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  • Django foreign key error

    - by Hulk
    In models the code is as, 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() In views, p_l=header.objects.filter(id=rid) for rows in row_data: row_query =criteria(details=rows,headerid=p_l) row_query.save() In row_query =criteria(details=rows,headerid=p_l) there is an error saying 'long' object is not callable in models.py in __unicode__, What is wrong in the code Thanks..

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  • 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?

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  • 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

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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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  • Django: customizing the message after a successful form save

    - by chiurox
    Hello, whenever I save a model in my Admin interface, it displays the usual "successfully saved message." However, I want to know if it's possible to customize this message because I have a situation where I want to warn the user about what he just saved and the implications of these actions. class PlanInlineFormset(forms.models.BaseInlineFormset): def clean(self): ### How can I detect the changes? ### (self.changed_data doesn't work because it's an inline) ### and display what he/she just changed at the top AFTER the successful save? class PlanInline(admin.TabularInline): model = Plan formset = PlanInlineFormset

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  • 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?

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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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  • querying for timestamp field in django

    - by Hulk
    In my views i have the date in the following format s_date=20090106 and e_date=20100106 The model is defined as class Activity(models.Model): timestamp = models.DateTimeField(auto_now_add=True) how to query for the timestamp filed with the above info. Activity.objects.filter(timestamp>=s_date and timestamp<=e_date) Thanks.....

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  • Possible form field types per model field type

    - by Jonathan
    Django's documentation specifies for each model field type the corresponding default form field type. Alas, I couldn't find in the documentation, or anywhere else, what form field types are possible per model field type. Not all combinations are possible, right? Same question for widgets...

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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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  • user inheritance in django

    - by amateur
    Hi guys, I saw a couple of ways extending user information of users and decided to adopt the model inheritance method. for instance, I have : class Parent(User): contact_means = models.IntegerField() is_staff = False objects = userManager() Now it is done, I've downloaded django_registration to help me out with sending emails to new users. The thing is, instead of using registration forms to register new user, I want to to invoke the email sending/acitvation capability of django_registration. So my workflow is: 1. add new Parent object in admin page. 2. send email My problem is, the django-registration creates a new registration profile together with a new user in the user table. how do I tweak this such that I am able to add the user entry into the custom user table. I have tried to create a modelAdmin and alter the save_model method to launch the create_inactive_user from django_registration, however I do not how to save the user object generated from django_registration into my Parent table when I have using model inheritance and I do not have a Foreign key attribute in my parent model.

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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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  • How to add manytomany field to flatpage admin

    - by valya
    Hello! I have a site with Flatpages, and now I need to be able to add galleries (Gallery model) to the page. This is going to be ManyToManyField, so there is no need to change the flatpages table. But I still can't define ManyToManyField in proxy class. I can define ManyToManyField in Gallery model, but it isn't comfortable for the client. How can I change FlatPage Admin to add a ManyToManyField from Galleries model?

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  • Are factors such as Intellisense support and strong typing enough to justify the use of an 'Anaemic Domain Model'?

    - by David Osborne
    It's easy to accept that objects should be used in all layers except a layer nominated as a data layer. However, it's just as easy to end-up with an 'anaemic domain model' that is just an object representation of data with no real functionality ( http://martinfowler.com/bliki/AnemicDomainModel.html ). However, using objects in this fashion brings the benefit of factors such as Intellisense support, strong typing, readability, discoverability, etc. Are these factors strong arguments for an otherwise, anaemic domain model?

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  • How to retrieve items from a django queryset?

    - by sharataka
    I'm trying to get the video element in a queryset but am having trouble retrieving it. user_channel = Everything.objects.filter(profile = request.user, playlist = 'Channel') print user_channel[0] #returns the first result without error print user_channel[0]['video'] #returns error Models.py: class Everything(models.Model): profile = models.ForeignKey(User) playlist = models.CharField('Playlist', max_length = 2000, null=True, blank=True) platform = models.CharField('Platform', max_length = 2000, null=True, blank=True) video = models.CharField('VideoID', max_length = 2000, null=True, blank=True) video_title = models.CharField('Title of Video', max_length = 2000, null=True, blank=True) def __unicode__(self): return u'%s %s %s %s %s' % (self.profile, self.playlist, self.platform, self.video, self.video_title)

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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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  • 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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  • 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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  • 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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  • Filter Queryset in Django inlineformset_factory

    - by Dave
    I am trying to use inlineformset_factory to generate a formset. My models are defined as: class Measurement(models.Model): subject = models.ForeignKey(Animal) experiment = models.ForeignKey(Experiment) assay = models.ForeignKey(Assay) values = models.CommaSeparatedIntegerField(blank=True, null=True) class Experiment(models.Model): date = models.DateField() notes = models.TextField(max_length = 500, blank=True) subjects= models.ManyToManyField(Subject) in my view i have: def add_measurement(request, experiment_id): experiment = get_object_or_404(Experiment, pk=experiment_id) MeasurementFormSet = inlineformset_factory(Experiment, Measurement, extra=10, exclude=('experiment')) if request.method == 'POST': formset = MeasurementFormSet(request.POST,instance=experiment) if formset.is_valid(): formset.save() return HttpResponseRedirect( experiment.get_absolute_url() ) else: formset = MeasurementFormSet(instance=experiment) return render_to_response("data_entry_form.html", {"formset": formset, "experiment": experiment }, context_instance=RequestContext(request)) but i want to restrict the Measurement.subject field to only subjects defined in the Experiment.subjects queryset. I have tried a couple of different ways of doing this but I am a little unsure what the best way to accomplish this is. I tried to over-ride the BaseInlineFormset class with a new queryset, but couldnt figure out how to correctly pass the experiment parameter.

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  • Sort list of items in Django

    - by mridang
    Hi Guys, I have a Django view called change_priority. I'm posting a request to this view with a commas separated list of values which is basically the order of the items in my model. The data looks like this: 1,4,11,31,2,4,7 I have a model called Items which has two values - id and priority. Upon getting this post request, how can I set the priority of the Item depending upon the list order. So my data in the db would look like. 1,1 4,2 11,3 31,4 2,5 4,6 7,7 Thanks guys.

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  • django-admin formfield_for_* change default value per/depending on instance

    - by Nick Ma.
    Hi, I'm trying to change the default value of a foreignkey-formfield to set a Value of an other model depending on the logged in user. But I'm racking my brain on it... This: Changing ForeignKey’s defaults in admin site would an option to change the empty_label, but I need the default_value. #Now I tried the following without errors but it didn't had the desired effect: class EmployeeAdmin(admin.ModelAdmin): ... def formfield_for_foreignkey(self, db_field, request=None, **kwargs): formfields= super(EmployeeAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs) if request.user.is_superuser: return formfields if db_field.name == "company": #This is the RELEVANT LINE kwargs["initial"] = request.user.default_company return db_field.formfield(**kwargs) admin.site.register(Employee, EmployeeAdmin) ################################################################## # REMAINING Setups if someone would like to know it but i think # irrelevant concerning the problem ################################################################## from django.contrib.auth.models import User, UserManager class CompanyUser(User): ... objects = UserManager() company = models.ManyToManyField(Company) default_company= models.ForeignKey(Company, related_name='default_company') #I registered the CompanyUser instead of the standard User, # thats all up and working ... class Employee(models.Model): company = models.ForeignKey(Company) ... Hint: kwargs["default"] ... doesn't exist. Thanks in advance, Nick

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