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• Select a subset of foreign key elements in inlineformset_factory in Django

  - by Enis Afgan

  Hello, I have a model with two foreign keys: class Model1(models.Model): model_a = models.ForeignKey(ModelA) model_b = models.ForeignKey(ModelB) value = models.IntegerField() Then, I create an inline formset class, like so: an_inline_formset = inlineformset_factory(ModelA, Model1, fk_name="model_a") and then instantiate it, like so: a_formset = an_inline_formset(request.POST, instance=model_A_object) Once this formset gets rendered in a template/page, there is ChoiceField associated with the model_b field. The problem I'm having is that the elements in the resulting drop down menu include all of the elements found in ModelB table. I need to select a subset of those based on some criteria from ModelB. At the same time, I need to keep the reference to the instance of model_A_object when instantiating inlineformset_factory and, therefore, I can't just use this example. Any suggestions?

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• How do I flag only one of the formsets in django admin ?

  - by azuer88

  I have these (simplified) models: class Question(models.Model): question = models.CharField(max_length=60) class Choices(models.Model): question = models.ForeignKey(Question) text = models.CharField(max_length=60) is_correct = models.BooleanField(default=False) I've made Choices as an inline of Question (in admin). Is there a way to make sure that only one Choice will have is_correct = True? Ideally, is_correct will be displayed as a radio button when it is displayed in the admin formset (TabularInline). My first solution was to override the validation of the formset but did understand how to do this. (is_correct is displayed as checkbox, and I'd display an error that only one is_correct should be selected.)

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• A Django form for entering a 0 to n email addresses

  - by Erik

  I have a Django application with some fairly common models in it: UserProfile and Organization. A UserProfile or an Organization can both have 0 to n emails, so I have an Email model that has a GenericForeignKey. UserProfile and Organization Models both have a GenericRelation called emails that points back to the Email model (summary code provided below). The question: what is the best way to provide an Organization form that allows a user to enter organization details including 0 to n email addresses? My Organization create view is a Django class-based view. I'm leading towards creating a dynamic form and an enabling it with Javascript to allow the user to add as many email addresses as necessary. I will render the form with django-crispy-forms. I've thought about doing this with a formset embedded within the form, but this seems like overkill for email addresses. Embedding a formset in a form delivered by a class-based view is cumbersome too. Note that the same issue occurs with the Organization fields phone_numbers and locations. emails.py: from django.db import models from parent_mixins import Parent_Mixin class Email(Parent_Mixin,models.Model): email_type = models.CharField(blank=True,max_length=100,null=True,default=None,verbose_name='Email Type') email = models.EmailField() class Meta: app_label = 'core' organizations.py: from emails import Email from locations import Location from phone_numbers import Phone_Number from django.contrib.contenttypes import generic from django.db import models class Organization(models.Model): active = models.BooleanField() duns_number = models.CharField(blank=True,default=None,null=True,max_length=9) # need to validate this emails = generic.GenericRelation(Email,content_type_field='parent_type',object_id_field='parent_id') legal_name = models.CharField(blank=True,default=None,null=True,max_length=200) locations = generic.GenericRelation(Location,content_type_field='parent_type',object_id_field='parent_id') name = models.CharField(blank=True,default=None,null=True,max_length=200) organization_group = models.CharField(blank=True,default=None,null=True,max_length=200) organization_type = models.CharField(blank=True,default=None,null=True,max_length=200) phone_numbers = generic.GenericRelation(Phone_Number,content_type_field='parent_type',object_id_field='parent_id') taxpayer_id_number = models.CharField(blank=True,default=None,null=True,max_length=9) # need to validate this class Meta: app_label = 'core' parent_mixins.py from django.contrib.contenttypes.models import ContentType from django.contrib.contenttypes import generic from django.db import models class Parent_Mixin(models.Model): parent_type = models.ForeignKey(ContentType,blank=True,null=True) parent_id = models.PositiveIntegerField(blank=True,null=True) parent = generic.GenericForeignKey('parent_type', 'parent_id') class Meta: abstract = True app_label = 'core'

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• Django formsets: make first required?

  - by Mark

  These formsets are exhibiting exactly the opposite behavior that I want. My view is set up like this: def post(request): # TODO: handle vehicle formset VehicleFormSetFactory = formset_factory(VehicleForm, extra=1) if request.POST: vehicles_formset = VehicleFormSetFactory(request.POST) else: vehicles_formset = VehicleFormSetFactory() And my template looks like this: <div id="vehicle_forms"> {{ vehicles_formset.management_form }} {% for form in vehicles_formset.forms %} <h4>Vehicle {{forloop.counter}}</h4> <table> {% include "form.html" %} </table> {% endfor %} </div> That way it initially generates only 1 form, like I want. But I want that one form to be required! When I dynamically add blank forms with JavaScript and vehicles_formset.empty_form all those extra forms are required, which I don't want. From the docs: The formset is smart enough to ignore extra forms that were not changed. This is the behavior the first form is exhibiting (not what I want) but not the behavior that the extra forms are exhibiting (what I do want). Is there some attribute I can can change to at least make one form required?

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• raw_id_fields for modelforms

  - by nbv4

  I have a modelform which has one field that is a ForeignKey value to a model which as 40,000 rows. The default modelform tries to create a select box with 40,000 options, which, to say the least is not ideal. Even more so when this modelform is used in a formset factory! In the admin, this is easiely avoidable by using "raw_id_fields", but there doesn't seem to be a modelform equivalent. How can I do this? Here is my modelform: class OpBaseForm(ModelForm): base = forms.CharField() class Meta: model = OpBase exclude = ['operation', 'routes'] extra = 0 raw_id_fields = ('base', ) #does nothing The first bolded line works by not creating the huge unwieldy selectbox, but when I try to save a fieldset of this form, I get the error: "OpBase.base" must be a "Base" instance. In order for the modelform to be saved, 'base' needs to be a Base instance. Apparently, a string representation of a Base primary key isn't enough (at least not automatically). I need some kind of mechanism to change the string that is given my the form, to a Base instance. And this mechanism has to work in a formset. Any ideas? If only raw_id_fields would work, this would be easy as cake. But as far as I can tell, it only is available in the admin.

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• Django unit testing: South-migrated DB works in MySQL, throws duplicate PK error in PostGreSQL. Am I

  - by unclaimedbaggage

  Hi folks, (Worth starting off with a disclaimer: I'm very new to PostGreSQL) I have a django site which involves a standard app/tests.py testing file. If I migrate the DB to MySQL (through South),, the tests all pass. However in PostGresQL, I'm getting the following error: IntegrityError: duplicate key value violates unique constraint "business_contact_pkey" Note this happens while unit testing only - the actual page runs fine in both MySQL & PostGresql. Really having a heckuva time figuring this one out. Anyone have ideas? Below are the Postgresql "\d business_contact" & offending tests.py method if they help. No changes made to either DB except the (same) South migrations Thanks first_name | character varying(200) | not null mobile_phone | character varying(100) | surname | character varying(200) | not null business_id | integer | not null created | timestamp with time zone | not null deleted | boolean | not null default false updated | timestamp with time zone | not null slug | character varying(150) | not null phone | character varying(100) | email | character varying(75) | id | integer | not null default nextval('business_contact_id_seq'::regclass) Indexes: "business_contact_pkey" PRIMARY KEY, btree (id) "business_contact_slug_key" UNIQUE, btree (slug) "business_contact_business_id" btree (business_id) Foreign-key constraints: "business_id_refs_id_772cc1b7b40f4b36" FOREIGN KEY (business_id) REFERENCES business(id) DEFERRABLE INITIALLY DEFERRED Referenced by: TABLE "business" CONSTRAINT "primary_contact_id_refs_id_dfaf59c4041c850" FOREIGN KEY (primary_contact_id) REFERENCES business_contact(id) DEFERRABLE INITIALLY DEFERRED TEST DEF: def test_add_business_contact(self): """ Add a business contact """ contact_slug = 'test-new-contact-added-new-adf' business_id = 1 business = Business.objects.get(id=business_id) postdata = { 'first_name': 'Test', 'surname': 'User', 'business': '1', 'slug': contact_slug, 'email': '[email protected]', 'phone': '12345678', 'mobile_phone': '9823452', 'business': 1, 'business_id': 1, } #Test to ensure contacts that should not exist are not returned contact_not_exists = Contact.objects.filter(slug=contact_slug) self.assertFalse(contact_not_exists) #Add the contact and ensure it is present in the DB afterwards """ contact_add_url = '%s%s/contact/add/' % (settings.BUSINESS_URL, business.slug) self.client.post(contact_add_url, postdata) added_contact = Contact.objects.filter(slug=contact_slug) print added_contact try: self.assertTrue(added_contact) except: formset = ContactForm(postdata) print formset.errors self.assertFalse(True, "Contact not found in the database - most likely, the post values in the test didn't validate against the form")

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• Django: customizing the message after a successful form save

  - by chiurox

  Hello, whenever I save a model in my Admin interface, it displays the usual "successfully saved message." However, I want to know if it's possible to customize this message because I have a situation where I want to warn the user about what he just saved and the implications of these actions. class PlanInlineFormset(forms.models.BaseInlineFormset): def clean(self): ### How can I detect the changes? ### (self.changed_data doesn't work because it's an inline) ### and display what he/she just changed at the top AFTER the successful save? class PlanInline(admin.TabularInline): model = Plan formset = PlanInlineFormset

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• Django: How to iterate over formsets and access cleaned data?

  - by Mark

  What if I want to do something with my formset other than immediately saving it? How can I do this? for form in vehicles_formset.forms: listing.id = None listing.vehicle_year = form.cleaned_data['year'] listing.vehicle_make = form.cleaned_data['make'] listing.vehicle_model = form.cleaned_data['model'] listing.vin = form.cleaned_data['vin'] listing.vehicle_runs = form.cleaned_data['runs'] listing.vehicle_convertible = form.cleaned_data['convertible'] listing.vehicle_modified = form.cleaned_data['modified'] listing.save() (Thus creating multiple listings) Apparently cleaned_data does not exist. There's a bunch of stuff in the data dict like form-0-year but it's pretty useless to me like that.

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• Django nested formsets

  - by dragoon

  Hi, I have an edit object view that contains a formset(one or many if this matters), now I want to create a page that can display multiple edit object forms and submit it in a single form. What is the correct way to achieve this task?

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• How to render a text box instead of a select box when using model form set

  - by iHeartDucks

  Hi, When I render my formset, one of the field renders as a select box because it is a foreign field in the model. Is there a way to change this to a text input? I want to populate that field by using Ajax auto complete. Adding a widget to the modelform is not working because the modelformset_factory takes a model and not a model form.

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• Django modelform ForeignKey List

  - by Harry

  How do you get each item in the ForeignKey field in a list, for example: class Delegate(models.Model): excursion = models.ForeignKey(Excursion, limit_choices_to = {'is_activity': False}, related_name='excursion', null=True, blank=True) Template: {% for object in formset.excursion_set.all %} {{ object.lable }} etc {% endfor %} My reason is that I don't want the options to display as a dropdown, but in a custom way that I will style etc.

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• UnicodeEncodeError when uploading files in Django admin

  - by Samuel Linde

  Note: I asked this question on StackOverflow, but I realize this might be a more proper place to ask this kind of question. I'm trying to upload a file called 'Testaråäö.txt' via the Django admin app. I'm running Django 1.3.1 with Gunicorn 0.13.4 and Nginx 0.7.6.7 on a Debian 6 server. Database is PostgreSQL 8.4.9. Other Unicode data is saved to the database with no problem, so I guess the problem must be with the filesystem somehow. I've set http { charset utf-8; } in my nginx.conf. LC_ALL and LANG is set to 'sv_SE.UTF-8'. Running 'locale' verifies this. I even tried setting LC_ALL and LANG in my nginx init script just to make sure locale is set properly. Here's the traceback: Traceback (most recent call last): File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/handlers/base.py", line 111, in get_response response = callback(request, *callback_args, **callback_kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 307, in wrapper return self.admin_site.admin_view(view)(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 93, in _wrapped_view response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/views/decorators/cache.py", line 79, in _wrapped_view_func response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/sites.py", line 197, in inner return view(request, *args, **kwargs) File "/srv/django/letebo/app/cms/admin.py", line 81, in change_view return super(PageAdmin, self).change_view(request, obj_id) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 28, in _wrapper return bound_func(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 93, in _wrapped_view response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 24, in bound_func return func(self, *args2, **kwargs2) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/transaction.py", line 217, in inner res = func(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 985, in change_view self.save_formset(request, form, formset, change=True) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 677, in save_formset formset.save() File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 482, in save return self.save_existing_objects(commit) + self.save_new_objects(commit) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 613, in save_new_objects self.new_objects.append(self.save_new(form, commit=commit)) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 717, in save_new obj.save() File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 460, in save self.save_base(using=using, force_insert=force_insert, force_update=force_update) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 504, in save_base self.save_base(cls=parent, origin=org, using=using) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 543, in save_base for f in meta.local_fields if not isinstance(f, AutoField)] File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/fields/files.py", line 255, in pre_save file.save(file.name, file, save=False) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/fields/files.py", line 92, in save self.name = self.storage.save(name, content) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 48, in save name = self.get_available_name(name) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 74, in get_available_name while self.exists(name): File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 218, in exists return os.path.exists(self.path(name)) File "/srv/.virtualenvs/letebo/lib/python2.6/genericpath.py", line 18, in exists st = os.stat(path) UnicodeEncodeError: 'ascii' codec can't encode characters in position 52-54: ordinal not in range(128) I tried running Gunicorn with debugging turned on, and the file uploads without any problem at all. I suppose this must mean that the issue is with Nginx. Still beats me where to look, though. Here are the raw response headers from Gunicorn and Nginx, if it makes any sense: Gunicorn: HTTP/1.1 302 FOUND Server: gunicorn/0.13.4 Date: Thu, 09 Feb 2012 14:50:27 GMT Connection: close Transfer-Encoding: chunked Expires: Thu, 09 Feb 2012 14:50:27 GMT Vary: Cookie Last-Modified: Thu, 09 Feb 2012 14:50:27 GMT Location: http://my-server.se:8000/admin/cms/page/15/ Cache-Control: max-age=0 Content-Type: text/html; charset=utf-8 Set-Cookie: messages="yada yada yada"; Path=/ Nginx: HTTP/1.1 500 INTERNAL SERVER ERROR Server: nginx/0.7.67 Date: Thu, 09 Feb 2012 14:50:57 GMT Content-Type: text/html; charset=utf-8 Transfer-Encoding: chunked Connection: close Vary: Cookie 500 UPDATE: Both locale.getpreferredencoding() and sys.getfilesystemencoding() outputs 'UTF-8'. locale.getdefaultlocale() outputs ('sv_SE', 'UTF8'). This seem correct to me, so I'm still not sure why I keep getting these errors.

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• why are read only form fields in django a bad idea?

  - by jamida

  I've been looking for a way to create a read-only form field and every article I've found on the subject comes with a statement that "this is a bad idea". Now for an individual form, I can understand that there are other ways to solve the problem, but using a read only form field in a modelformset seems like a completely natural idea. Consider a teacher grade book application where the teacher would like to be able to enter all the students' (note the plural students) grades with a single SUBMIT. A modelformset could iterate over all the student-grades in such a way that the student name is read-only and the grade is the editable field. I like the power and convenience of the error checking and error reporting you get with a modelformset but leaving the student name editable in such a formset is crazy. Since the expert django consensus is that read-only form fields are a bad idea, I was wondering what the standard django best practice is for the example student-grade example above?

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• django-admin - how to modify ModelAdmin to create multiple objects at once?

  - by skrobul

  let's assume that I have very basic model class Message(models.Model): msg = models.CharField(max_length=30) this model is registered with admin module: class MessageAdmin(admin.ModelAdmin): pass admin.site.register(Message, MessageAdmin) Currently when I go into the admin interface, after clicking "Add message" I have only one form where I can enter the msg. I would like to have multiple forms (formset perhaps) on the "Add page" so I can create multiple messages at once. It's really annoying having to click "Save and add another" every single time. Ideally I would like to achieve something like InlineModelAdmin but it turns out that you can use it only for the models that are related to the object which is edited. What would you recommend to use to resolve this problem?

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• How to manually render a Django template for an inlineformset_factory with can_delete = True / False

  - by chefsmart

  I have an inlineformset with a custom Modelform. So it looks something like this: MyInlineFormSet = inlineformset_factory(MyMainModel, MyInlineModel, form=MyCustomInlineModelForm) I am rendering this inlineformset manually in a template so that I have more control over widgets and javascript. So I go in a loop like {% for form in myformset.forms %} and then manually render each field as described on this page http://docs.djangoproject.com/en/dev/topics/forms/#customizing-the-form-template The formset has can_delete = True or can_delete = False depending on whether the user is creating new objects or editing existing ones. Question is, how do I manually render the can_delete checkbox?

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• How do I flag only one of the formsets in django admin?

  - by azuer88

  I have these (simplified) models: class Question(models.Model): question = models.CharField(max_length=60) class Choices(models.Model): question = models.ForeignKey(Question) text = models.CharField(max_length=60) is_correct = models.BooleanField(default=False) I've made Choices as an inline of Question (in admin). Is there a way to make sure that only one Choice will have is_correct = True? Ideally, is_correct will be displayed as a radio button when it is displayed in the admin formset (TabularInline). my admin.py has: from django.contrib import admin class OptionInline(admin.TabularInline): model = Option extra = 5 max_num = 5 class QuestionAdmin(admin.ModelAdmin): inlines = [OptionInline, ] admin.site.register(QType) admin.site.register(Question, QuestionAdmin)

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• Inline Form Validation in Django

  - by allanhenderson

  Newbie request that seems difficult to implement. I would like to make an entire inline formset within an admin change form compulsory.. so in my current scenario when I hit save on an Invoice form (in Admin) the inline Order form is blank. I'd like to stop people creating invoices with no orders associated. Anyone know an easy way to do that? Normal validation like (required=True) on the model field doesn't appear to work in this instance. Thanks!!

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• Deleting inline model in Django admin

  - by Overclocked

  I am trying to use Django admin to remove an inline model. I get an error saying [u'Select a valid choice. That choice is not one of the available choices.'] I check the location where the exception is raised. It seems like the inline model was removed, then a form validation happened on the parent model with the deleted inline model still as value of the form. That caused the failure. In my admin model, I had a save_formset method defined, that called formset.save(commit=False). Is the commit=False not telling Django to not delete the related models? Thanks.

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• Custom Django admin URL + changelist view for custom list filter by Tags

  - by Botondus

  In django admin I wanted to set up a custom filter by tags (tags are introduced with django-tagging) I've made the ModelAdmin for this and it used to work fine, by appending custom urlconf and modifying the changelist view. It should work with URLs like: http://127.0.0.1:8000/admin/reviews/review/only-tagged-vista/ But now I get 'invalid literal for int() with base 10: 'only-tagged-vista', error which means it keeps matching the review edit page instead of the custom filter page, and I cannot figure out why since it used to work and I can't find what change might have affected this. Any help appreciated. Relevant code: class ReviewAdmin(VersionAdmin): def changelist_view(self, request, extra_context=None, **kwargs): from django.contrib.admin.views.main import ChangeList cl = ChangeList(request, self.model, list(self.list_display), self.list_display_links, self.list_filter, self.date_hierarchy, self.search_fields, self.list_select_related, self.list_per_page, self.list_editable, self) cl.formset = None if extra_context is None: extra_context = {} if kwargs.get('only_tagged'): tag = kwargs.get('tag') cl.result_list = cl.result_list.filter(tags__icontains=tag) extra_context['extra_filter'] = "Only tagged %s" % tag extra_context['cl'] = cl return super(ReviewAdmin, self).changelist_view(request, extra_context=extra_context) def get_urls(self): from django.conf.urls.defaults import patterns, url urls = super(ReviewAdmin, self).get_urls() def wrap(view): def wrapper(*args, **kwargs): return self.admin_site.admin_view(view)(*args, **kwargs) return update_wrapper(wrapper, view) info = self.model._meta.app_label, self.model._meta.module_name my_urls = patterns('', # make edit work from tagged filter list view # redirect to normal edit view url(r'^only-tagged-\w+/(?P<id>.+)/$', redirect_to, {'url': "/admin/"+self.model._meta.app_label+"/"+self.model._meta.module_name+"/%(id)s"} ), # tagged filter list view url(r'^only-tagged-(P<tag>\w+)/$', self.admin_site.admin_view(self.changelist_view), {'only_tagged':True}, name="changelist_view"), ) return my_urls + urls Edit: Original issue fixed. I now receive 'Cannot filter a query once a slice has been taken.' for line: cl.result_list = cl.result_list.filter(tags__icontains=tag) I'm not sure where this result list is sliced, before tag filter is applied. Edit2: It's because of the self.list_per_page in ChangeList declaration. However didn't find a proper solution yet. Temp fix: if kwargs.get('only_tagged'): list_per_page = 1000000 else: list_per_page = self.list_per_page cl = ChangeList(request, self.model, list(self.list_display), self.list_display_links, self.list_filter, self.date_hierarchy, self.search_fields, self.list_select_related, list_per_page, self.list_editable, self)

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• Multiple model forms with some pre-populated fields

  - by jimbocooper

  Hi! Hope somebody can help me, since I've been stuck for a while with this... I switched to another task, but now back to the fight I still can't figure out how to come out from the black hole xD The thing is as follows: Let's say I've got a product model, and then a set of Clients which have rights to submit data for the products they've been subscribed (Many to Many from Client to Product). Whenever my client is going to submit data, I need to create as many forms as products he's subscribed, and pre-populate each one of them with the "product" field as long as perform a quite simple validation (some optional fields have to be completed if it's client's first submission). I would like one form "step" for each product submission, so I've tried formWizards... but the problem is you can't pre-assign values to the forms... this can be solved afterwards when submitting, though... but not the problem that it doesn't allow validation either, so at the end of each step I can check some data before rendering next step. Then I've tried model formsets, but then there's no way to pre-populate the needed fields. I came across some django plugins, but I'm not confident yet if any of them will make it.... Did anybody has a similar problem so he can give me a ray of light? Thanks a lot in advance!! :) edit: The code I used in the formsets way is as follows: prods = Products.objects.filter(Q(start_date__lte=today) & Q(end_date__gte=today), requester=client) num = len(prods) PriceSubmissionFormSet = modelformset_factory(PriceSubmission, extra=num) formset = PriceSubmissionFormSet(queryset=PriceSubmission.objects.none())

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• Limit foreign key choices in select in an inline form in admin

  - by mightyhal

  Edited :-) Hopefully a bit clearer now. The logic is of the model is: A Building has many Rooms A Room may be inside another Room (a closet, for instance--ForeignKey on 'self') A Room can only in inside of another Room in the same building (this is the tricky part) Here's the code I have: #spaces/models.py from django.db import models class Building(models.Model): name=models.CharField(max_length=32) def __unicode__(self): return self.name class Room(models.Model): number=models.CharField(max_length=8) building=models.ForeignKey(Building) inside_room=models.ForeignKey('self',blank=True,null=True) def __unicode__(self): return self.number and: #spaces/admin.py from ex.spaces.models import Building, Room from django.contrib import admin class RoomAdmin(admin.ModelAdmin): pass class RoomInline(admin.TabularInline): model = Room extra = 2 class BuildingAdmin(admin.ModelAdmin): inlines=[RoomInline] admin.site.register(Building, BuildingAdmin) admin.site.register(Room) The inline will display only rooms in the current building (which is what I want). The problem, though, is that for the inside_room drop down, it displays all of the rooms in the Rooms table (including those in other buildings). In the inline of rooms, I need to limit the inside_room choices to only rooms which are in the current building being displayed by the main form. I can't figure out a way to do it with either a limit_choices_to in the model, nor can I figure out how exactly to override the admin's inline formset properly (I feel like I should be somehow create a custom inline form, pass the building_id of the main form to the custom inline, then limit the queryset for the field's choices based on that--but I just can't wrap my head around how to do it). Maybe this is too complex for the admin site, but it seems like something that would be generally useful... Thanks again for your help!

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• Django form linking 2 models by many to many field.

  - by Ed

  I have two models: class Actor(models.Model): name = models.CharField(max_length=30, unique = True) event = models.ManyToManyField(Event, blank=True, null=True) class Event(models.Model): name = models.CharField(max_length=30, unique = True) long_description = models.TextField(blank=True, null=True) I want to create a form that allows me to identify the link between the two models when I add a new entry. This works: class ActorForm(forms.ModelForm): class Meta: model = Actor The form includes both name and event, allowing me to create a new Actor and simultaneous link it to an existing Event. On the flipside, class EventForm(forms.ModelForm): class Meta: model = Event This form does not include an actor association. So I am only able to create a new Event. I can't simultaneously link it to an existing Actor. I tried to create an inline formset: EventFormSet = forms.models.inlineformset_factory(Event, Actor, can_delete = False, extra = 2, form = ActorForm) but I get an error <'class ctg.dtb.models.Actor'> has no ForeignKey to <'class ctg.dtb.models.Event'> This isn't too surprising. The inlineformset worked for another set of models I had, but this is a different example. I think I'm going about it entirely wrong. Overall question: How can I create a form that allows me to create a new Event and link it to an existing Actor?

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• Moving from MySQL to MySQLi? I have the code here but I don't get it

  - by MuqMan

  I have posted the code there, please help me out as I am a newbie, I don't know much in terms of deprecation and stuff. <?php session_start(); include('settings.php'); $issub = $_POST['issub']; if($issub == "yes") { require('settings.php'); $dbcon = mysql_connect($dbhost, $dbuser, $dbpword); if(!dbcon) { die('Could not connect'.mysql_error()); } $selectdb = mysql_select_db($db, $dbcon); $formset = 'yes'; $val = 0; $user = trim($_POST['username'], ' '); $luser = mysql_real_escape_string($user); $password = $_POST['password']; $lpassword = mysql_real_escape_string($password); $selectdb; $userq = mysql_query("SELECT user FROM users WHERE user='".$luser."'"); $userresult = @mysql_result($userq, 0); //echo $userresult; if($userresult == $user) { $val = $val + 1; $usercorrect = 'yes'; } else { $usercorrect = 'no'; } $dbselect; $passwordq = mysql_query("SELECT password FROM users where user='".$luser."'"); $passresult = @mysql_result($passwordq, 0); if($passresult == sha1($password)) { $val = $val + 1; $passcorrect = 'yes'; } else { $passcorrect = 'no'; } if ($val == 2) { $_SESSION['loggedin'] = 'yes'; $_SESSION['uloggedin'] = $user; header('location: logged.php'); } }?> <?php ini_set('display_errors', 1); require('testinclude.php'); ?> <body> <div id="loginform"> <form action="/login.php" method="post" > <input type="hidden" name="issub" value="yes" /> <?php if($usercorrect == 'no') { echo '<span class="required"><i><small>The email address or password you entered is incorrect, please try again.</a></small></i></span>'; } ?> <br /> email: <?php if ($issub == 'yes') { if($user == null){ echo '<br /><span class="required"><i><small>Please enter your email address</a></small></i></span>'; } } ?> <br /><input type="text" name="username" id="usename" /> <br /> password: <br /><input type="password" name="password" id="password" /> <br /> <input type="submit" value="login" /> </form> <div> </body>

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