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  • Django query filter a set of data

    - by dana
    if a have a query like following = Relations.objects.filter(initiated_by = request.user) in which i'm having all the users followed by the currently logged in user, and i want to display those user's blog posts. Using a query like: blog = New.objects.filter(created_by = following) it only shows me the blog posts of the user with the id = 1 (though the currently logged in user doesn't actually follow him) in template i have : {% for object in blog %} <a href='/accounts/profile_view/{{object.created_by}}/'> {{object.created_by}}</a> <br /> {{object.post}}<br /> {% endfor %} Where am i wrong?

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  • django url - link problem

    - by dana
    i have an application, and in my urls.py i have something like that: urlpatterns = patterns('', url(r'^profile_view/(?P<id>\d+)/$', profile_view, name='profile_view'),) meaning that the profile_view function has id as a parameter. Now, i want to call that function from my template, using a link like Reply The problem is that i don't know how to use the above url as a link, meaning how can i 'pass the id parameter to a link'? thank you

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  • django redirect url containing id

    - by dana
    hello, i want to add in my settings.py a declaration like: LOGIN_REDIRECT_URL='^private_profile/(?P<id>\d+)/$' #or LOGIN_REDIRECT_URL='/accounts/private_profile/id/' so that when the user with the id 1, for example,is logging in, he will be redirected to LOGIN_REDIRECT_URL='/accounts/private_profile/1/' but both alternatives, LOGIN_REDIRECT_URL='^private_profile/(?P<id>\d+)/$' #or LOGIN_REDIRECT_URL='/accounts/private_profile/id/' are wrong, because in my browser i don't see the current user id, where am i wrong? Thanks

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  • Django: Applying Calculations To A Query Set

    - by TheLizardKing
    I have a QuerySet that I wish to pass to a generic view for pagination: links = Link.objects.annotate(votes=Count('vote')).order_by('-created')[:300] This is my "hot" page which lists my 300 latest submissions (10 pages of 30 links each). I want to now sort this QuerySet by an algorithm that HackerNews uses: (p - 1) / (t + 2)^1.5 p = votes minus submitter's initial vote t = age of submission in hours Now because applying this algorithm over the entire database would be pretty costly I am content with just the last 300 submissions. My site is unlikely to be the next digg/reddit so while scalability is a plus it is required. My question is now how do I iterate over my QuerySet and sort it by the above algorithm? For more information, here are my applicable models: class Link(models.Model): category = models.ForeignKey(Category, blank=False, default=1) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add=True) modified = models.DateTimeField(auto_now=True) url = models.URLField(max_length=1024, unique=True, verify_exists=True) name = models.CharField(max_length=512) def __unicode__(self): return u'%s (%s)' % (self.name, self.url) class Vote(models.Model): link = models.ForeignKey(Link) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add=True) def __unicode__(self): return u'%s vote for %s' % (self.user, self.link) Notes: I don't have "downvotes" so just the presence of a Vote row is an indicator of a vote or a particular link by a particular user.

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  • Saving model object in django throws no error but attribute value doesn't change

    - by Pilgrim
    Hi. I get a model object, change an attribute, save it and it still has the old attribute: >>> g = SiteProfile.objects.get(pk=3) >>> g.renew_date datetime.date(2010, 4, 11) >>> g.renew_date = date.today()+timedelta(days=365) >>> g.renew_date datetime.date(2011, 4, 11) >>> g.save() >>> g.renew_date datetime.datetime(2010, 4, 11, 16, 57, 4, 192684) Anyone know if this is an issue with the database or something else?

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  • How to save to Django Model that Have Mulitple Foreign Keys Fields

    - by Spikie
    I have Models for business Apps class staff_name(models.Model): TITLE_CHOICES = ( ('Mr', 'Mr'), ('Miss', 'Miss'), ( 'Mrs', 'Mrs'), ( 'chief', 'chief'), ) titlename = models.CharField(max_length=10,choices=TITLE_CHOICES) firstname = models.CharField(max_length=150) surname = models.CharField(max_length=150) date = models.DateTimeField(auto_now=True) class meta: ordering = ["date"] get_latest_by = "date" class inventory_transaction(models.Model): stock_in = models.DecimalField(blank=True, null=True,max_digits=8, decimal_places=2) stock_out = models.DecimalField(blank=True,null=True,max_digits=8, decimal_places=2) Number_container = models.ForeignKey(container_identity) staffs = models.ForeignKey(staff_name) goods_details = models.ForeignKey(departments) balance = models.DecimalField(max_digits=8, decimal_places=2) date = models.DateTimeField(auto_now=True) What i want to accomplish is check if the staff have made entry to the table before if yes add the value for the stock in plus (last) balance field and assign to balance if no just assign stock in value to balance field and save these are my codes These are my codes: try: s = staffname.staffs_set.all().order_by("-date").latest() # staffname is the instant of the class model staff_name e = s.staffs_set.create(stockin=vdataz,balance=s.balance + vdataz ) # e is the instant of the class model inventory_transaction e.save e.staffs.add(s) e.from_container.add(containersno) e.goods_details.add(department) except ObjectDoesNotExist: e = staff_name.objects.create(stockin=vdataz,balance=vdataz ) e.save e.staffs.add(staffname) e.from_container.add(containersno) e.goods_details.add(department) I will really appreciate a solution Thanks I hope it make more sense now. iam on online if you need more explanation just ask in the comment.Thank you for your help

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  • django getting current user id

    - by dana
    hello, i have a mini app where users can login, view their profile, and follow each other. 'Follow' is a relation like a regular 'friend' relationship in virtual communities, but it is not necessarily reciprocal, meaning that one can follow a user, without the need that the user to be following back that person who follows him. my problem is: if i am a logged in user, and i navigate to a profile X, and push the button follow, how can i take the current profile id ?(current profile meaning the profile that I, the logged in user, am viewing right now.) the view: def follow(request): if request.method == 'POST': form = FollowForm(request.POST) if form.is_valid(): new_obj = form.save(commit=False) new_obj.initiated_by = request.user u = User.objects. what here? new_obj.follow = u new_obj.save() return HttpResponseRedirect('.') else: form = FollowForm() return render_to_response('followme/follow.html', { 'form': form, }, context_instance=RequestContext(request)) thanks in advance!

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  • Django Admin: Many-to-Many listbox doesn't show up with a through parameter

    - by NP
    Hi All, I have the following models: class Message(models.Model): date = models.DateTimeField() user = models.ForeignKey(User) thread = models.ForeignKey('self', blank=True, null=True) ... class Forum(models.Model): name = models.CharField(max_length=24) messages = models.ManyToManyField(Message, through="Message_forum", blank=True, null=True) ... class Message_forum(models.Model): message = models.ForeignKey(Message) forum = models.ForeignKey(Forum) status = models.IntegerField() position = models.IntegerField(blank=True, null=True) tags = models.ManyToManyField(Tag, blank=True, null=True) In the admin site, when I go to add/change a forum, I don't see the messages listbox as you'd expect. However, it shows up if I remove the 'through' parameter in the ManyToManyField declaration. What's up with that? I've registered all three models (plus Tag) to the admin site in admin.py. TIA

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  • Django - foreignkey problem

    - by realshadow
    Hey, Imagine you have this model: class Category(models.Model): node_id = models.IntegerField(primary_key = True) type_id = models.IntegerField(max_length = 20) parent_id = models.IntegerField(max_length = 20) sort_order = models.IntegerField(max_length = 20) name = models.CharField(max_length = 45) lft = models.IntegerField(max_length = 20) rgt = models.IntegerField(max_length = 20) depth = models.IntegerField(max_length = 20) added_on = models.DateTimeField(auto_now = True) updated_on = models.DateTimeField(auto_now = True) status = models.IntegerField(max_length = 20) node = models.ForeignKey(Category_info, verbose_name = 'Category_info', to_field = 'node_id' The important part is the foreignkey. When I try: Category.objects.filter(type_id = type_g.type_id, parent_id = offset, status = 1) I get an error that get returned more than category, which is fine, because it is supposed to return more than one. But I want to filter the results trough another field, which would be type id (from the second Model) Here it is: class Category_info(models.Model): objtree_label_id = models.AutoField(primary_key = True) node_id = models.IntegerField(unique = True) language_id = models.IntegerField() label = models.CharField(max_length = 255) type_id = models.IntegerField() The type_id can be any number from 1 - 5. I am desparately trying to get only one result where the type_id would be number 1. Here is what I want in sql: SELECT n.*, l.* FROM objtree_nodes n JOIN objtree_labels l ON (n.node_id = l.node_id) WHERE n.type_id = 15 AND n.parent_id = 50 AND l.type_id = 1 Any help is GREATLY appreciated. Regards

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  • Self Authenticating Links in Django

    - by awolf
    In my web app I would like to be able to email self-authenticating links to users. These links will contain a unique token (uuid). When they click the link the token being present in the query string will be enough to authenticate them and they won't have to enter their username and password. What's the best way to do this?

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  • Django form and i18n

    - by madewulf
    I have forms that I want to display in different languages : I used the label parameter to set a parameter, and used ugettext() on the labels : agreed_tos = forms.BooleanField(label=ugettext('I agree to the terms of service and to the privacy policy.')) But when I am rendering the form in my template, using {{form.as_p}} The labels are not translated. Does somebody have a solution for this problem ?

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  • ahow can I resolve Django Error: str' object has no attribute 'autoescape'?

    - by Angelbit
    Hi have tried to create a inclusion tag on Django but don't work and return str' object has no attribute 'autoescape' this is the code of custom tag: from django import template from quotes.models import Quotes register = template.Library() def show_quote(): quote = Quotes.objects.values('quote', 'author').get(id=0) return {'quote': quote['quote']} register.inclusion_tag('quotes.html')(show_quote) EDIT: Quote class from django.db import models class Quotes(models.Model): quote = models.CharField(max_length=255) author = models.CharField(max_length=100) class Meta: db_table = 'quotes' quotes.html <blockquote id="quotes">{{ quote }}</blockquote>

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  • How do I write this URL in Django?

    - by alex
    (r'^/(?P<the_param>[a-zA-z0-9_-]+)/$','myproject.myapp.views.myview'), How can I change this so that "the_param" accepts a URL(encoded) as a parameter? So, I want to pass a URL to it. mydomain.com/http%3A//google.com

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  • Django: Update order attribute for objects in a queryset

    - by lazerscience
    I'm having a attribute on my model to allow the user to order the objects. I have to update the element's order depending on a list, that contains the object's ids in the new order; right now I'm iterating over the whole queryset and set one objects after the other. What would be the easiest/fastest way to do the same with the whole queryset? def update_ordering(model, order): """ order is in the form [id,id,id,id] for example: [8,4,5,1,3] """ id_to_order = dict((order[i], i) for i in range(len(order))) for x in model.objects.all(): x.order = id_to_order[x.id] x.save()

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  • Limit number of views per day in Django

    - by ariddell
    Is there an easy way to limit the number of times a view can be accessed by a given IP address per day/week? A simplified version of the technique used by some booksellers to limit the number of pages of a book you can preview? There's only one view that this limit need apply to--i.e. it's not a general limit--and it would be nice if I could just have a variable overlimit in the template context. The solution need not be terribly robust, but limiting by IP address seemed like a better idea than using a cookie. I've looked into the session middleware but it doesn't make any references to tracking IP addresses as far as I can tell. Has anyone encountered this problem?

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  • input_formats in django admin has no effect

    - by pablo
    I'm trying to use input_foramts in the admin but it has no effect. What am I doing wrong? # model class Feedback(models.Model): created_at = models.DateTimeField(auto_now_add=True) # admin form class FeedbackAdminForm(forms.ModelForm): created_at = forms.DateTimeField(input_formats=('%d/%m/%Y',)) class Meta: model = Feedback # admin class FeedbackAdmin(admin.ModelAdmin): form = FeedbackAdminForm admin.site.register(Feedback, FeedbackAdmin) Thanks

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  • How to host 50 domains/sites with common Django code base

    - by Off Rhoden
    I have 50 different websites that use the same layout and code base, but mostly non-overlapping data (regional support sites, not link farm). Is there a way to have a single installation of the code and run all 50 at the same time? When I have a bug to fix (or deploy new feature), I want to deploy ONE time + 1 restart and be done with it. Also: Code needs to know what domain the request is coming to so the appropriate data is displayed.

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  • Django: order by count of a ForeignKey field?

    - by AP257
    This is almost certainly a duplicate question, in which case apologies, but I've been searching for around half an hour on SO and can't find the answer here. I'm probably using the wrong search terms, sorry. I have a User model and a Submission model. Each Submission has a ForeignKey field called user_submitted for the User who uploaded it. class Submission(models.Model): uploaded_by = models.ForeignKey('User') class User(models.Model): name = models.CharField(max_length=250 ) My question is pretty simple: how can I get a list of the three users with the most Submissions? I trued creating a num_submissions method on the User model: def num_submissions(self): num_submissions = Submission.objects.filter(uploaded_by=self).count() return num_submissions and then doing: top_users = User.objects.filter(problem_user=False).order_by('num_submissions')[:3] but this fails, as do all the other things I've tried. Can I actually do it using a smart database query? Or should I just do something more hacky in the views file?

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  • what is this 'map' mean..in django

    - by zjm1126
    this is the code: def create(request, form_class=MapForm, template_name="maps/create.html"): map_form = form_class(request.POST or None) if map_form.is_valid(): map = map_form.save(commit=False) and the map_form is : class MapForm(forms.ModelForm): slug = forms.SlugField(max_length=20, help_text = _("a short version of the name consisting only of letters, numbers, underscores and hyphens."), #error_message = _("This value must contain only letters, numbers, underscores and hyphens.")) ) def clean_slug(self): if Map.objects.filter(slug__iexact=self.cleaned_data["slug"]).count() > 0: raise forms.ValidationError(_("A Map already exists with that slug.")) return self.cleaned_data["slug"].lower() def clean_name(self): if Map.objects.filter(name__iexact=self.cleaned_data["name"]).count() > 0: raise forms.ValidationError(_("A Map already exists with that name.")) return self.cleaned_data["name"] class Meta: model = Map fields = ('name', 'slug', 'description')

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  • Writing a custom auth system (like the default django auth system), use it to generate tables in DB

    - by dotty
    Hay all, I've been reading up on middleware and how to use it with a context object. I want to write a simple middleware class which i can use on my own applications, it will essentially be a cut down version of the django one. The problem i seem to have is that if i have INSTALLED_APPS = ('django.contrib.my_auth') in the settings file, all is well. I've also added MIDDLEWARE_CLASSES = ('django.contrib.my_auth.middleware.MyAuthMiddleware') in it and everything is fine. My question is, how would i make my middleware automatically generate tables from a models.py module, much like how the django auth does when i run manage.py syncdb? thanks

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