,required=True, widget=forms.Select(attrs={'required':'required'}))
Note: It requires to link to bootstrap files in your .html page of that form.
You will need to do this is JS. This app integrates forms with parsley.js to tag the forms with correct data-* attributes automatically.

Categories