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  • How to do custom display and auto-select in django admin multi-select field?

    - by rsp
    I'm new to django, so please feel free to tell me if I'm doing this incorrectly. I am trying to create a django ordering system. My order model: class Order(models.Model): ordered_by = models.ForeignKey(User, limit_choices_to = {'groups__name': "Managers", 'is_active': 1}) in my admin ANY user can enter an order, but ordered_by must be someone in the group "managers" (this is the behavior I want). Now, if the logged in user happens to be a manager I want it to automatically fill in the field with that logged in user. I have accomplished this by: class OrderAdmin(admin.ModelAdmin): def formfield_for_foreignkey(self, db_field, request, **kwargs): if db_field.name == "ordered_by": if request.user in User.objects.filter(groups__name='Managers', is_active=1): kwargs["initial"] = request.user.id kwargs["empty_label"] = "-------------" return db_field.formfield(**kwargs) return super(OrderAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs) This also works, but the admin puts the username as the display for the select box by default. It would be nice to have the user's real name listed. I was able to do it with this: class UserModelMultipleChoiceField(forms.ModelMultipleChoiceField): def label_from_instance(self, obj): return obj.first_name + " " + obj.last_name class OrderForm(forms.ModelForm): ordered_by = UserModelChoiceField(queryset=User.objects.all().filter(groups__name='Managers', is_active=1)) class OrderAdmin(admin.ModelAdmin): form = OrderForm My problem: I can't to both of these. If I put in the formfield_for_foreignkey function and add form = OrderForm to use my custom "UserModelChoiceField", it puts the nice name display but it won't select the currently logged in user. I'm new to this, but my guess is that when I use UserModelChoiceField it "erases" the info passed in via formfield_for_foreignkey. Do I need to use the super() function somehow to pass on this info? or something completely different?

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  • Django: Summing values of records grouped by foreign key

    - by Dan0
    Hi there In django, given the following models (slightly simplified), I'm struggling to work out how I would get a view including sums of groups class Client(models.Model): api_key = models.CharField(unique=True, max_length=250, primary_key=True) name = models.CharField(unique=True, max_length=250) class Purchase(models.Model): purchase_date = models.DateTimeField() client = models.ForeignKey(SavedClient, to_field='api_key') amount_to_invoice = models.FloatField(null=True) For a given month, I'd like to see e.g. April 2010 For Each Client that purchased this month: * CLient: Name * Total amount of Purchases for this month * Total cost of purchases for this month For each Purchase made by client: * Date * Amount * Etc I've been looking into django annotation, but can't get my head around how to sum values of a field for a particular group over a particular month and send the information to a template as a variable/tag. Any info would be appreciated

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  • Django: reverse lookup URL of feeds?

    - by Santa
    I am having trouble doing a reverse URL lookup for Django-generated feeds. I have the following setup in urls.py: feeds = { 'latest': LatestEntries, } urlpatterns = patterns('', # ... # enable feeds (RSS) url(r'^feeds/(?P<url>.*)/$', 'django.contrib.syndication.views.feed', {'feed_dict': feeds}, name='feeds_view'), ) I have tried using the following template tag: <a href="{% url feeds_view latest %}">RSS feeds</a> But the resulting link is not what want (http://my.domain.com/feeds//). It should be http://my.domain.com/feeds/latest/. For now, I am using a hack to generate the URL for the template: <a href="http://{{ request.META.HTTP_HOST }}/feeds/latest">RSS feeds</a> But, as you can see, it clearly is not DRY. Is there something I am missing?

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  • Django: Country drop down list?

    - by User
    I have a form for address information. One of the fields is for the address country. Currently this is just a textbox. I would like a drop down list (of ISO 3166 countries) for this. I'm a django newbie so I haven't even used a Django Select widget yet. What is a good way to do this? Hard-code the choices in a file somewhere? Put them in the database? In the template?

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  • Project name inserted automatically in url when using django template url tag

    - by thebossman
    I am applying the 'url' template tag to all links in my current Django project. I have my urls named like so... url(r'^login/$', 'login', name='site_login'), This allows me to access /login at my site's root. I have my template tag defined like so... <a href="{% url site_login %}"> It works fine, except that Django automatically resolves that url as /myprojectname/login, not /login. Both urls are accessible. Why? Is there an option to remove the projectname? This occurs for all url tags, not just this one.

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  • django cross-site reverse a url

    - by tutuca
    I have a similar question than django cross-site reverse. But i think I can't apply the same solution. I'm creating an app that lets the users create their own site. After completing the signup form the user should be redirected to his site's new post form. Something along this lines: new_post_url = 'http://%s.domain:9292/manage/new_post %site.domain' logged_user = authenticate(username=user.username, password=user.password) if logged_user is not None: login(request, logged_user) return redirect(new_product_url) Now, I know that "new_post_url" is awful and makes babies cry so I need to reverse it in some way. I thought in using django.core.urlresolvers.reverse to solve this but that only returns urls on my domain, and not in the user's newly created site, so it doesn't works for me. So, do you know a better/smarter way to solve this?

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  • Load django template from the database

    - by Björn Lindqvist
    Hello, Im trying to render a django template from a database outside of djangos normal request-response structure. But it appears to be non-trivial due to the way django templates are compiled. I want to do something like this: >>> s = Template.objects.get(pk = 123).content >>> some_method_to_render(s, {'a' : 123, 'b' : 456}) >>> ... the rendered output here ... How do you do this?

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  • How to use multiple flatpages models in a django app?

    - by the_drow
    I have multiple models that can be converted to flatpages but have to have some extra information (For example I have an about us page but I also have a blog). However I understand that there must be only one flatpages model since the middleware only returns the flatpages instance and does not resolve the child models. What do I have to do? EDIT: It seems I need to change the views. Here's the current code: from django.contrib.flatpages.models import FlatPage from django.template import loader, RequestContext from django.shortcuts import get_object_or_404 from django.http import HttpResponse, HttpResponseRedirect from django.conf import settings from django.core.xheaders import populate_xheaders from django.utils.safestring import mark_safe from django.views.decorators.csrf import csrf_protect DEFAULT_TEMPLATE = 'flatpages/default.html' # This view is called from FlatpageFallbackMiddleware.process_response # when a 404 is raised, which often means CsrfViewMiddleware.process_view # has not been called even if CsrfViewMiddleware is installed. So we need # to use @csrf_protect, in case the template needs {% csrf_token %}. # However, we can't just wrap this view; if no matching flatpage exists, # or a redirect is required for authentication, the 404 needs to be returned # without any CSRF checks. Therefore, we only # CSRF protect the internal implementation. def flatpage(request, url): """ Public interface to the flat page view. Models: `flatpages.flatpages` Templates: Uses the template defined by the ``template_name`` field, or `flatpages/default.html` if template_name is not defined. Context: flatpage `flatpages.flatpages` object """ if not url.endswith('/') and settings.APPEND_SLASH: return HttpResponseRedirect("%s/" % request.path) if not url.startswith('/'): url = "/" + url # Here instead of getting the flat page it needs to find if it has a page with a child model. f = get_object_or_404(FlatPage, url__exact=url, sites__id__exact=settings.SITE_ID) return render_flatpage(request, f) @csrf_protect def render_flatpage(request, f): """ Internal interface to the flat page view. """ # If registration is required for accessing this page, and the user isn't # logged in, redirect to the login page. if f.registration_required and not request.user.is_authenticated(): from django.contrib.auth.views import redirect_to_login return redirect_to_login(request.path) if f.template_name: t = loader.select_template((f.template_name, DEFAULT_TEMPLATE)) else: t = loader.get_template(DEFAULT_TEMPLATE) # To avoid having to always use the "|safe" filter in flatpage templates, # mark the title and content as already safe (since they are raw HTML # content in the first place). f.title = mark_safe(f.title) f.content = mark_safe(f.content) # Here I need to be able to configure what I am passing in the context c = RequestContext(request, { 'flatpage': f, }) response = HttpResponse(t.render(c)) populate_xheaders(request, response, FlatPage, f.id) return response

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  • Create Django formset wihtout multiple queries

    - by Martin
    I need to display multiple forms (up to 10) of a model on a page. This is the code I use for to accomplish this. TheFormSet = formset_factory(SomeForm, extra=10) ... formset = TheFormSet(prefix='party') return render_to_response('template.html', { 'formset' : formset, }) The problem is, that it seems to me that Django queries the database for each of the forms in the formset, even though the data displayed in them is the same. Is this the way Formsets work or am I doing something wrong? Is there a way around it inside django or would I have to use JavaScript for a workaround?

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  • django auth_views.login and redirects

    - by Zayatzz
    Hello I could not understand why after logging in from address: http://localhost/en/accounts/login/?next=/en/test/ I get refirected to http://localhost/accounts/profile/ So i ran search in django files and found that this address is the default LOGIN_REDIRECT_URL for django. What i did not understand is why it gets redirected to there. I guessed, that my login form's post address should be : /accounts/login/?next=/en/test/ instead of /accounts/login/ I wrote it into template and it worked. But since the redirect url changes dynamically, how can i make this login post forms address change dynamically too? is there a templatetag for that or something? Alan

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  • Django QuerySet filter + order_by + limit

    - by handsofaten
    So I have a Django app that processes test results, and I'm trying to find the median score for a certain assessment. I would think that this would work: e = Exam.objects.all() total = e.count() median = int(round(total / 2)) median_exam = Exam.objects.filter(assessment=assessment.id).order_by('score')[median:1] median_score = median_exam.score But it always returns an empty list. I can get the result I want with this: e = Exam.objects.all() total = e.count() median = int(round(total / 2)) exams = Exam.objects.filter(assessment=assessment.id).order_by('score') median_score = median_exam[median].score I would just prefer not to have to query the entire set of exams. I thought about just writing a raw MySQL query that looks something like: SELECT score FROM assess_exam WHERE assessment_id = 5 ORDER BY score LIMIT 690,1 But if possible, I'd like to stay within Django's ORM. Mostly, it's just bothering me that I can't seem to use order_by with a filter and a limit. Any ideas?

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  • Is it possible to change the model name in the django admin site?

    - by luc
    Hello, I am translating a django app and I would like to translate also the homepage of the django admin site. On this page are listed the application names and the model class names. I would like to translate the model class name but I don't find how to give a user-friendly name for a model class. Does anybody know how to do that?

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  • Django 1.5 custom User model error. "Manager isn't available; User has been swapped"

    - by bpetit
    I extend the django user model as described in the dev doc. I wan't to keep most of the original User model features so I extend the AbstractUser class. I've defined in settings.py: AUTH_USER_MODEL = 'myapp.CustomUser' My user class: class CustomUser(AbstractUser): custom_field = models.ForeignKey('OtherModel') objects = UserManager() Everything seems to work fine but when I try to make it managed by the admin site: admin.site.register(CustomUser, UserAdmin) I get this error on the admin CustomUser creation page (after validation of the password confirmation form): AttributeError: Manager isn't available; User has been swapped for 'myapp.CustomUser' The point is that I need this model managed by the admin site in order to have the same creation process as with the original User model (two step process with password validation). Thanks for any reply

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  • Django | django-socialregistration error

    - by MMRUser
    I'm trying to add the facebook connect feature to my site, I decided to use django socialregistration.All are setup including pyfacebook, here is my source code. settings.py MIDDLEWARE_CLASSES = ( 'django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'facebook.djangofb.FacebookMiddleware', 'socialregistration.middleware.FacebookMiddleware', ) urls.py (r'^callback/$', 'fbproject.fbapp.views.callback'), views.py def callback(request): return render_to_response('canvas.fbml') Template <html> <body> {% load facebook_tags %} {% facebook_button %} {% facebook_js %} </body> </html> but when I point to the URL, I'm getting this error Traceback (most recent call last): File "C:\Python26\lib\site-packages\django\core\servers\basehttp.py", line 279, in run self.result = application(self.environ, self.start_response) File "C:\Python26\lib\site-packages\django\core\servers\basehttp.py", line 651, in __call__ return self.application(environ, start_response) File "C:\Python26\lib\site-packages\django\core\handlers\wsgi.py", line 241, in __call__ response = self.get_response(request) File "C:\Python26\lib\site-packages\django\core\handlers\base.py", line 73, in get_response response = middleware_method(request) File "build\bdist.win32\egg\socialregistration\middleware.py", line 13, in process_request request.facebook.check_session(request) File "C:\Python26\lib\site-packages\facebook\__init__.py", line 1293, in check_session self.session_key_expires = int(params['expires']) ValueError: invalid literal for int() with base 10: 'None' Django 1.1.1 *Python 2.6.2*

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  • How to deal with multiple sub-type of one super-type in Django admin

    - by Henri
    What would be the best solution for adding/editing multiple sub-types. E.g a super-type class Contact with sub-type class Client and sub-type class Supplier. The way shown here works, but when you edit a Contact you get both inlines i.e. sub-type Client AND sub-type Supplier. So even if you only want to add a Client you also get the fields for Supplier of vice versa. If you add a third sub-type , you get three sub-type field groups, while you actually only want one sub-type group, in the mentioned example: Client. E.g.: class Contact(models.Model): contact_name = models.CharField(max_length=128) class Client(models.Model): contact = models.OneToOneField(Contact, primary_key=True) user_name = models.CharField(max_length=128) class Supplier(models.Model): contact.OneToOneField(Contact, primary_key=True) company_name = models.CharField(max_length=128) and in admin.py class ClientInline(admin.StackedInline): model = Client class SupplierInline(admin.StackedInline): model = Supplier class ContactAdmin(admin.ModelAdmin): inlines = (ClientInline, SupplierInline,) class ClientAdmin(admin.ModelAdmin): ... class SupplierAdmin(admin.ModelAdmin): ... Now when I want to add a Client, i.e. only a Client I edit Contact and I get the inlines for both Client and Supplier. And of course the same for Supplier. Is there a way to avoid this? When I want to add/edit a Client that I only see the Inline for Client and when I want to add/edit a Supplier that I only see the Inline for Supplier, when adding/editing a Contact? Or perhaps there is a different approach. Any help or suggestion will be greatly appreciated.

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  • Manditory read-only fields in django

    - by jamida
    I'm writing a test "grade book" application. The models.py file is shown below. class Student(models.Model): name = models.CharField(max_length=50) parent = models.CharField(max_length=50) def __unicode__(self): return self.name class Grade(models.Model): studentId = models.ForeignKey(Student) finalGrade = models.CharField(max_length=3) I'd like to be able to change the final grade for several students in a modelformset but for now I'm just trying one student at a time. I'm also trying to create a form for it that shows the student name as a field that can not be changed, the only thing that can be changed here is the finalGrade. So I used this trick to make the studentId read-only. class GradeROForm(ModelForm): studentId = forms.ModelChoiceField(queryset=Student.objects.all()) def __init__(self, *args, **kwargs): super(GradeROForm,self).__init__(*args, **kwargs) instance = getattr(self, 'instance', None) if instance and instance.id: self.fields['studentId'].widget.attrs['disabled']='disabled' def clean_studentId(self): instance = getattr(self,'instance',None) if instance: return instance.studentId else: return self.cleaned_data.get('studentId',None) class Meta: model=Grade And here is my view: def modifyGrade(request,student): student = Student.objects.get(name=student) mygrade = Grade.objects.get(studentId=student) if request.method == "POST": myform = GradeROForm(data=request.POST, instance=mygrade) if myform.is_valid(): grade = myform.save() info = "successfully updated %s" % grade.studentId else: myform=GradeROForm(instance=mygrade) return render_to_response('grades/modifyGrade.html',locals()) This displays the form like I expect, but when I hit "submit" I get a form validation error for the student field telling me this field is required. I'm guessing that, since the field is "disabled", the value is not being reported in the POST and for reasons unknown to me the instance isn't being used in its place. I'm a new Django/Python programmer, but quite experienced in other languages. I can't believe I've stumbled upon such a difficult to solve problem in my first significant django app. I figure I must be missing something. Any ideas?

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  • Django Upload form to S3 img and form validation

    - by citadelgrad
    I'm fairly new to both Django and Python. This is my first time using forms and upload files with django. I can get the uploads and saves to the database to work fine but it fails to valid email or check if the users selected a file to upload. I've spent a lot of time reading documentation trying to figure this out. Thanks! views.py def submit_photo(request): if request.method == 'POST': def store_in_s3(filename, content): conn = S3Connection(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY) bucket = conn.create_bucket(AWS_STORAGE_BUCKET_NAME) mime = mimetypes.guess_type(filename)[0] k = Key(bucket) k.key = filename k.set_metadata("Content-Type", mime) k.set_contents_from_file(content) k.set_acl('public-read') if imghdr.what(request.FILES['image_url']): qw = request.FILES['image_url'] filename = qw.name image = filename content = qw.file url = "http://bpd-public.s3.amazonaws.com/" + image data = {image_url : url, user_email : request.POST['user_email'], user_twittername : request.POST['user_twittername'], user_website : request.POST['user_website'], user_desc : request.POST['user_desc']} s = BeerPhotos(data) if s.is_valid(): #import pdb; pdb.set_trace() s.save() store_in_s3(filename, content) return HttpResponseRedirect(reverse('photos.views.thanks')) return s.errors else: return errors else: form = BeerPhotoForm() return render_to_response('photos/submit_photos.html', locals(),context_instance=RequestContext(request) forms.py class BeerPhotoForm(forms.Form): image_url = forms.ImageField(widget=forms.FileInput, required=True,label='Beer',help_text='Select a image of no more than 2MB.') user_email = forms.EmailField(required=True,help_text='Please type a valid e-mail address.') user_twittername = forms.CharField() user_website = forms.URLField(max_length=128,) user_desc = forms.CharField(required=True,widget=forms.Textarea,label='Description',) template.html <div id="stylized" class="myform"> <form action="." method="post" enctype="multipart/form-data" width="450px"> <h1>Photo Submission</h1> {% for field in form %} {{ field.errors }} {{ field.label_tag }} {{ field }} {% endfor %} <label><span>Click here</span></label> <input type="submit" class="greenbutton" value="Submit your Photo" /> </form> </div>

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  • Invalidating Memcached Keys on save() in Django

    - by Zack
    I've got a view in Django that uses memcached to cache data for the more highly trafficked views that rely on a relatively static set of data. The key word is relatively: I need invalidate the memcached key for that particular URL's data when it's changed in the database. To be as clear as possible, here's the meat an' potatoes of the view (Person is a model, cache is django.core.cache.cache): def person_detail(request, slug): if request.is_ajax(): cache_key = "%s_ABOUT_%s" % settings.SITE_PREFIX, slug # Check the cache to see if we've already got this result made. json_dict = cache.get(cache_key) # Was it a cache hit? if json_dict is None: # That's a negative Ghost Rider person = get_object_or_404(Person, display = True, slug = slug) json_dict = { 'name' : person.name, 'bio' : person.bio_html, 'image' : person.image.extra_thumbnails['large'].absolute_url, } cache.set(cache_key) # json_dict will now exist, whether it's from the cache or not response = HttpResponse() response['Content-Type'] = 'text/javascript' response.write(simpljson.dumps(json_dict)) # Make sure it's all properly formatted for JS by using simplejson return response else: # This is where the fully templated response is generated What I want to do is get at that cache_key variable in it's "unformatted" form, but I'm not sure how to do this--if it can be done at all. Just in case there's already something to do this, here's what I want to do with it (this is from the Person model's hypothetical save method) def save(self): # If this is an update, the key will be cached, otherwise it won't, let's see if we can't find me try: old_self = Person.objects.get(pk=self.id) cache_key = # Voodoo magic to get that variable old_key = cache_key.format(settings.SITE_PREFIX, old_self.slug) # Generate the key currently cached cache.delete(old_key) # Hit it with both barrels of rock salt # Turns out this doesn't already exist, let's make that first request even faster by making this cache right now except DoesNotExist: # I haven't gotten to this yet. super(Person, self).save() I'm thinking about making a view class for this sorta stuff, and having functions in it like remove_cache or generate_cache since I do this sorta stuff a lot. Would that be a better idea? If so, how would I call the views in the URLconf if they're in a class?

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  • Math on Django Templates

    - by Leandro Abilio
    Here's another question about Django. I have this code: views.py cursor = connections['cdr'].cursor() calls = cursor.execute("SELECT * FROM cdr where calldate > '%s'" %(start_date)) result = [SQLRow(cursor, r) for r in cursor.fetchall()] return render_to_response("cdr_user.html", {'calls':result }, context_instance=RequestContext(request)) I use a MySQL query like that because the database is not part of a django project. My cdr table has a field called duration, I need to divide that by 60 and multiply the result by a float number like 0.16. Is there a way to multiply this values using the template tags? If not, is there a good way to do it in my views? My template is like this: {% for call in calls %} <tr class="{% cycle 'odd' 'even' %}"><h3> <td valign="middle" align="center"><h3>{{ call.calldate }}</h3></td> <td valign="middle" align="center"><h3>{{ call.disposition }}</h3></td> <td valign="middle" align="center"><h3>{{ call.dst }}</h3></td> <td valign="middle" align="center"><h3>{{ call.billsec }}</h3></td> <td valign="middle" align="center">{{ (call.billsec/60)*0.16 }}</td></h3> </tr> {% endfor %} The last is where I need to show the value, I know the "(call.billsec/60)*0.16" is impossible to be done there. I wrote it just to represent what I need to show.

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  • Django ModelForm is giving me a validation error that doesn't make sense

    - by River Tam
    I've got a ModelForm based on a Picture. class Picture(models.Model): name = models.CharField(max_length=100) pub_date = models.DateTimeField('date published') tags = models.ManyToManyField('Tag', blank=True) content = models.ImageField(upload_to='instaton') def __unicode__(self): return self.name class PictureForm(forms.ModelForm): class Meta: model = Picture exclude = ('pub_date','tags') That's the model and the ModelForm, of course. def submit(request): if request.method == 'POST': # if the form has been submitted form = PictureForm(request.POST) if form.is_valid(): return HttpResponseRedirect('/django/instaton') else: form = PictureForm() # blank form return render_to_response('instaton/submit.html', {'form': form}, context_instance=RequestContext(request)) That's the view (which is being correctly linked to by urls.py) Right now, I do nothing when the form submits. I just check to make sure it's valid. If it is, I forward to the main page of the app. <form action="/django/instaton/submit/" method="post"> {% csrf_token %} {{ form.as_p }} <input type="submit" value"Submit" /> </form> And there's my template (in the correct location). When I try to actually fill out the form and just validate it, even if I do so correctly, it sends me back to the form and says "This field is required" between Name and Content. I assume it's referring to Content, but I'm not sure. What's my problem? Is there a better way to do this?

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  • Django internationalization for admin pages - translate model name and attributes

    - by geekQ
    Django's internationalization is very nice (gettext based, LocaleMiddleware), but what is the proper way to translate the model name and the attributes for admin pages? I did not find anything about this in the documentation: http://docs.djangoproject.com/en/dev/topics/i18n/internationalization/ http://www.djangobook.com/en/2.0/chapter19/ I would like to have "???????? ????? ??? ?????????" instead of "???????? order ??? ?????????". Note, the 'order' is not translated. First, I defined a model, activated USE_I18N = True in settings.py, run django-admin makemessages -l ru. No entries are created by default for model names and attributes. Grepping in the Django source code I found: $ ack "Select %s to change" contrib/admin/views/main.py 70: self.title = (self.is_popup and ugettext('Select %s') % force_unicode(self.opts.verbose_name) or ugettext('Select %s to change') % force_unicode(self.opts.verbose_name)) So the verbose_name meta property seems to play some role here. Tried to use it: class Order(models.Model): subject = models.CharField(max_length=150) description = models.TextField() class Meta: verbose_name = _('order') Now the updated po file contains msgid 'order' that can be translated. So I put the translation in. Unfortunately running the admin pages show the same mix of "???????? order ??? ?????????". I'm currently using Django 1.1.1. Could somebody point me to the relevant documentation? Because google can not. ;-) In the mean time I'll dig deeper into the django source code...

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  • Django admin output extra HTML in ModelSite

    - by VoteyDisciple
    Ultimately, I want to add an <iframe> to the display of a particular model on Django's admin page. Django is already rendering the form for this model correctly, but I want to add this <iframe> in addition to Django's form. The src attribute needs to involve the primary key for the currently-displayed record. I've learned how to properly override the change_form.html template through Django's documentation, and I can add markup to the right block, but I can't figure out how to access the primary key value. (No amount of determined Googling has helped at all.) Alternatively, is there a direct way to specify that I want to produce extra output in my ModelSite definition?

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  • django views question

    - by Hulk
    In my django views i have the following def create(request): query=header.objects.filter(id=a)[0] a=query.criteria_set.all() logging.debug(a.details) I get an error saying 'QuerySet' object has no attribute 'details' in the debug statement .What is this error and what should be the correct statemnt to query this.And the model corresponding to this is as follows where as the models has the following: class header(models.Model): title = models.CharField(max_length = 255) created_by = models.CharField(max_length = 255) def __unicode__(self): return self.id() class criteria(models.Model): details = models.CharField(max_length = 255) headerid = models.ForeignKey(header) def __unicode__(self): return self.id() Thanks..

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