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  • Django formset doesn't validate

    - by tsoporan
    Hello, I am trying to save a formset but it seems to be bypassing is_valid() even though there are required fields. To test this I have a simple form: class AlbumForm(forms.Form): name = forms.CharField(required=True) The view: @login_required def add_album(request, artist): artist = Artist.objects.get(slug__iexact=artist) AlbumFormSet = formset_factory(AlbumForm) if request.method == 'POST': formset = AlbumFormSet(request.POST, request.FILES) if formset.is_valid(): return HttpResponse('worked') else: formset = AlbumFormSet() return render_to_response('submissions/addalbum.html', { 'artist': artist, 'formset': formset, }, context_instance=RequestContext(request)) And the template: <form action="" method="post" enctype="multipart/form-data">{% csrf_token %} {{ formset.management_form }} {% for form in formset.forms %} <ul class="addalbumlist"> {% for field in form %} <li> {{ field.label_tag }} {{ field }} {{ field.errors }} </li> {% endfor %} </ul> {% endfor %} <div class="inpwrap"> <input type="button" value="add another"> <input type="submit" value="add"> </div> </form> What ends up happening is I hit "add" without entering a name then HttpResponse('worked') get's called seemingly assuming it's a valid form. I might be missing something here, but I can't see what's wrong. What I want to happen is, just like any other form if the field is required to spit out an error if its not filled in. Any ideas?

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  • Website stress test in Python - Django

    - by RadiantHex
    Hi folks, I'm trying to build a small stress test script to test how quickly a set of requests gets done. Need to measure speed for 100 requests. Problem is that I wouldn't know how to implement it, as it would require parallel url requests to be called. Any ideas?

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  • Check request type in Django

    - by Art
    While it is recommended to use the following construct to check whether request is POST, if request.method == 'POST': pass It is likely that people will find if request.POST: pass to be more elegant and concise. Are there any reasons not to use it, apart from personal preference?

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  • Getting an entry before and after a given entry in a Django Queryset

    - by Vernon
    I am creating a simple blog as part of a website and I am getting stuck on something that I am assuming is simple. If I call any blog post, say by it's title, from a queryset, how can I get the entry before and after the post in it's published order. I can iterate over the whole thing, get the position of the entry I have and use that to call the one before and the one after. But that is a long bit of code for something that I am sure I can do more simply. What I want would be something like this: next_post = Posts.object.filter(title=current_title).order_by("-published")[-1] Of course because of the filter, it is not going to work, but just to give you the idea of what I am looking for.

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  • get average from set of objects in django

    - by dotty
    Hay, i have a simple rating system for a property. You give it a mark out of 5 (stars). The models are defined like this def Property(models.Model) # stuff here def Rating(models.Model) property = models.ForeignKey(Property) stars = models.IntegerField() What i want to do is get a property, find all the Rating objects, collect them, then get the average 'stars' from them. any ideas how to do this?

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  • Django Grouping Query

    - by Matt
    I have the following (simplified) models: class Donation(models.Model): entry_date = models.DateTimeField() class Category(models.Model): name = models.CharField() class Item(models.Model): donation = models.ForeignKey(Donation) category = models.ForeignKey(Category) I'm trying to display the total number of items, per category, grouped by the donation year. I've tried this: Donation.objects.extra(select={'year': "django_date_trunc('year', %s.entry_date)" % Donation._meta.db_table}).values('year', 'item__category__name').annotate(items=Sum('item__quantity')) But I get a Field Error on item__category__name. I've also tried: Item.objects.extra(select={"year": "django_date_trunc('year', entry_date)"}, tables=["donations_donation"]).values("year", "category__name").annotate(items=Sum("quantity")).order_by() Which generally gets me what I want, but the item quantity count is multiplied by the number of donation records. Any ideas? Basically I want to display this: 2010 - Category 1: 10 items - Category 2: 17 items 2009 - Category 1: 5 items - Category 3: 8 items

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  • Django/jQuery - read file and pass to browser as file download prompt

    - by danspants
    I've previously asked a question regarding passing files to the browser so a user receives a download prompt. However these files were really just strings creatd at the end of a function and it was simple to pass them to an iframe's src attribute for the desired effect. Now I have a more ambitious requirement, I need to pass pre existing files of any format to the browser. I have attempted this using the following code: def return_file(request): try: bob=open(urllib.unquote(request.POST["file"]),"rb") response=HttpResponse(content=bob,mimetype="application/x-unknown") response["Content-Disposition"] = "attachment; filename=nothing.xls" return HttpResponse(response) except: return HttpResponse(sys.exc_info()) With my original setup the following jQuery was sufficient to give the desired download prompt: jQuery('#download').attr("src","/return_file/"); However this won't work anymore as I need to pass POST values to the function. my attempt to rectify that is below, but instead of a download prompt I get the file displayed as text. jQuery.get("/return_file/",{"file":"c:/filename.xls"},function(data) { jQuery(thisButton).children("iframe").attr("src",data); }); Any ideas as to where I'm going wrong? Thanks!

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  • Django multiple generic_inline_formset in a view

    - by Can Burak Cilingir
    We have a bunch of formsets: EmailAddressInlineFormSet = generic_inlineformset_factory( EmailAddress, extra=1, exclude=["created_by","last_modified_by"]) emailaddressformset = EmailAddressInlineFormSet( instance=person, prefix="emailaddress") # [ more definitions ] and, in the view, we process them as: emailaddressformset = EmailAddressInlineFormSet( request.POST, instance=person, prefix="emailaddress") # [ more definitions ] So, nothing fancy or unordinary. The unfortunate or unordinary fact is, we have 15 of these formsets, one for email addresses, other for phone numbers etc. so the view code is ugly and not-so-manageable. What would be the most unhackish way to handle this number of formsets in a single view? At the end -i guess- I'm looking for a class or a functionality like multiple_generic_inline_formset and open to all kind of suggestions or discussions.

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  • django m2m how can i get m2m table elements in a view

    - by dana
    i have a model using m2m feature: class Classroom(models.Model): user = models.ForeignKey(User, related_name = 'classroom_creator') classname = models.CharField(max_length=140, unique = True) date = models.DateTimeField(auto_now=True) open_class = models.BooleanField(default=True) members = models.ManyToManyField(User,related_name="list of invited members", through = 'Membership') and i want to take all members of one class in a view and display them using the template system. In the view, i'm trying to take all the members from a classroom like that: def inside_classroom(request,classname): try: theclass = Classroom.objects.get(classname = classname) members = Members.objects.all() etc but it doesn't work,(though the db_table is named Classroom_Members) i guess i have to use another query for getting all the members from the classroom classname. also, i want to verify if the request.user is a member using (if request.user in members) how can i het those members? Thanks in advance!

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  • django handling file uploads - target different than media folder

    - by Tom Tom
    Hi, I want to enable the user to upload media which will not be saved in the media folder. When I use the following line of code data will be uploaded to media/upload/logo . logo_img = models.FileField(upload_to='upload/logo', blank=True) I'm wondering how I can change this behaviour. I would try to write a custom FileField and a view that serves the data based on the database entries. I do not want to place the data the user uploads to the media folder, since it is no public data. Is this approach correct? Are there solutions out there which do exactly what I want and I would reinvent the wheel with implementing this by myself? Would appreciate any help!

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  • [Django] Change state of obiects

    - by gameboy
    hi I have following problem. I have model: class Towar(models.Model): nrSeryjny=models.CharField(max_length=100) opis=models.CharField(max_length=255) naStanie=models.NullBooleanField(null=True) def __unicode__(self): return "%s" % self.opis def lowerName(self): return self.__class__.__name__.lower() def checkState(self): return self.naStanie def changeState(self,state): self.naStanie=state class Meta: ordering=['nrSeryjny'] app_label = 'baza' permissions=(("view_towar","mozna miec podglad dla towar"),) and model : class Wypozyczenie(models.Model): dataPobrania=models.DateField() pracownik=models.ForeignKey(User,null=True) kontrahent=models.ForeignKey(Kontrahenci,null=True) towar=models.ForeignKey(Towar,null=True) objects=WypozyczenieManager() default_objects=models.Manager() ZwrotyObjects=WypozyczenieZwrotyManager() def lowerName(self): return self.__class__.__name__.lower() def __unicode__(self): if self.towar == None: return "Dla:%s -- Kto:%s -- Kiedy:%s -- Co:%s" % (self.kontrahent,self.pracownik,self.dataPobrania,"Brak") else: return "Dla:%s -- Kto:%s -- Kiedy:%s -- Co:%s" % (self.kontrahent,self.pracownik,self.dataPobrania,self.towar) class Meta: ordering=['dataPobrania'] app_label = 'baza' permissions=(("view_wypozyczenie","mozna miec podglad dla wypozyczenie"),) and view to adding models: def modelAdd(request,model,modelForm): mod=model() if request.user.has_perm('baza.add_%s' % mod.lowerName()): if request.method=='POST': form=modelForm(request.POST) if form.is_valid(): form.save() return HttpResponseRedirect('/'+ mod.lowerName() + '/') else: form=modelForm() v=RequestContext(request,{'form':form}) return render_to_response('add_form.html',v) and i whant do that, when i add Wypozyczenie and save it then the Towar that is stored by Wypozyczenie change his na stanie from True to False Greets

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  • django: How to make one form from multiple models containing foreignkeys

    - by Tim
    I am trying to make a form on one page that uses multiple models. The models reference each other. I am having trouble getting the form to validate because I cant figure out how to get the id of two of the models used in the form into the form to validate it. I used a hidden key in the template but I cant figure out how to make it work in the views My code is below: views: def the_view(request, a_id,): if request.method == 'POST': b_form= BForm(request.POST) c_form =CForm(request.POST) print "post" if b_form.is_valid() and c_form.is_valid(): print "valid" b_form.save() c_form.save() return HttpResponseRedirect(reverse('myproj.pro.views.this_page')) else: b_form= BForm() c_form = CForm() b_ide = B.objects.get(pk=request.b_id) id_of_a = A.objects.get(pk=a_id) return render_to_response('myproj/a/c.html', {'b_form':b_form, 'c_form':c_form, 'id_of_a':id_of_a, 'b_id':b_ide }) models class A(models.Model): name = models.CharField(max_length=256, null=True, blank=True) classe = models.CharField(max_length=256, null=True, blank=True) def __str__(self): return self.name class B(models.Model): aid = models.ForeignKey(A, null=True, blank=True) number = models.IntegerField(max_length=1000) other_number = models.IntegerField(max_length=1000) class C(models.Model): bid = models.ForeignKey(B, null=False, blank=False) field_name = models.CharField(max_length=15) field_value = models.CharField(max_length=256, null=True, blank=True) forms from mappamundi.mappa.models import A, B, C class BForm(forms.ModelForm): class Meta: model = B exclude = ('aid',) class CForm(forms.ModelForm): class Meta: model = C exclude = ('bid',) B has a foreign key reference to A, C has a foreign key reference to B. Since the models are related, I want to have the forms for them on one page, 1 submit button. Since I need to fill out fields for the forms for B and C & I dont want to select the id of B from a drop down list, I need to somehow get the id of the B form into the form so it will validate. I have a hidden field in the template, I just need to figure how to do it in the views

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  • I am having issues with django test

    - by Mohamed
    I have this test case def test_loginin_student_control_panel(self): c = Client() c.login(username="tauri", password="gaul") response = c.get('/student/') self.assertEqual(response.status_code, 200) the view associated with the test case is this @login_required def student(request): return render_to_response('student/controlpanel.html') so my question is why the above test case redirects user to login page? should not c.login suppose to take care authenticating user?

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  • How to create instances of related models in Django

    - by sevennineteen
    I'm working on a CMSy app for which I've implemented a set of models which allow for creation of custom Template instances, made up of a number of Fields and tied to a specific Customer. The end-goal is that one or more templates with a set of custom fields can be defined through the Admin interface and associated to a customer, so that customer can then create content objects in the format prescribed by the template. I seem to have gotten this hooked up such that I can create any number of Template objects, but I'm struggling with how to create instances - actual content objects - in those templates. For example, I can define a template "Basic Page" for customer "Acme" which has the fields "Title" and "Body", but I haven't figured out how to create Basic Page instances where these fields can be filled in. Here are my (somewhat elided) models... class Customer(models.Model): ... class Field(models.Model): ... class Template(models.Model): label = models.CharField(max_length=255) clients = models.ManyToManyField(Customer, blank=True) fields = models.ManyToManyField(Field, blank=True) class ContentObject(models.Model): label = models.CharField(max_length=255) template = models.ForeignKey(Template) author = models.ForeignKey(User) customer = models.ForeignKey(Customer) mod_date = models.DateTimeField('Modified Date', editable=False) def __unicode__(self): return '%s (%s)' % (self.label, self.template) def save(self): self.mod_date = datetime.datetime.now() super(ContentObject, self).save() Thanks in advance for any advice!

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  • Model objects versions in Django

    - by pablo
    Hi I'm building an e-commerce website. I have a Product and Order models. It's possible that a customer order a product and then the admin change its price or other fields before the customer actually get it. A possible solution is to add a 'version' field to the Product model. When the admin update a product field I'll add a timestamp and create a new object instead of updating the old one. An Order will have a reference to a specific product version. Does this make sense? Will overriding the Product Save method be sufficient to make it work? Thanks

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  • Django notification get date one accesses a link

    - by dana
    hi there, i'm making a notification system, so that a user in a virtual community to be announced when someone sends him a message, or starts following him (the follow relation is like a friend relation, but it is not necessarily reciprocal) my view function: def notification_view(request, last_checked): u = Relation.objects.filter(date_follow>Notification.objects.get(last_checked=last_checked)) v = Message.objects.filter(date>Notification.objects.get(last_checked=last_checked)) if u: notification_type = follow if notice_settings == receive_notification or notice_settings == only_follow following = u if v: notification_type = message if notice_settings == receive_notification or notice_settings == only_messages message = v return render_to_response('notification/notification.html', { 'following': following, 'message':message, }, context_instance=RequestContext(request)) the models.py: class NoticeType(models.Model): follow = models.ForeignKey(Relations, editable = False) message = models.ForeignKey(Messages) classroom_invitation = models.ForeignKey(Classroom) class Notification(models.Model): receiver = models.ForeignKey(User, editable=False) date = models.DateTimeField(auto_now=True, editable = False) notice_type = models.ForeignKey(NoticeType, editable = False, related_name = "notification_type") sent = models.BooleanField(default = True) last_checked = models.DateTimeField(auto_now=True, editable = False) class NotificationSettings(models.Model): user = models.ForeignKey(User) receive_notifications = models.BooleanField(default = True) only_follow = models.BooleanField(default = False) only_message = models.BooleanField(default = False) only_classroom = models.BooleanField(default = False) #receive_on_email = models.BooleanField(default = False) my problem is: i want last_checked to be the time when someone acceses a link (the notification link). How can i possibily save that time? how can i get it? thanks in avance!

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  • Django: problem with merging querysets after annotation

    - by Björn Lilja
    Hi I have a manager for "Dialog" looking like this: class AnnotationManager(models.Manager): def get_query_set(self): return super(AnnotationManager, self).get_query_set().annotate( num_votes=Count('vote', distinct=True), num_comments=Count('comment', distinct=True), num_commentators = Count('comment__user', distinct=True), ) Votes and Comments has a ForeignKey to Dialog. Comments has a ForeignKey to User. When I do this: dialogs_queryset = Dialog.public.filter(organization=organization) dialogs_popularity = dialogs_queryset.exclude(num_comments=0) | dialogs_queryset.exclude(num_votes=0) ...dialogs_popularity will never returned the combination, but only the dialogs with more than 0 comments, or if I change the order of the OR, the dialogs with more than 0 votes! To me, the expected behavior would be to get the dialogs with more than 0 votes AND the dialogs with more than 0 comments. What am I missing? Or is there a bug in the annotation behavior here?

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  • Create hyperlink in django template of object that has a space

    - by Ed
    I am trying to create a dynamic hyperlink that depends on a value passed from a function: {% for item in field_list %} <a href={% url index_view %}{{ item }}/> {{ item }} </a> <br> {% endfor %} The problem is that one of the items in field_list is "Hockey Player". The link for some reason is dropping everything after the space, so it creates the hyperlink on the entire "Hockey Player", but the address is http://126.0.0.1:8000/Hockey How can I get it to go to http://126.0.0.1:8000/Hockey Player/ instead?

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  • how to number t-rows ,when table generated using nested forloop in django templates

    - by stackover
    Hi, This part is from views.py results=[(A,[stuObj1,stuObj2,stuObj3]),(B,[stuObj4,stuObj5,stuObj6]),(C,[stuObj7,stuObj8])] for tup in results: total = tot+len(tup[1]) render_to_response(url,{'results':res , 'total':str(tot),}) this is template code: <th class="name">Name</th> <th class="id">Student ID</th> <th class="grade">Grade</th> {% for tup in results %} {% for student in tup|last %} {% with forloop.parentloop.counter as parentid%} {% with forloop.counter as centerid%} <tbody class="results-body"> <tr> <td>{{student.fname|lower|capfirst}} {{student.lname|lower|capfirst}}</td> <td>{{student.id}}</td> <td>{{tup|first}}</td> </tr> {% endfor %} {% endfor %} Now the problems am having are 1. numbering the rows. Here my problem is am not sure if i can do things like total=total-1 in the templates to get the numbered rows like <td>{{total}}</td> 2.applying css to tr:ever or odd. Whats happening in this case is everytime the loop is running the odd/even ordering is lost. these seems related problems. Any ideas would be great :)

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  • Combining regroup with get_foo_display in Django templates

    - by shacker
    I'm using the regroup template tag to group queryset output on a Choices field. In the model: RESOURCE_TYPES = ( ('tut','External tutorial'), ('read','Additional reading'), ('org','Company or organization'), ) restype = models.CharField('Resource type',max_length=6,choices=RESOURCE_TYPES) in the view: resources = Resource.objects.filter(tutorial=tutorial) in the template: {% regroup resources by restype as resource_list %} {% for type in resource_list %} <h3>{{type.grouper}}</h3> So type.grouper renders as 'tut' or 'org' on the page, rather than the long form. Normally you would use the get_foo_display syntax to get at the value of the choice, rather than the key. But the value doesn't seem to be available after going through regroup. There's no way I can find to use get_foo_display on {{type.grouper}}. It makes sense when you think about it, but what's the workaround? Thanks.

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