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  • Django | how to append form field to the urlconf

    - by MMRUser
    I want to pass a form's field value to the next page (template) after user submit the page, the field could be user name, consider the following setup def form_submit(request): if request.method == 'POST': form = UsersForm(request.POST) if form.is_valid(): cd = form.cleaned_data try: newUser = form.save() return HttpResponseRedirect('/mysite/nextpage/') except Exception, ex: return HttpResponse("Ane apoi %s" % str(ex)) else: return HttpResponse('Error') "nextpage" is the template that renders after user submit the form, so I want to know how to append the form's field (user name) to the url and get that value from the view in order to pass it to the next page.. thanks.

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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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  • Per instance dynamic fields django model

    - by Roberto Rosario
    I have a model with a JSON field or a link to a CouchDB document. I can currently access the dynamic informaction in a way such as: genericdocument.objects.get(pk=1) == genericdocument.json_field['sample subfield'] instead I would like genericdocument.sample_subfield to maintain compatibility with all the apps the project currently shares.

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  • Django: get count of ForeignKey item in template?

    - by AP257
    Straightforward question - apologies if it is a duplicate, but I can't find the answer if so. I have a User model and a Submission model, like this: class Submission(models.Model): uploaded_by = models.ForeignKey('User') class User(models.Model): name = models.CharField(max_length=250 ) How can I show the number of Submissions made by each user in the template? I've tried {{ user.submission.count }}, like this: for user in users: {{ user.name }} ({{ user.submission.count }} submissions) but no luck...

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  • Wordpress & Django -- One domain, two servers. Possible?

    - by DomoDomo
    My question is about hosting Django and Wordpress under one domain, but two physical machines (actually, they are VMs but same diff). Let's say I have a Django webapp at example.com. I'd like to start a Wordpress blog about my webapp, so any blog page rank mojo flows back to my webapp, I'd like the blog address t be example.com/blog. My understanding is blog.example.com would not transfer said page rank mojo. Because I'm worried about Wordpress security flaws compromising my Django webapp, I want to host Django and Wordpress on two physically separate machines. Given all that, is it possible using re-write rules or a reverse proxy server to do this? I know the easy way is to make my Wordpress blog a subdomain, but I really don't want to do that. Has anyone done this in the past, is it stable? If I need a third server to be a dedicated reverse proxy, that's totally fine. Thanks!

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  • Creating a Group of Groups in Django

    - by Greg
    I'm creating my own Group model; I'm not referring to the builtin Group model. I want each hroup to be a member of another group (it's parent), but there is the one "top" group that doesn't have a parent group. The admin interface won't let me create a group without entering a parent. I get the error personnel_group.parent_id may not be NULL. My Group model looks like this: class Group(models.Model): name = models.CharField(max_length=50) parent = models.ForeignKey('self', blank=True, null=True) order = models.IntegerField() icon = models.ImageField(upload_to='groups', blank=True, null=True) description = models.TextField(blank=True, null=True) How can I accomplish this? Thanks.

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  • django model relation definition

    - by Laurent Luce
    Hello, Let say I have 3 models: A, B and C with the following relations. A can have many B and many C. B can have many C Is the following correct: class A(models.Model): ... class B(models.Model): ... a = ForeignKey(A) class C(models.Model): ... a = ForeignKey(A) b = ForeignKey(B)

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  • How to make django test framework read from live database?

    - by lfborjas
    I realize there's a similar question here, but this one has a different approach: I have a django app that does queries over data indexed with djapian ; I'd like to write unit tests for this app's search component, and, obviously, I'd need the django settings module and all connections with the database active, so the test runner that django provides seems ideal. however, the django testing framework creates a dummy database and I'd hate to dump all my data to a fixture and then index it (the tests would take forever!); My data isn't at risk because the tests would only read from the database, so, how could this be achieved? -I'm new at this whole unit testing thing, so the solution of writing a new test runner I read in that similar question doesn't enlighten me a bit, at least not without some details

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  • Django | capture sub domain as a string

    - by MMRUser
    How do I capture a part of sub-domain name and get that name as a string in my views through a request. ex: user.domain.com developer.domain.com I want to capture the user part of this domain name through a request (lets say when the first time user hits the page). Thanks.

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  • Django Aggregation Across Reverse Relationship

    - by Tom
    Given these two models: class Profile(models.Model): user = models.ForeignKey(User, unique=True, verbose_name=_('user')) about = models.TextField(_('about'), blank=True) zip = models.CharField(max_length=10, verbose_name='zip code', blank=True) website = models.URLField(_('website'), blank=True, verify_exists=False) class ProfileView(models.Model): profile = models.ForeignKey(Profile) viewer = models.ForeignKey(User, blank=True, null=True) created = models.DateTimeField(auto_now_add=True) I want to get all profiles sorted by total views. I can get a list of profile ids sorted by total views with: ProfileView.objects.values('profile').annotate(Count('profile')).order_by('-profile__count') But that's just a dictionary of profile ids, which means I then have to loop over it and put together a list of profile objects. Which is a number of additional queries and still doesn't result in a QuerySet. At that point, I might as well drop to raw SQL. Before I do, is there a way to do this from the Profile model? ProfileViews are related via a ForeignKey field, but it's not as though the Profile model knows that, so I'm not sure how to tie the two together. As an aside, I realize I could just store views as a property on the Profile model and that may turn out to be what I do here, but I'm still interested in learning how to better use the Aggregation functions.

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  • django ifequal naturalday

    - by Scott Willman
    I'm not sure why, but this condition will never evaluate True for me. I'm feeding it datetime.today() in the urls file. Am I missing something? Template: {% load humaize %} {{ entry.date|naturalday }} {# Evals to "today" #} {% ifequal entry.date|naturalday "today" %} True {{ entry.date|date:"fA"|lower }} {{ entry.date|naturalday|title }} {% else %} False {{ entry.date|naturalday|title }} {% endifequal %}

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  • Django Template Inheritance -- Missing Images?

    - by user367817
    Howdy, I have got the following file heirarchy: project   other stuff   templates       images           images for site       app1           templates for app1       registration           login template       base.html (base for entire site)       style.css (for base.html) In the login template, I am extending 'base.html.' 'base.html' uses 'style.css' along with all of the images in the 'templates/images' directory. For some reason, none of the CSS styles or images will show up in the login template, even though I'm extending it. Does this missing image issue have something to do with screwed up "media" settings somewhere? I never understood those, but this is a major roadblock in my proof-of-concept, so any help is appreciated. Thanks!

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  • optimizing the jquery in django

    - by sankar
    I have done the following code which actually dynamically generate the values in the third list box using ajax and jquery concepts. Thoug it works, i need to optimize it. Below is the code i am working on now. <html> <head> <title>Comparison based on ID Number</title> <script src="http://code.jquery.com/jquery-latest.js"></script> </head> <body> {% if form.errors %} <p style="color: red;"> Please correct the error{{ form.errors|pluralize }} below. </p> {% endif %} <form action="/idnumber/" method="post" align= "center">{% csrf_token %} <p>Select different id numbers and the name of the <b> Measurement Group </b>to start the comparison</p> <table align = "center"> <tr> <th><label for="id_criteria_1">Id Number 1:</label></th> <th> {{ form.idnumber_1 }} </th> </tr> <tr> <th><label for="id_criteria_2">Id Number 2:</label></th> <th> {{ form.idnumber_2 }} </th> </tr> <tr> <th><label for="group_name_list">Group Name:</label></th> <th><select name="group_name_list" id="id_group_name_list"> <option>Select</option> </select> </th> </tr> <script> $('#id_idnumber_2').change( function get_group_names() { var value1 = $('#id_idnumber_1').attr('value'); var value2 = $(this).attr('value'); alert(value1); alert(value2); var request = $.ajax({ url: "/getGroupnamesforIDnumber/", type: "GET", data: {idnumber_1 : value1,idnumber_2 : value2}, dataType: "json", success: function(data) { alert(data); var myselect = document.getElementById("group_name_list"); document.getElementById("group_name_list").options.length = 1; var length_of_data = data.length; alert(length_of_data); try{ for(var i = 0;i < length_of_data; i++) { myselect.add(new Option(data[i].group_name, "i"), myselect.options[i]) //add new option to beginning of "sample" } } catch(e){ //in IE, try the below version instead of add() for(var i = 0;i < length_of_data; i++) { myselect.add(new Option(data[i].group_name, data[i].group_name)) //add new option to end of "sample" } } } }); }); </script> <tr align = "center"><th><input type="submit" value="Submit"></th></tr> </table> </form> </body> </html> everything works fine but there is a little problem in my code. (ie) my ajax function calls only when there is a change in the select list 2 (ie) 'id_number_2'. I want to make it call in such a way that which ever select box, the third list box should be updated automatically. Can anyone please help me on this with a possible logical solution

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  • Asynchronous database update in Django?

    - by Mark
    I have a big form on my site. When the users fill it out and submit it, most of the data just gets dumped to the database, and then they get redirected to a new page. However, I'd also like to use the data to query another site, and then parse the results. That might take a bit longer. It's not essential that the user sees these results right away, so I was wondering if it's possible to asynchronously call a function that will handle this, and then return an HttpResponse from my view like usual without making them wait? If so... how? Any particular libraries I should look at?

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  • django many to many validation when add()

    - by Julien
    Hi i have a Category model with parent/child self relation For primary category and sub categories : class Place(models.Model): name = models.CharField(_("name"), max_length=100) categories = models.ManyToManyField("Category", verbose_name=_("categories")) class Category(models.Model): name = models.CharField(_("name"), max_length=100) parent = models.ForeignKey('self', blank=True, null=True, related_name='child_set') i need to prevent orphans, to prevent this kind of errors (in admin web interface) c_parent = Category(name='Restaurant') c_parent.save() c_child = Category(name="Japanese restaurant", parent=c_parent) c_child.save() place1 = Place (name="Planet sushi") place1.save() place1.categories.add(c_parent) place1.categories.add(c_child) So now we have a new Place called "Planet sushi", it's a Restaurant (root category), and a Japanese Restaurant (sub category) but i want to prevent this kind of things : place2 = Place (name="Tokyofood") place2.save() place2.categories.add(c_child) because parent is not set, or is not the correct parent category where can i do form validation for the admin ? and other forms (because any user can add a new place and will have to choose correct categories for)

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  • Django forms I cannot save picture file

    - by dana
    i have the model: class OpenCv(models.Model): created_by = models.ForeignKey(User, blank=True) first_name = models.CharField(('first name'), max_length=30, blank=True) last_name = models.CharField(('last name'), max_length=30, blank=True) url = models.URLField(verify_exists=True) picture = models.ImageField(help_text=('Upload an image (max %s kilobytes)' %settings.MAX_PHOTO_UPLOAD_SIZE),upload_to='jakido/avatar',blank=True, null= True) bio = models.CharField(('bio'), max_length=180, blank=True) date_birth = models.DateField(blank=True,null=True) domain = models.CharField(('domain'), max_length=30, blank=True, choices = domain_choices) specialisation = models.CharField(('specialization'), max_length=30, blank=True) degree = models.CharField(('degree'), max_length=30, choices = degree_choices) year_last_degree = models.CharField(('year last degree'), max_length=30, blank=True,choices = year_last_degree_choices) lyceum = models.CharField(('lyceum'), max_length=30, blank=True) faculty = models.ForeignKey(Faculty, blank=True,null=True) references = models.CharField(('references'), max_length=30, blank=True) workplace = models.ForeignKey(Workplace, blank=True,null=True) the form: class OpencvForm(ModelForm): class Meta: model = OpenCv fields = ['first_name','last_name','url','picture','bio','domain','specialisation','degree','year_last_degree','lyceum','references'] and the view: def save_opencv(request): if request.method == 'POST': form = OpencvForm(request.POST, request.FILES) # if 'picture' in request.FILES: file = request.FILES['picture'] filename = file['filename'] fd = open('%s/%s' % (MEDIA_ROOT, filename), 'wb') fd.write(file['content']) fd.close() if form.is_valid(): new_obj = form.save(commit=False) new_obj.picture = form.cleaned_data['picture'] new_obj.created_by = request.user new_obj.save() return HttpResponseRedirect('.') else: form = OpencvForm() return render_to_response('opencv/opencv_form.html', { 'form': form, }, context_instance=RequestContext(request)) but i don't seem to save the picture in my database... something is wrong, and i can't figure out what :(

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  • Putting links into text in Django

    - by Dane Larsen
    I have a notifications app that generates notifications for users. The notification class has to be really general, because notifications are generated by all sorts of different things. My question is this: How do I insert links into the text of the notifications? What I tried was this: note = Notification(..., notification="""%s %s has accepted the task: <a href="/tasks/%d/">%s</a>.""" % (request.user.first_name, request.user.last_name, task.id, task.name), ...) In retrospect, it's obvious this wouldn't work. How should I go about this? Thanks in advance!

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