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  • Advanced Django query with subselects and custom JOINS

    - by Bryan Ward
    I have been investigating this number theoretic function (found in the Height model) and I need to query for things based on the prime factorization of the primary key, or id. I have created a model for Factors of the id which maintains all of the prime factors. class Height(models.Model): b = models.IntegerField(null=True, blank=True) c = models.IntegerField(null=True, blank=True) d = models.FloatField(null=True, blank=True) class Factors(models.Model): height = models.ForeignKey(Height, null=True, blank=True) factor = models.IntegerField(null=True, blank=True) degree = models.IntegerField(null=True, blank=True) prime_id = models.IntegerField(null=True, blank=True) For example, if id=24, then the associated entries in the factors table would be height_id=24,factor=2,degree=3,prime_id=0 height_id=24,factor=3,degree=1,prime_id=1 the prime_id keep track of the relative order of the primes. Now let p < q < r < s all be prime numbers and a,b,c,d be positive integers. Then I want to be able to query for all Heights of the form id=(p**a)*(q**b)*(r**c)*(s**d). Now this is simple in the case that all of p,q,r,s,a,b,c,d are known in that I can just run Height.objects.get(id=(p**a)*(q**b)*(r**c)*(s**d)) But I need to be able to query for something like (2**a)*(3**2)*(r**c)*(s**d) where r,s,a,d are unknown and all Heights of such form will be returned. Furthermore, not all of the rows in Height will have exactly four prime factors, so I need to make sure that I am not matching rows of the form id=(p**a)*(q**b)*(r**c)*(s**d)*(t**e)... From what I can tell, the following MySQL query accomplishes this, but I would like to do it through the Django ORM. I also don't know if this MySQL query is the proper way to go about doing things. SELECT h.*,count(f.height_id) AS factorsCount FROM height AS h LEFT JOIN factors AS f ON ( f.height_id = h.id AND f.height_id IN (SELECT height_id FROM factors where prime_id=1 AND factor=2 AND degree=1) AND f.height_id IN (SELECT height_id FROM factors where prime_id=2 AND factor=3 AND degree=2) AND f.height_id IN (SELECT height_id FROM factors where prime_id=3 AND factor=5 AND degree=1) AND f.height_id IN (SELECT height_id FROM factors where prime_id=4 AND factor=7 ANd degree=1) ) GROUP BY h.id HAVING factorsCount=4 ORDER BY h.id; Any ideas or suggestions for things to try?

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  • Django shows too many warnings when deleting an object

    - by valya
    Hello! I have two models: class Account(models.Model): main_request = models.ForeignKey('JournalistRequest', related_name='main_request') key = models.CharField(_('Key'), max_length=100) class JournalistRequest(models.Model): account = models.ForeignKey(Account, blank=True, null=True) When I try to delete a JournalistRequest, It shows warning with a lot of nesting, like Are you sure you want to delete the selected ?????? ??? objects? All of the following objects and their related items will be deleted: Journalist Request: some request Account: some account Journalist Request: some request Account: some account Journalist Request: some request Account: some account Journalist Request: some request Account: some account Journalist Request: some request All accounts are the same one (ids are same), and all requests are the same one, so I think it becaues of a recursion. But I have no idea how to solve this problem in Django 1.1.1! Can you help me?

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  • Django, making a page activate for a fixed time

    - by Hellnar
    Greetings I am hacking Django and trying to test something such as: Like woot.com , I want to sell "an item per day", so only one item will be available for that day (say the default www.mysite.com will be redirected to that item), Assume my urls for calling these items will be such: www.mysite.com/item/<number> my model for item: class Item(models.Model): item_name = models.CharField(max_length=30) price = models.FloatField() content = models.TextField() #keeps all the html content start_time = models.DateTimeField() end_time = models.DateTimeField() And my view for rendering this: def results(request, item_id): item = get_object_or_404(Item, pk=item_id) now = datetime.now() if item.start_time > now: #render and return some "not started yet" error templete elif item.end_time < now: #render and return some "item selling ended" error templete else: # render the real templete for selling this item What would be the efficient and clever model & templete for achieving this ?

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  • Annotate and Aggregate function in django

    - by thesteve
    In django I have the following tables and am trying to count the number of votes by item. class Votes(models.Model): user = models.ForeignKey(User) item = models.ForeignKey(Item) class Item(models.Model): name = models.CharField() description = models.TextField() I have the following queryset queryset = Votes.objects.values('item__name').annotate(Count('item')) that returns a list with item name and view count but not the item object. How can I set it up so that the object is returned instead of just the string value? I have been messing around with Manager and Queryset methods, that the right track? Any advice would be appreciated.

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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 urls on json request

    - by Hulk
    When making a django request through json as, var info=id + "##" +name+"##" $.post("/supervise/activity/" + info ,[] , function Handler(data,arr) { } In urls.py (r'^activity/(?P<info>\d+)/$, 'activity'), In views, def activity(request,info): print info The request does not go through.info is a string.How can this be resolved Thanks..

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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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  • django {% tag %} problem

    - by Sevenearths
    I don't know if its me but {% tag ??? %} has bee behaving a bit sporadically round me (django ver 1.2.3). I have the following main.html file: <html> {% include 'main/main_css.html' %} <body> test! <a href="{% url login.views.logout_view %}">logout</a> test! <a href="{% url client.views.client_search_last_name_view %}">logout</a> </body> </html> with the urls.py being: from django.conf.urls.defaults import * import settings from login.views import * from mainapp.views import * from client.views import * # Uncomment the next two lines to enable the admin: from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', # Example: # (r'^weclaim/', include('weclaim.foo.urls')), (r'^login/$', 'login.views.login_view'), (r'^logout/$', 'login.views.logout_view'), (r'^$', 'mainapp.views.main_view'), (r'^client/search/last_name/(A-Za-z)/$', 'client.views.client_search_last_name_view'), #(r'^client/search/post_code/(A-Za-z)/$', 'client.views.client_search_last_name_view'), # Uncomment the next line to enable the admin: (r'^admin/', include(admin.site.urls)), (r'^static/(?P<path>.*)$', 'django.views.static.serve',{'document_root': settings.MEDIA_ROOT}), ) and the views.py for login being: from django.shortcuts import render_to_response, redirect from django.template import RequestContext from django.contrib import auth import mainapp.views def login_view(request): if request.method == 'POST': uname = request.POST.get('username', '') psword = request.POST.get('password', '') user = auth.authenticate(username=uname, password=psword) # if the user logs in and is active if user is not None and user.is_active: auth.login(request, user) return redirect(mainapp.views.main_view) else: return render_to_response('loginpage.html', {'login_failed': '1',}, context_instance=RequestContext(request)) else: return render_to_response('loginpage.html', {'dave': '1',}, context_instance=RequestContext(request)) def logout_view(request): auth.logout(request) return render_to_response('loginpage.html', {'logged_out': '1',}, context_instance=RequestContext(request)) and the views.py for clients being: from django.shortcuts import render_to_response, redirect from django.template import RequestContext import login.views def client_search_last_name_view(request): if request.user.is_authenticated(): return render_to_response('client/client_search_last_name.html', {}, context_instance=RequestContext(request)) else: return redirect(login.views.login_view) Yet when I login it django raises an 'NoReverseMatch' for {% url client.views.client_search_last_name_view %} but not for {% url login.views.logout_view %} Now why would this be?

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  • Using {% url ??? %} in django templates

    - by user563247
    I have looked a lot on google for answers of how to use the 'url' tag in templates only to find many responses saying 'You just insert it into your template and point it at the view you want the url for'. Well no joy for me :( I have tried every permutation possible and have resorted to posting here as a last resort. So here it is. My urls.py looks like this: from django.conf.urls.defaults import * from login.views import * from mainapp.views import * import settings # Uncomment the next two lines to enable the admin: from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', # Example: # (r'^weclaim/', include('weclaim.foo.urls')), (r'^login/', login_view), (r'^logout/', logout_view), ('^$', main_view), # Uncomment the admin/doc line below and add 'django.contrib.admindocs' # to INSTALLED_APPS to enable admin documentation: # (r'^admin/doc/', include('django.contrib.admindocs.urls')), # Uncomment the next line to enable the admin: (r'^admin/', include(admin.site.urls)), #(r'^static/(?P<path>.*)$', 'django.views.static.serve',{'document_root': '/home/arthur/Software/django/weclaim/templates/static'}), (r'^static/(?P<path>.*)$', 'django.views.static.serve',{'document_root': settings.MEDIA_ROOT}), ) My 'views.py' in my 'login' directory looks like: from django.shortcuts import render_to_response, redirect from django.template import RequestContext from django.contrib import auth def login_view(request): if request.method == 'POST': uname = request.POST.get('username', '') psword = request.POST.get('password', '') user = auth.authenticate(username=uname, password=psword) # if the user logs in and is active if user is not None and user.is_active: auth.login(request, user) return render_to_response('main/main.html', {}, context_instance=RequestContext(request)) #return redirect(main_view) else: return render_to_response('loginpage.html', {'box_width': '402', 'login_failed': '1',}, context_instance=RequestContext(request)) else: return render_to_response('loginpage.html', {'box_width': '400',}, context_instance=RequestContext(request)) def logout_view(request): auth.logout(request) return render_to_response('loginpage.html', {'box_width': '402', 'logged_out': '1',}, context_instance=RequestContext(request)) and finally the main.html to which the login_view points looks like: <html> <body> test! <a href="{% url logout_view %}">logout</a> </body> </html> So why do I get 'NoReverseMatch' every time? *(on a slightly different note I had to use 'context_instance=RequestContext(request)' at the end of all my render-to-response's because otherwise it would not recognise {{ MEDIA_URL }} in my templates and I couldn't reference any css or js files. I'm not to sure why this is. Doesn't seem right to me)*

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  • ProgrammingError when aggregating over an annotated & grouped Django ORM query

    - by ento
    I'm trying to construct a query to get the "average, maximum, minimum number of items purchased by a single user". The data source is this simple sales record table: class SalesRecord(models.Model): id = models.IntegerField(primary_key=True) user_id = models.IntegerField() product_code = models.CharField() price = models.IntegerField() created_at = models.DateTimeField() A new record is inserted into this table for every item purchased by a user. Here's my attempt at building the query: q = SalesRecord.objects.all() q = q.values('user_id').annotate( # group by user and count the # of records count=Count('id'), # (= # of items) ).order_by() result = q.aggregate(Max('count'), Min('count'), Avg('count')) When I try to execute the code, a ProgrammingError is raised at the last line: (1064, "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM (SELECT sales_records.user_id AS user_id, COUNT(sales_records.`' at line 1") Django's error screen shows that the SQL is SELECT FROM (SELECT `sales_records`.`player_id` AS `player_id`, COUNT(`sales_records`.`id`) AS `count` FROM `sales_records` WHERE (`sales_records`.`created_at` >= %s AND `sales_records`.`created_at` <= %s ) GROUP BY `sales_records`.`player_id` ORDER BY NULL) subquery It's not selecting anything! Can someone please show me the right way to do this? Hacking Django I've found that clearing the cache of selected fields in django.db.models.sql.BaseQuery.get_aggregation() seems to solve the problem. Though I'm not really sure this is a fix or a workaround. @@ -327,10 +327,13 @@ # Remove any aggregates marked for reduction from the subquery # and move them to the outer AggregateQuery. + self._aggregate_select_cache = None + self.aggregate_select_mask = None for alias, aggregate in self.aggregate_select.items(): if aggregate.is_summary: query.aggregate_select[alias] = aggregate - del obj.aggregate_select[alias] + if alias in obj.aggregate_select: + del obj.aggregate_select[alias] ... yields result: {'count__max': 267, 'count__avg': 26.2563, 'count__min': 1}

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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 1.1 beta issue

    - by ha22109
    Hello all, I m using django 1.1 beta.I m facing porblem in case of list_editable.First it was throughing exception saying need ordering in case of list_editable" then i added ordering in model but know it is giving me error.The code is working fine with django1.1 final. here is my code model.py class User(models.Model): advertiser = models.ForeignKey(WapUser,primary_key=True) status = models.CharField(max_length=20,choices=ADVERTISER_INVITE_STATUS,default='invited') tos_version = models.CharField(max_length=5) contact_email = models.EmailField(max_length=80) contact_phone = models.CharField(max_length=15) contact_mobile = models.CharField(max_length=15) contact_person = models.CharField(max_length=80) feedback=models.BooleanField(choices=boolean_choices,default=0) def __unicode__(self): return self.user.login class Meta: db_table = u'roi_advertiser_info' managed=False ordering=['feedback',] admin.py class UserAdmin(ReadOnlyAdminFields, admin.ModelAdmin): list_per_page = 15 fields = ['advertiser','contact_email','contact_phone','contact_mobile','contact_person'] list_display = ['advertiser','contact_email','contact_phone','contact_mobile','contact_person','status','feedback'] list_editable=['feedback'] readonly = ('advertiser',) search_fields = ['advertiser__login_id'] radio_fields={'approve_auto': admin.HORIZONTAL} list_filter=['status','feedback'] admin.site.register(User,UserADmin)

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  • django - dynamic form fieldsets

    - by user110029
    A form will be spitting out an unknown number of questions to be answered. each question contains a prompt, a value field, and a unit field. The form is built at runtime in the formclass's init method. I'd like each question rendered on the form as an inline: prompt, value(input-text), units (select). this seems a case perfect for iterable form fieldsets, which could be easily styled. but since fieldsets - such as those in django-form-utils are defined as tuples, they are immutable... and I can't find a way to define them at runtime. is this possible, or perhaps another solution?

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  • How to display Django SelectDateWidget on one line using crispy forms

    - by Scott Johnson
    I am trying to display the 3 select fields that are rendered out using Django SelectDateWidget on one line. When I use crispy forms, they are all on separate rows. Is there a way to use the Layout helper to achieve this? Thank you! class WineAddForm(forms.ModelForm): hold_until = forms.DateField(widget=SelectDateWidget(years=range(1950, datetime.date.today().year+50)), required=False) drink_before = forms.DateField(widget=SelectDateWidget(years=range(1950, datetime.date.today().year+50)), required=False) helper = FormHelper() helper.form_method = 'POST' helper.form_class = 'form-horizontal' helper.label_class = 'col-lg-2' helper.field_class = 'col-lg-8' helper.add_input(Submit('submit', 'Submit', css_class='btn-wine')) helper.layout = Layout( 'name', 'year', 'description', 'country', 'region', 'sub_region', 'appellation', 'wine_type', 'producer', 'varietal', 'label_pic', 'hold_until', 'drink_before', ) class Meta: model = wine exclude = ('user', 'slug', 'likes')

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  • Prepopulating inlines based on the parent model in the Django Admin

    - by Alasdair
    I have two models, Event and Series, where each Event belongs to a Series. Most of the time, an Event's start_time is the same as its Series' default_time. Here's a stripped down version of the models. #models.py class Series(models.Model): name = models.CharField(max_length=50) default_time = models.TimeField() class Event(models.Model): name = models.CharField(max_length=50) date = models.DateField() start_time = models.TimeField() series = models.ForeignKey(Series) I use inlines in the admin application, so that I can edit all the Events for a Series at once. If a series has already been created, I want to prepopulate the start_time for each inline Event with the Series' default_time. So far, I have created a model admin form for Event, and used the initial option to prepopulate the time field with a fixed time. #admin.py ... import datetime class OEventInlineAdminForm(forms.ModelForm): start_time = forms.TimeField(initial=datetime.time(18,30,00)) class Meta: model = OEvent class EventInline(admin.TabularInline): form = EventInlineAdminForm model = Event class SeriesAdmin(admin.ModelAdmin): inlines = [EventInline,] I am not sure how to proceed from here. Is it possible to extend the code, so that the initial value for the start_time field is the Series' default_time?

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  • Odd behavior in Django Form (readonly field/widget)

    - by jamida
    I'm having a problem with a test app I'm writing to verify some Django functionality. The test app is a small "grade book" application that is currently using Alex Gaynor's readonly field functionality http://lazypython.blogspot.com/2008/12/building-read-only-field-in-django.html There are 2 problems which may be related. First, when I flop the comment on these 2 lines below: # myform = GradeForm(data=request.POST, instance=mygrade) myform = GradeROForm(data=request.POST, instance=mygrade) it works like I expect, except of course that the student field is changeable. When the comments are the shown way, the "studentId" field is displayed as a number (not the name, problem 1) and when I hit submit I get an error saying that studentId needs to be a Student instance. I'm at a loss as to how to fix this. I'm not wedded to Alex Gaynor's code. ANY code will work. I'm relatively new to both Python and Django, so the hints I've seen on websites that say "making a read-only field is easy" are still beyond me. // models.py 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) # testbed.grades.readonly is alex gaynor's code from testbed.grades.readonly import ReadOnlyField class GradeROForm(ModelForm): studentId = ReadOnlyField() class Meta: model=Grade class GradeForm(ModelForm): class Meta: model=Grade // views.py def modifyGrade(request,student): student = Student.objects.get(name=student) mygrade = Grade.objects.get(studentId=student) if request.method == "POST": # myform = GradeForm(data=request.POST, instance=mygrade) myform = GradeROForm(data=request.POST, instance=mygrade) if myform.is_valid(): grade = myform.save() info = "successfully updated %s" % grade.studentId else: # myform=GradeForm(instance=mygrade) myform=GradeROForm(instance=mygrade) return render_to_response('grades/modifyGrade.html',locals()) // template <p>{{ info }}</p> <form method="POST" action=""> <table> {{ myform.as_table }} </table> <input type="submit" value="Submit"> </form> // Alex Gaynor's code from django import forms from django.utils.html import escape from django.utils.safestring import mark_safe from django.forms.util import flatatt class ReadOnlyWidget(forms.Widget): def render(self, name, value, attrs): final_attrs = self.build_attrs(attrs, name=name) if hasattr(self, 'initial'): value = self.initial return mark_safe("<span %s>%s</span>" % (flatatt(final_attrs), escape(value) or '')) def _has_changed(self, initial, data): return False class ReadOnlyField(forms.FileField): widget = ReadOnlyWidget def __init__(self, widget=None, label=None, initial=None, help_text=None): forms.Field.__init__(self, label=label, initial=initial, help_text=help_text, widget=widget) def clean(self, value, initial): self.widget.initial = initial return initial

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  • Django unique_together error and validation

    - by zubinmehta
    class Votes(models.Model): field1 = models.ForeignKey(Blah1) field2 = models.ForeignKey(Blah2) class Meta: unique_together = (("field1","field2"),) I am using this code as one of my models. Now i wanted to know two things: 1. It doesn't show any error and it saved an entry which wasn't unique together; So is the piece of code correct? 2. How can the unique_together constraint be be validated?

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  • Accessing django choice field

    - by Hulk
    there is a module as header , from test.models import SEL_VALUES class rubrics_header(models.Model): sel_values = models.IntegerField(choices=SEL_VALUES) So when SEL_VALUES is imported from test.modules.What is the code that has to go in views to get the choices in sel_values . And the test.modules has the following, class SEL_VALUES: vaue = 0 value2 = 1 class Entries(forms.Form) : models.IntegerField(choices=SEL_VALUES) SEL_VALUES = ((ACCESS.value,'NAME'),(ACCESS.value2,'DESIGNATION'))

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  • model not showing up in django admin.

    - by Zayatzz
    Hi. I have ceated several django apps and stuffs for my own fund and so far everything has been working fine. Now i just created new project (django 1.2.1) and have run into trouble from 1st moments. I created new app - game and new model Game. i created admin.py and put related stuff into it. Ran syncdb and went to check into admin. Model did not show up there. I proceeded to check and doublecheck and read through previous similar threads: http://stackoverflow.com/questions/1839927/registered-models-do-not-show-up-in-admin http://stackoverflow.com/questions/1694259/django-app-not-showing-up-in-admin-interface But as far as i can tell, they dont help me either. Perhaps someone else can point this out for me. models.py in game app: # -*- coding: utf-8 -*- from django.db import models class Game(models.Model): type = models.IntegerField(blank=False, null=False, default=1) teamone = models.CharField(max_length=100, blank=False, null=False) teamtwo = models.CharField(max_length=100, blank=False, null=False) gametime = models.DateTimeField(blank=False, null=False) admin.py in game app: # -*- coding: utf-8 -*- from jalka.game.models import Game from django.contrib import admin class GameAdmin(admin.ModelAdmin): list_display = ['type', 'teamone', 'teamtwo', 'gametime'] admin.site.register(Game, GameAdmin) project settings.py: MIDDLEWARE_CLASSES = ( 'django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.middleware.csrf.CsrfViewMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'django.contrib.messages.middleware.MessageMiddleware', ) ROOT_URLCONF = 'jalka.urls' TEMPLATE_DIRS = ( "/home/projects/jalka/templates/" ) INSTALLED_APPS = ( 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'django.contrib.messages', 'django.contrib.admin', 'game', ) urls.py: from django.conf.urls.defaults import * # Uncomment the next two lines to enable the admin: from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', # Example: # (r'^jalka/', include('jalka.foo.urls')), (r'^admin/', include(admin.site.urls)), ) Alan.

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  • Django Admin Page missing CSS

    - by super9
    I saw this question and recommendation from Django Projects here but still can't get this to work. My Django Admin pages are not displaying the CSS at all. This is my current configuration. settings.py ADMIN_MEDIA_PREFIX = '/media/admin/' httpd.conf <VirtualHost *:80> DocumentRoot /home/django/sgel ServerName ec2-***-**-***-***.ap-**********-1.compute.amazonaws.com ErrorLog /home/django/sgel/logs/apache_error.log CustomLog /home/django/sgel/logs/apache_access.log combined WSGIScriptAlias / /home/django/sgel/apache/django.wsgi <Directory /home/django/sgel/media> Order deny,allow Allow from all </Directory> <Directory /home/django/sgel/apache> Order deny,allow Allow from all </Directory> LogLevel warn Alias /media/ /home/django/sgel/media/ </VirtualHost> <VirtualHost *:80> ServerName sgel.com Redirect permanent / http://www.sgel.com/ </VirtualHost> In addition, I also ran the following to create (I think) the symbolic link ln -s /home/djangotest/sgel/media/admin/ /usr/lib/python2.6/site-packages/django/contrib/admin/media/ UPDATE In my httpd.conf file, User django Group django When I run ls -l in my /media directory drwxr-xr-x 2 root root 4096 Apr 4 11:03 admin -rw-r--r-- 1 root root 9 Apr 8 09:02 test.txt Should that root user be django instead? UPDATE 2 When I enter ls -la in my /media/admin folder total 12 drwxr-xr-x 2 root root 4096 Apr 13 03:33 . drwxr-xr-x 3 root root 4096 Apr 8 09:02 .. lrwxrwxrwx 1 root root 60 Apr 13 03:33 media -> /usr/lib/python2.6/site-packages/django/contrib/admin/media/ The thing is, when I navigate to /usr/lib/python2.6/site-packages/django/contrib/admin/media/, the folder was empty. So I copied the CSS, IMG and JS folders from my Django installation into /usr/lib/python2.6/site-packages/django/contrib/admin/media/ and it still didn't work

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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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  • Why is django giving me an attribute error when I call _set.all() for its children models?

    - by user1876508
    I have two models defined from django.db import models class Blog(models.Model): title = models.CharField(max_length=144) @property def posts(self): self.Post_set.all() class Post(models.Model): title = models.CharField(max_length=144) text = models.TextField() blog = models.ForeignKey('Blog') but the problem is, when I run shell, and enter >>> blog = Blog(title="My blog") >>> post = Post(title="My first post", text="Here is the main text for my blog post", blog=blog) >>> blog.posts I get the error Traceback (most recent call last): File "<console>", line 1, in <module> File "/home/lucas/Programming/Python/Django/djangorestfun/blog/models.py", line 9, in posts self.Post_set.all() AttributeError: 'Blog' object has no attribute 'Post_set' >>> Now I am having the following problem >>> from blog.models import * >>> blog = Blog(title="gewrhter") >>> blog.save() >>> blog.__dict__ {'_state': <django.db.models.base.ModelState object at 0x259be10>, 'id': 1, 'title': 'gewrhter'} >>> blog._state.__dict__ {'adding': False, 'db': 'default'} >>> post = Post(title="sdhxcvb", text="hdbfdgb", blog=blog) >>> post.save() >>> post.__dict__ {'blog_id': 1, 'title': 'sdhxcvb', 'text': 'hdbfdgb', '_blog_cache': <Blog: Blog object>, '_state': <django.db.models.base.ModelState object at 0x259bed0>, 'id': 1} >>> blog.posts >>> print blog.posts None Second update So I followed your guide, but I am still getting nothing. In addition, blog.posts gives me an error. >>> from blog.models import * >>> blog = Blog(title="asdf") >>> blog.save() >>> post = Post(title="asdf", text="sdxcvb", blog=blog) >>> post.save() >>> blog.posts Traceback (most recent call last): File "<console>", line 1, in <module> AttributeError: 'Blog' object has no attribute 'posts' >>> print blog.all_posts None

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  • Order in many to many relation in Django model

    - by Pietro Speroni
    I am writing a small website to store the papers I have written. The relation papers<- author is important, but the order of the name of the authors (which one is First Author, which one is second order, and so on) is also important. I am just learning Django so I don't know much. In any case so far I have done: from django.db import models class author(models.Model): Name = models.CharField(max_length=60) URLField = models.URLField(verify_exists=True, null=True, blank=True) def __unicode__(self): return self.Name class topic(models.Model): TopicName = models.CharField(max_length=60) def __unicode__(self): return self.TopicName class publication(models.Model): Title = models.CharField(max_length=100) Authors = models.ManyToManyField(author, null=True, blank=True) Content = models.TextField() Notes = models.TextField(blank=True) Abstract = models.TextField(blank=True) pub_date = models.DateField('date published') TimeInsertion = models.DateTimeField(auto_now=True) URLField = models.URLField(verify_exists=True,null=True, blank=True) Topic = models.ManyToManyField(topic, null=True, blank=True) def __unicode__(self): return self.Title This work fine in the sense that I now can define who the authors are. But I cannot order them. How should I do that? Of course I could add a series of relations: first author, second author,... but it would be ugly, and would not be flexible. Any better idea? Thanks

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  • django-admin: creating,saving and relating a m2m model

    - by pastylegs
    I have two models: class Production(models.Model): gallery = models.ManyToManyField(Gallery) class Gallery(models.Model): name = models.CharField() I have the m2m relationship in my productions admin, but I want that functionality that when I create a new Production, a default gallery is created and the relationship is registered between the two. So far I can create the default gallery by overwriting the productions save: def save(self, force_insert=False, force_update=False): if not ( Gallery.objects.filter(name__exact="foo").exists() ): g = Gallery(name="foo") g.save() self.gallery.add(g) This creates and saves the model instance (if it doesn't already exist), but I don't know how to register the relationship between the two?

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  • Django QuerySet filter method returns multiple entries for one record

    - by Yaroslav
    Trying to retrieve blogs (see model description below) that contain entries satisfying some criteria: Blog.objects.filter(entries__title__contains='entry') The results is: [<Blog: blog1>, <Blog: blog1>] The same blog object is retrieved twice because of JOIN performed to filter objects on related model. What is the right syntax for filtering only unique objects? Data model: class Blog(models.Model): name = models.CharField(max_length=100) def __unicode__(self): return self.name class Entry(models.Model): title = models.CharField(max_length=100) blog = models.ForeignKey(Blog, related_name='entries') def __unicode__(self): return self.title Sample data: b1 = Blog.objects.create(name='blog1') e1 = Entry.objects.create(title='entry 1', blog=b1) e1 = Entry.objects.create(title='entry 2', blog=b1)

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