Search Results

Search found 15298 results on 612 pages for 'django template tags'.

Page 7/612 | < Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >

  • Decrease DB requests number from Django templates

    - by Andrew
    I publish discount offers for my city. Offer models are passed to template ( ~15 offers per page). Every offer has lot of items(every item has FK to it's offer), thus i have to make huge number of DB request from template. {% for item in offer.1 %} {{item.descr}} {{item.start_date}} {{item.price|floatformat}} {%if not item.tax_included %}{%trans "Without taxes"%}{%endif%} <a href="{{item.offer.wwwlink}}" >{%trans "Buy now!"%}</a> </div> <div class="clear"></div> {% endfor %} So there are ~200-400 DB requests per page, that's abnormal i expect. In django code it is possible to use select_related to prepopulate needed values, how can i decrease number of requests in template?

    Read the article

  • Django snippet with logic

    - by etam
    Hi, is there a way to create a Django snippet that has logic? I think about something like contact template tag: {% contact_form %} with template: <form action="send_contact_form" method="POST">...</form> with logic: def send_contact_form(): ... I want to be able to use it anywhere in my projects. It should work only by specifying 1 template tag... Do you know what I mean? Is it possible? Thanks in advance, Etam.

    Read the article

  • Django templates check condition

    - by Hulk
    If there are are no values in the table how can should the code be to indicate no name found else show the drop down box in the below code {% for name in dict.names %} <option value="{{name.id}}" {% for selected_id in selected_name %}{% ifequal name.id selected_id %} {{ selected }} {% endifequal %} {% endfor %}>{{name.firstname}}</option>{% endfor %} </select> Thanks..

    Read the article

  • Physical address and contact details in Meta tags

    - by Steve
    I remember someone once saying it was useful (for SEO purposes) to include your address and contact details in the site's Meta tags. I don't recall if he meant in in the meta description or in another type of meta tag. Have you hear of this strategy before? I believe the person implied that Google would tie all of your Local Listings together for a better SEO score if you listed matching contact details in your meta tags.

    Read the article

  • Login URL using authentication information in Django

    - by fuSi0N
    I'm working on a platform for online labs registration for my university. Login View [project views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib import auth def index(request): return render_to_response('index.html', {}, context_instance = RequestContext(request)) def login(request): if request.method == "POST": post = request.POST.copy() if post.has_key('username') and post.has_key('password'): usr = post['username'] pwd = post['password'] user = auth.authenticate(username=usr, password=pwd) if user is not None and user.is_active: auth.login(request, user) if user.get_profile().is_teacher: return HttpResponseRedirect('/teachers/'+user.username+'/') else: return HttpResponseRedirect('/students/'+user.username+'/') else: return render_to_response('index.html', {'msg': 'You don\'t belong here.'}, context_instance = RequestContext(request) return render_to_response('login.html', {}, context_instance = RequestContext(request)) def logout(request): auth.logout(request) return render_to_response('index.html', {}, context_instance = RequestContext(request)) URLS #========== PROJECT URLS ==========# urlpatterns = patterns('', (r'^media/(?P<path>.*)$', 'django.views.static.serve', {'document_root': settings.MEDIA_ROOT }), (r'^admin/', include(admin.site.urls)), (r'^teachers/', include('diogenis.teachers.urls')), (r'^students/', include('diogenis.students.urls')), (r'^login/', login), (r'^logout/', logout), (r'^$', index), ) #========== TEACHERS APP URLS ==========# urlpatterns = patterns('', (r'^(?P<username>\w{0,50})/', labs), ) The login view basically checks whether the logged in user is_teacher [UserProfile attribute via get_profile()] and redirects the user to his profile. Labs View [teachers app views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib.auth.decorators import user_passes_test from django.contrib.auth.models import User from accounts.models import * from labs.models import * def user_is_teacher(user): return user.is_authenticated() and user.get_profile().is_teacher @user_passes_test(user_is_teacher, login_url="/login/") def labs(request, username): q1 = User.objects.get(username=username) q2 = u'%s %s' % (q1.last_name, q1.first_name) q2 = Teacher.objects.get(name=q2) results = TeacherToLab.objects.filter(teacher=q2) return render_to_response('teachers/labs.html', {'results': results}, context_instance = RequestContext(request)) I'm using @user_passes_test decorator for checking whether the authenticated user has the permission to use this view [labs view]. The problem I'm having with the current logic is that once Django authenticates a teacher user he has access to all teachers profiles basically by typing the teachers username in the url. Once a teacher finds a co-worker's username he has direct access to his data. Any suggestions would be much appreciated.

    Read the article

  • CodeIgniter & Datamapper as frontend, Django Admin as backend, database tables inconsistent

    - by Rasiel
    I created a database for a site i'm doing using Django as the admin backend. However because the server where the site is hosted on, won't be able to support Python, I find myself needing to do the front end in PHP and as such i've decided to use CodeIgniter along with Datamapper to map the models/relationship. However DataMapper requires the tables to be in a specific format for it to work, and Django maps its tables differently, using the App name as the prefix in the table. I've tried using the prefix & join_prefix vars in datamapper but still doesn't map them correctly. Has anyone used a combination of this? and if so how have the fixed the issue of db table names being inconsistent? Is there anything out there that i can use to make them work together?

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • How to customize a many-to-many inline model in django admin

    - by Jonathan
    I'm using the admin interface to view invoices and products. To make things easy, I've set the products as inline to invoices, so I will see the related products in the invoice's form. As you can see I'm using a many-to-many relationship. In models.py: class Product(models.Model): name = models.TextField() price = models.DecimalField(max_digits=10,decimal_places=2) class Invoice(models.Model): company = models.ForeignKey(Company) customer = models.ForeignKey(Customer) products = models.ManyToManyField(Product) In admin.py: class ProductInline(admin.StackedInline): model = Invoice.products.through class InvoiceAdmin(admin.ModelAdmin): inlines = [FilteredApartmentInline,] admin.site.register(Product, ProductAdmin) The problem is that django presents the products as a table of drop down menus (one per associated product). Each drop down contains all the products listed. So if I have 5000 products and 300 are associated with a certain invoice, django actually loads 300x5000 product names. Also the table is not aesthetic. How can I change it so that it'll just display the product's name in the inline table? Which form should I override, and how?

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

  • Calling a method from within a django model save() override

    - by Jonathan
    I'm overriding a django model save() method. Within the override I'm calling another method of the same class and instance which calculates one of the instance's fields based on other fields of the same instance. class MyClass(models.Model): field1 = models.FloatField() field2 = models.FloatField() field3 = models.FloatField() def calculateField1(self) self.field1 = self.field2 + self.field3 def save(self, *args, **kwargs): self.calculateField1() super(MyClass, self).save(*args, **kwargs) The override method is called when I change the model in admin. Alas I've discovered that within calculateField1() field2 and field3 have the values of the instance from before I edited them in admin. If I enter the instance again in admin and save again, only then field1 receives the correct value as field2 and field3 are already updated. Is this the correct behavior on django's side? If yes, then how can I use the new values within calculateField1? I cannot implement the calculation within the save() as calculateField1() actually quite long and I need it to be called from elsewhere.

    Read the article

  • Django - markup parser in template or view?

    - by Amit Ron
    I am building a website where my pages are written in MediaWiki Markup, for which I have a working parser function in Python. Where exactly do I parse my markup: in the view's code, or in the template? My first guess would be something like: return render_to_response( 'blog/post.html', {'post': post, 'content': parseMyMarkup(post.content) }) Is this the usual convention, or should I do something different?

    Read the article

  • 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?

    Read the article

  • How to save links with tags and parameters in TextField

    - by xRobot
    I have this simple Post model: class Post(models.Model): title = models.CharField(_('title'), max_length=60, blank=True, null=True) body = models.TextField(_('body')) blog = models.ForeignKey(Blog, related_name="posts") user = models.ForeignKey(User) I want that when I insert in the form the links, then these links are saved in the body from this form: http://www.example.com or www.example.com to this form ( with tag and rel="nofollow" parameter ): <a href="http://www.example.com" rel="nofollow">www.example.com</a> How can I do this ? Thanks ^_^

    Read the article

  • Django DRY Feeds

    - by Mandx
    I'm using the Django Feeds Framework and it's really nice, very intuitive and easy to use. But, I think there is a problem when creating links to feeds in HTML. For example: <link rel="alternate" type="application/rss+xml" title="{{ feed_title }}" href="{{ url_of_feed }}" /> Link's HREF attribute can be easily found out, just use reverse() But, what about the TITLE attribute? Where the template engine should look for this? Even more, what if the feed is build up dinamically and the title depends on parameters (like this)? I can't come up with a solution that "seems" DRY to me... All that I can come up with is using context processors o template tags, but it gets messy when the context procesor/template tag has to find parameters to construct the Feed class, and writing this I realize I don't even know how to create a Feed instance myself within the view. If I put all this logic in the view, it would not be just one view. Also, the value for TITLE would be in the view AND in the feed.

    Read the article

  • 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*

    Read the article

  • 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?

    Read the article

  • 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>

    Read the article

  • 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?

    Read the article

< Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >