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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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  • 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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  • Formatting inline many-to-many related models presented in django admin

    - by Jonathan
    I've got two django models (simplified): class Product(models.Model): name = models.TextField() price = models.IntegerField() class Invoice(models.Model): company = models.TextField() customer = models.TextField() products = models.ManyToManyField(Product) I would like to see the relevant products as a nice table (of product fields) in an Invoice page in admin and be able to link to the individual respective Product pages. My first thought was using the admin's inline - but django used a select box widget per related Product. This isn't linked to the Product pages, and also as I have thousands of products, and each select box independently downloads all the product names, it quickly becomes unreasonably slow. So I turned to using ModelAdmin.filter_horizontal as suggested here, which used a single instance of a different widget, where you have a list of all Products and another list of related Products and you can add\remove products in the later from the former. This solved the slowness, but it still doesn't show the relevant Product fields, and it ain't linkable. So, what should I do? tweak views? override ModelForms? I Googled around and couldn't find any example of such code...

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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 templates array assignment

    - by Hulk
    The following is in views: rows=query.evaluation_set.all() row_arr = [] for row in rows: row_arr.append(row.row_details) dict.update({'row_arr' : row_arr ,'col_arr' : col_arr}) return render_to_response('valuemart/show.html',context_instance=RequestContext(request,{'dict': dict})) How to extract the row_Arr array in the templates in javascript and list out all its values.row_Arr contains data of a column <script> var row_arr = '{{dict.row_arr}}'; //extract values here </script> Thanks..

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  • Django and floatformat tag

    - by Hellnar
    Hello, I want to modify / change the way the floatformat works. By default it changes the input decimal as such: {{ 1.00|floatformat }} -> 1 {{ 1.50|floatformat }} -> 1.5 {{ 1.53|floatformat }} -> 1.53 I want to change this abit as such: If there is a floating part, it should keep the first 2 floating digits. If no floating (which means .00) it should simply cut out the floating part. IE: {{ 1.00|floatformat }} -> 1 {{ 1.50|floatformat }} -> 1.50 {{ 1.53|floatformat }} -> 1.53

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  • 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..

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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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  • Django url parameters

    - by Hulk
    How to pass two paramters in urls in django <script> url=/toolbox/display/" + id + "2"; window.location=url; </script> Also how is this handeled in urls.py (r'^display/(?P<rid>\d+)/(?P<param>\d+)/$', 'table_display'), In views, def table_display(request,rid,param): print param //This should print 2

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  • Django forms "not" using forms from models

    - by zubinmehta
    I have a form generated from various models and the various values filled go and sit in some other table. Hence, in this case I haven't used the inbuilt Django forms(i.e. I am not creating forms from models ). Now the data which is posted from the self made form is handled by view1 which should clean the data accordingly. How do I go about it and use the various functions clean and define validation errors (and preferably not do validation logic in the view itself!)

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  • Extended Django base-class with multiple instances

    - by Gijs
    I'm modeling a simple movie database using Django. models.py defines a base model Person. I extend Person into Actor and Director, which works as I imagined. Persons must be unique. When (in the Admin) I create an instance of Actor, and this person is also a Director, it won't save because of the unique = True. Any ideas how to solve this problem? (generic foreign keys?) Thx

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  • Django: How to get current user in admin forms

    - by lazerscience
    In Django's ModelAdmin I need to display forms customized according to the permissions an user has. Is there a way of getting the current user object into the form class, so that i can customize the form in its __init__ method? I think saving the current request in a thread local would be a possibility but this would be my last resort think I'm thinking it is a bad design approach....

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  • Django filter vs exclude

    - by Enrico
    Is there a difference between filter and exclude in django? If I have self.get_query_set().filter(modelField=x) and I want to add another criteria, is there a meaningful difference between to following two lines of code? self.get_query_set().filter(user__isnull=False, modelField=x) self.get_query_set().filter(modelField=x).exclude(user__isnull=True) is one considered better practice or are they the same in both function and performance?

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  • Django User model, adding function

    - by Hellnar
    Hello, I want to add a new function to the default User model of Django for retrieveing a related list of Model type. Such Foo model: class Foo(models.Model): owner = models.ForeignKey(User, related_name="owner") likes = models.ForeignKey(User, related_name="likes") ........ #at some view user = request.user foos= user.get_related_foo_models() Hwo can this be achieved ?

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  • How to store an integer leaded by zeros in django

    - by Oscar Carballal
    Hello, I'm trying to store a number in django that looks like this: 000001 My problem is that if I type this inside an IntegerField it gets converted to "1" without the leading zeros. I've tried also with a DecimalField with the same result. How can I store the leading zeros whithout using a CharField? (I need to manipulate that number in it's integer form)

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  • How to make custom join query with Django ?

    - by xRobot
    I have these 2 models: genre = ( ('D', 'Dramatic'), ('T', 'Thriller'), ('L', 'Love'), ) class Book(models.Model): title = models.CharField(max_length=100) genre = models.CharField(max_length=1, choices=genre) class Author(models.Model): user = models.ForeignKey(User, unique=True) born = models.DateTimeField('born') book = models.ForeignKey(Book) I need to retrieve first_name and last_name of all authors of dramatic's books. How can I do this in django ?

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  • Django: name of many to many items in the admin interface

    - by Adam
    I have a many to many field, which I'm displaying in the django admin panel. When I add multiple items, they all come up as "ASGGroup object" in the display selector. Instead, I want them to come up as whatever the ASGGroup.name field is set to. How do I do this? My models looks like: class Thing(Model): read_groups = ManyToManyField('ASGGroup', related_name="thing_read", blank=True) class ASGGroup(Model): name = CharField(max_length=63, null=True) But what I'm seeing the m2m widget display is:

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  • Accessing updated M2M fields in overriden save() in django's admin

    - by Jonathan
    I'd like to use the user updated values of a ManyToManyField in a model's overriden save() method when I save an instance in admin. It turns out that by design, django does not update the M2M field before calling save(), but only after the save() is complete as part of the form save... How can I access the new values of this field in the override save() ?

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  • Django: Inherit Permssions from abstract models?

    - by lazerscience
    Is it possible to inherit permissions from an abstract model in Django? I can not really find anything about that. For me this doesn't work! class PublishBase(models.Model): class Meta: abstract = True get_latest_by = 'created' permissions = (('change_foreign_items', "Can change other user's items"),) EDIT: Not working means it fails silently. Permission is not created, as it wouldn't exist on the models inheriting from this class.

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