Search Results

Search found 20523 results on 821 pages for 'django query'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 5/821 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Insert django form into template dynamically using javascript??

    - by qulzam
    I want to add same django form instance on same template. i already add one before and other add dynamically using javascript. for example 'form" is a django form: newcell.innerHTML = {{ form.firstname }}; The problem is that when i submit the form, in view the request object has only one value (that is not add using javascript). how can i get the values of other form elements values that is added dynamically runtime.

    Read the article

  • Django : debugging templatetags

    - by interstar
    How on earth do people debug Django templatetags? I created one, based on a working example, my new tag looks the same to me as the existing one. But I just get a 'my_lib' is not a valid tag library: Could not load template library from django.templatetags.my_lib, No module named my_lib I know that this is probably because of something failing when defining the lib. But how do I see what's going on? What do you use to debug this situation?

    Read the article

  • Django models query

    - by Hulk
    Code: class criteria(models.Model): details = models.CharField(max_length = 512) Headerid = models.ForeignKey(Header) def __unicode__(self): return self.id() the details corresponds to a textarea in the UI and a validation is done for 512 characters but when this is saved. /home/project/django/django/core/handlers/base.py in get_response, line 109 Is this any thing related with schema or number of characters entered from UI

    Read the article

  • Django: Generating a queryset from a GET request

    - by Nimmy Lebby
    I have a Django form setup using GET method. Each value corresponds to attributes of a Django model. What would be the most elegant way to generate the query? Currently this is what I do in the view: def search_items(request): if 'search_name' in request.GET: query_attributes = {} query_attributes['color'] = request.GET.get('color', '') if not query_attributes['color']: del query_attributes['color'] query_attributes['shape'] = request.GET.get('shape', '') if not query_attributes['shape']: del query_attributes['shape'] items = Items.objects.filter(**query_attributes) But I'm pretty sure there's a better way to go about it.

    Read the article

  • Tying in to Django Admin's Model History

    - by akdom
    The Setup: I'm working on a Django application which allows users to create an object in the database and then go back and edit it as much as they desire. Django's admin site keeps a history of the changes made to objects through the admin site. The Question: How do I hook my application in to the admin site's change history so that I can see the history of changes users make to their "content" ?

    Read the article

  • 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..

    Read the article

  • Django attribute error: 'module' object has no attribute 'is_usable'

    - by Robert A Henru
    Hi, I got the following error when calling the url in Django. It's working before, I guess it's related with some accidental changes I made, but I have no idea what they are. Thanks before for the help, Robert Environment: Request Method: GET Request URL: http://localhost:8000/time/ Django Version: 1.2 Python Version: 2.6.1 Installed Applications: ['django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'django.contrib.messages', 'django.contrib.admin', 'djlearn.books'] Installed Middleware: ('django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.middleware.csrf.CsrfViewMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'django.contrib.messages.middleware.MessageMiddleware') Traceback: File "/Library/Python/2.6/site-packages/django/core/handlers/base.py" in get_response 100. response = callback(request, *callback_args, **callback_kwargs) File "/Users/rhenru/Workspace/django/djlearn/src/djlearn/../djlearn/views.py" in current_datetime 16. return render_to_response('current_datetime.html',{'current_date':now,}) File "/Library/Python/2.6/site-packages/django/shortcuts/__init__.py" in render_to_response 20. return HttpResponse(loader.render_to_string(*args, **kwargs), **httpresponse_kwargs) File "/Library/Python/2.6/site-packages/django/template/loader.py" in render_to_string 181. t = get_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in get_template 157. template, origin = find_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template 128. loader = find_template_loader(loader_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template_loader 111. if not func.is_usable: Exception Type: AttributeError at /time/ Exception Value: 'module' object has no attribute 'is_usable'

    Read the article

  • django crispy-forms inline forms

    - by abolotnov
    I'm trying to adopt crispy-forms and bootstrap and use as much of their functionality as possible instead of inventing something over and over again. Is there a way to have inline forms functionality with crispy-forms/bootstrap like django-admin forms have? Here is an example: class NewProjectForm(forms.Form): name = forms.CharField(required=True, label=_(u'???????? ???????'), widget=forms.TextInput(attrs={'class':'input-block-level'})) group = forms.ModelChoiceField(required=False, queryset=Group.objects.all(), label=_(u'?????? ????????'), widget=forms.Select(attrs={'class':'input-block-level'})) description = forms.CharField(required=False, label=_(u'???????? ???????'), widget=forms.Textarea(attrs={'class':'input-block-level'})) class Meta: model = Project fields = ('name','description','group') def __init__(self, *args, **kwargs): self.helper = FormHelper() self.helper.form_class = 'horizontal-form' self.helper.form_action = 'submit_new_project' self.helper.layout = Layout( Field('name', css_class='input-block-level'), Field('group', css_class='input-block-level'), Field('description',css_class='input-block-level'), ) self.helper.add_input(Submit('submit',_(u'??????? ??????'))) self.helper.add_input(Submit('cancel',_(u'? ?????????'))) super(NewProjectForm, self).__init__(*args, **kwargs) it will display a decent form: How do I go about adding a form that basically represents this model: class Link(models.Model): name = models.CharField(max_length=255, blank=False, null=False, verbose_name=_(u'????????')) url = models.URLField(blank=False, null=False, verbose_name=_(u'??????')) project = models.ForeignKey('Project') So there will be a project and name/url links and way to add many, like same thing is done in django-admin where you are able to add extra 'rows' with data related to your main model. On the sreenshot below you are able to fill out data for 'Question' object and below that you are able to add data for QuestionOption objects -you are able to click the '+' icon to add as many QuestionOptions as you want. I'm not looking for a way to get the forms auto-generated from models (that's nice but not the most important) - is there a way to construct a form that will let you add 'rows' of data like django-admin does?

    Read the article

  • 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}

    Read the article

  • SQL SERVER – SSMS Automatically Generates TOP (100) PERCENT in Query Designer

    - by pinaldave
    Earlier this week, I was surfing various SQL forums to see what kind of help developer need in the SQL Server world. One of the question indeed caught my attention. I am here regenerating complete question as well scenario to illustrate the point in a precise manner. Additionally, I have added added second part of the question to give completeness. Question: I am trying to create a view in Query Designer (not in the New Query Window). Every time I am trying to create a view it always adds  TOP (100) PERCENT automatically on the T-SQL script. No matter what I do, it always automatically adds the TOP (100) PERCENT to the script. I have attempted to copy paste from notepad, build a query and a few other things – there is no success. I am really not sure what I am doing wrong with Query Designer. Here is my query script: (I use AdventureWorks as a sample database) SELECT Person.Address.AddressID FROM Person.Address INNER JOIN Person.AddressType ON Person.Address.AddressID = Person.AddressType.AddressTypeID ORDER BY Person.Address.AddressID This script automatically replaces by following query: SELECT TOP (100) PERCENT Person.Address.AddressID FROM Person.Address INNER JOIN Person.AddressType ON Person.Address.AddressID = Person.AddressType.AddressTypeID ORDER BY Person.Address.AddressID However, when I try to do the same from New Query Window it works totally fine. However, when I attempt to create a view of the same query it gives following error. Msg 1033, Level 15, State 1, Procedure myView, Line 6 The ORDER BY clause is invalid in views, inline functions, derived tables, subqueries, and common table expressions, unless TOP, OFFSET or FOR XML is also specified. It is pretty clear to me now that the script which I have written seems to need TOP (100) PERCENT, so Query . Why do I need it? Is there any work around to this issue. I particularly find this question pretty interesting as it really touches the fundamentals of the T-SQL query writing. Please note that the query which is automatically changed is not in New Query Editor but opened from SSMS using following way. Database >> Views >> Right Click >> New View (see the image below) Answer: The answer to the above question can be very long but I will keep it simple and to the point. There are three things to discuss in above script 1) Reason for Error 2) Reason for Auto generates TOP (100) PERCENT and 3) Potential solutions to the above error. Let us quickly see them in detail. 1) Reason for Error The reason for error is already given in the error. ORDER BY is invalid in the views and a few other objects. One has to use TOP or other keywords along with it. The way semantics of the query works where optimizer only follows(honors) the ORDER BY in the same scope or the same SELECT/UPDATE/DELETE statement. There is a possibility that one can order after the scope of the view again the efforts spend to order view will be wasted. The final resultset of the query always follows the final ORDER BY or outer query’s order and due to the same reason optimizer follows the final order of the query and not of the views (as view will be used in another query for further processing e.g. in SELECT statement). Due to same reason ORDER BY is now allowed in the view. For further accuracy and clear guidance I suggest you read this blog post by Query Optimizer Team. They have explained it very clear manner the same subject. 2) Reason for Auto Generated TOP (100) PERCENT One of the most popular workaround to above error is to use TOP (100) PERCENT in the view. Now TOP (100) PERCENT allows user to use ORDER BY in the query and allows user to overcome above error which we discussed. This gives the impression to the user that they have resolved the error and successfully able to use ORDER BY in the View. Well, this is incorrect as well. The way this works is when TOP (100) PERCENT is used the result is not guaranteed as well it is ignored in our the query where the view is used. Here is the blog post on this subject: Interesting Observation – TOP 100 PERCENT and ORDER BY. Now when you create a new view in the SSMS and build a query with ORDER BY to avoid the error automatically it adds the TOP 100 PERCENT. Here is the connect item for the same issue. I am sure there will be more connect items as well but I could not find them. 3) Potential Solutions If you are reading this post from the beginning in that case, it is clear by now that ORDER BY should not be used in the View as it does not serve any purpose unless there is a specific need of it. If you are going to use TOP 100 PERCENT with ORDER BY there is absolutely no need of using ORDER BY rather avoid using it all together. Here is another blog post of mine which describes the same subject ORDER BY Does Not Work – Limitation of the Views Part 1. It is valid to use ORDER BY in a view if there is a clear business need of using TOP with any other percentage lower than 100 (for example TOP 10 PERCENT or TOP 50 PERCENT etc). In most of the cases ORDER BY is not needed in the view and it should be used in the most outer query for present result in desired order. User can remove TOP 100 PERCENT and ORDER BY from the view before using the view in any query or procedure. In the most outer query there should be ORDER BY as per the business need. I think this sums up the concept in a few words. This is a very long topic and not easy to illustrate in one single blog post. I welcome your comments and suggestions. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Server Management Studio, SQL Tips and Tricks, SQL View, T SQL, Technology

    Read the article

  • What is causing this OverflowError in Django?

    - by orokusaki
    I'm using a normal ModelForm.save() to create an object, and this exception comes up. It worked fine before until I added commit_manually, transaction.rollback() and transaction.commit() to my view. Has anyone else ran into this? Is this because of sqlite3? OverflowError: long too big to convert C:\Python26\Lib\site-packages\django-trunk\django\db\backends\sqlite3\base.py in execute, line 197 params: (203866156270872165269663274649746494334L,) query: u'SELECT (1) AS "a", "auth_user"."id", "auth_user"."username", "auth_user"."first_name", "auth_user"."last_name", "auth_user"."email", "auth_user"."password", "auth_user"."is_staff", "auth_user"."is_active", "auth_user"."is_superuser", "auth_user"."last_login", "auth_user"."date_joined" FROM "auth_user" WHERE "auth_user"."id" = ? LIMIT 1' self <django.db.backends.sqlite3.base.SQLiteCursorWrapper object at 0x015D5A98> Why would that L param be passed in, and

    Read the article

  • Django - provide additional information in template

    - by Ninefingers
    Hi all, I am building an app to learn Django and have started with a Contact system that currently stores Contacts and Addresses. C's are a many to many relationship with A's, but rather than use Django's models.ManyToManyField() I've created my own link-table providing additional information about the link, such as what the address type is to the that contact (home, work etc). What I'm trying to do is pass this information out to a view, so in my full view of a contact I can do this: def contact_view_full(request, contact_id): c = get_object_or_404(Contact, id=contact_id) a = [] links = ContactAddressLink.objects.filter(ContactID=c.id) for link in links: b = Address.objects.get(id=link.AddressID_id) a.append(b) return render_to_response('contact_full.html', {'contact_item': c, 'addresses' : a }, context_instance=RequestContext(request)) And so I can do the equivalent of c.Addresses.all() or however the ManyToManyField works. What I'm interested to know is how can I pass out information about the link in the link object with the 'addresses' : a information, so that when my template does this: {% for address in addresses %} <!-- ... --> {% endfor %} and properly associate the correct link object data with the address. So what's the best way to achieve this? I'm thinking a union of two objects might be an idea but I haven't enough experience with Django to know if that's considered the best way of doing it. Suggestions? Thanks in advance. Nf

    Read the article

  • Django model operating on a queryset

    - by jmoz
    I'm new to Django and somewhat to Python as well. I'm trying to find the idiomatic way to loop over a queryset and set a variable on each model. Basically my model depends on a value from an api, and a model method must multiply one of it's attribs by this api value to get an up-to-date correct value. At the moment I am doing it in the view and it works, but I'm not sure it's the correct way to achieve what I want. I have to replicate this looping elsewhere. Is there a way I can encapsulate the looping logic into a queryset method so it can be used in multiple places? I have this atm (I am using django-rest-framework): class FooViewSet(viewsets.ModelViewSet): model = Foo serializer_class = FooSerializer bar = # some call to an api def get_queryset(self): # Dynamically set the bar variable on each instance! foos = Foo.objects.filter(baz__pk=1).order_by('date') for item in foos: item.needs_bar = self.bar return items I would think something like so would be better: def get_queryset(self): bar = # some call to an api # Dynamically set the bar variable on each instance! return Foo.objects.filter(baz__pk=1).order_by('date').set_bar(bar) I'm thinking the api hit should be in the controller and then injected to instances of the model, but I'm not sure how you do this. I've been looking around querysets and managers but still can't figure it out nor decided if it's the best method to achieve what I want. Can anyone suggest the correct way to model this with django? Thanks.

    Read the article

  • Asynchronous daemon processing / ORM interaction with Django

    - by perrierism
    I'm looking for a way to do asynchronous data processing with a daemon that uses Django ORM. However, the ORM isn't thread-safe; it's not thread-safe to try to retrieve / modify django objects from within threads. So I'm wondering what the correct way to achieve asynchrony is? Basically what I need to accomplish is taking a list of users in the db, querying a third party api and then making updates to user-profile rows for those users. As a daemon or background process. Doing this in series per user is easy, but it takes too long to be at all scalable. If the daemon is retrieving and updating the users through the ORM, how do I achieve processing 10-20 users at a time? I would use a standard threading / queue system for this but you can't thread interactions like models.User.objects.get(id=foo) ... Django itself is an asynchronous processing system which makes asynchronous ORM calls(?) for each request, so there should be a way to do it? I haven't found anything in the documentation so far. Cheers

    Read the article

  • In Django-pagination Paginate does not working...

    - by mosg
    Hello. Python 2.6.2 django-pagination 1.0.5 Question: How to force pagination work correctly? The problem is that {% paginate %} does not work, but other {% load pagination_tags %} and {% autopaginate object_list 10 %} works! Error message appeared, when I add {% paginate %} into html page: TemplateSyntaxError at /logging Caught an exception while rendering: pagination/pagination.html What I have done: Install django-pagination without any problems. When I do in python import pagination, it's work well. Added pagination to INSTALLED_APP in settings.py: INSTALLED_APPS = ( # ..., 'pagination', ) Added in settings.py: TEMPLATE_CONTEXT_PROCESSORS = ( "django.core.context_processors.auth", "django.core.context_processors.debug", "django.core.context_processors.i18n", "django.core.context_processors.media", "django.core.context_processors.request" ) Also add to settings.py middleware: MIDDLEWARE_CLASSES = ( # ... 'pagination.middleware.PaginationMiddleware', ) Add to top in views.py: from django.template import RequestContext And finally add to my HTML template page lines: {% load pagination_tags %} ... {% autopaginate item_list 50 %} {% for item in item_list %} ... {% endfor %} {% paginate %} Thanks. PS: some edits required, because I can't django code style work well here :)

    Read the article

  • 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?

    Read the article

  • Login URL using authentication information in Django

    - by fuSi0N
    I'm working on a platform for online labs registration for my university. Login View [project views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib import auth def index(request): return render_to_response('index.html', {}, context_instance = RequestContext(request)) def login(request): if request.method == "POST": post = request.POST.copy() if post.has_key('username') and post.has_key('password'): usr = post['username'] pwd = post['password'] user = auth.authenticate(username=usr, password=pwd) if user is not None and user.is_active: auth.login(request, user) if user.get_profile().is_teacher: return HttpResponseRedirect('/teachers/'+user.username+'/') else: return HttpResponseRedirect('/students/'+user.username+'/') else: return render_to_response('index.html', {'msg': 'You don\'t belong here.'}, context_instance = RequestContext(request) return render_to_response('login.html', {}, context_instance = RequestContext(request)) def logout(request): auth.logout(request) return render_to_response('index.html', {}, context_instance = RequestContext(request)) URLS #========== PROJECT URLS ==========# urlpatterns = patterns('', (r'^media/(?P<path>.*)$', 'django.views.static.serve', {'document_root': settings.MEDIA_ROOT }), (r'^admin/', include(admin.site.urls)), (r'^teachers/', include('diogenis.teachers.urls')), (r'^students/', include('diogenis.students.urls')), (r'^login/', login), (r'^logout/', logout), (r'^$', index), ) #========== TEACHERS APP URLS ==========# urlpatterns = patterns('', (r'^(?P<username>\w{0,50})/', labs), ) The login view basically checks whether the logged in user is_teacher [UserProfile attribute via get_profile()] and redirects the user to his profile. Labs View [teachers app views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib.auth.decorators import user_passes_test from django.contrib.auth.models import User from accounts.models import * from labs.models import * def user_is_teacher(user): return user.is_authenticated() and user.get_profile().is_teacher @user_passes_test(user_is_teacher, login_url="/login/") def labs(request, username): q1 = User.objects.get(username=username) q2 = u'%s %s' % (q1.last_name, q1.first_name) q2 = Teacher.objects.get(name=q2) results = TeacherToLab.objects.filter(teacher=q2) return render_to_response('teachers/labs.html', {'results': results}, context_instance = RequestContext(request)) I'm using @user_passes_test decorator for checking whether the authenticated user has the permission to use this view [labs view]. The problem I'm having with the current logic is that once Django authenticates a teacher user he has access to all teachers profiles basically by typing the teachers username in the url. Once a teacher finds a co-worker's username he has direct access to his data. Any suggestions would be much appreciated.

    Read the article

  • CodeIgniter & Datamapper as frontend, Django Admin as backend, database tables inconsistent

    - by Rasiel
    I created a database for a site i'm doing using Django as the admin backend. However because the server where the site is hosted on, won't be able to support Python, I find myself needing to do the front end in PHP and as such i've decided to use CodeIgniter along with Datamapper to map the models/relationship. However DataMapper requires the tables to be in a specific format for it to work, and Django maps its tables differently, using the App name as the prefix in the table. I've tried using the prefix & join_prefix vars in datamapper but still doesn't map them correctly. Has anyone used a combination of this? and if so how have the fixed the issue of db table names being inconsistent? Is there anything out there that i can use to make them work together?

    Read the article

  • Django: Summing values of records grouped by foreign key

    - by Dan0
    Hi there In django, given the following models (slightly simplified), I'm struggling to work out how I would get a view including sums of groups class Client(models.Model): api_key = models.CharField(unique=True, max_length=250, primary_key=True) name = models.CharField(unique=True, max_length=250) class Purchase(models.Model): purchase_date = models.DateTimeField() client = models.ForeignKey(SavedClient, to_field='api_key') amount_to_invoice = models.FloatField(null=True) For a given month, I'd like to see e.g. April 2010 For Each Client that purchased this month: * CLient: Name * Total amount of Purchases for this month * Total cost of purchases for this month For each Purchase made by client: * Date * Amount * Etc I've been looking into django annotation, but can't get my head around how to sum values of a field for a particular group over a particular month and send the information to a template as a variable/tag. Any info would be appreciated

    Read the article

  • Django: Country drop down list?

    - by User
    I have a form for address information. One of the fields is for the address country. Currently this is just a textbox. I would like a drop down list (of ISO 3166 countries) for this. I'm a django newbie so I haven't even used a Django Select widget yet. What is a good way to do this? Hard-code the choices in a file somewhere? Put them in the database? In the template?

    Read the article

  • Django: reverse lookup URL of feeds?

    - by Santa
    I am having trouble doing a reverse URL lookup for Django-generated feeds. I have the following setup in urls.py: feeds = { 'latest': LatestEntries, } urlpatterns = patterns('', # ... # enable feeds (RSS) url(r'^feeds/(?P<url>.*)/$', 'django.contrib.syndication.views.feed', {'feed_dict': feeds}, name='feeds_view'), ) I have tried using the following template tag: <a href="{% url feeds_view latest %}">RSS feeds</a> But the resulting link is not what want (http://my.domain.com/feeds//). It should be http://my.domain.com/feeds/latest/. For now, I am using a hack to generate the URL for the template: <a href="http://{{ request.META.HTTP_HOST }}/feeds/latest">RSS feeds</a> But, as you can see, it clearly is not DRY. Is there something I am missing?

    Read the article

  • Project name inserted automatically in url when using django template url tag

    - by thebossman
    I am applying the 'url' template tag to all links in my current Django project. I have my urls named like so... url(r'^login/$', 'login', name='site_login'), This allows me to access /login at my site's root. I have my template tag defined like so... <a href="{% url site_login %}"> It works fine, except that Django automatically resolves that url as /myprojectname/login, not /login. Both urls are accessible. Why? Is there an option to remove the projectname? This occurs for all url tags, not just this one.

    Read the article

  • How to customize a many-to-many inline model in django admin

    - by Jonathan
    I'm using the admin interface to view invoices and products. To make things easy, I've set the products as inline to invoices, so I will see the related products in the invoice's form. As you can see I'm using a many-to-many relationship. In models.py: class Product(models.Model): name = models.TextField() price = models.DecimalField(max_digits=10,decimal_places=2) class Invoice(models.Model): company = models.ForeignKey(Company) customer = models.ForeignKey(Customer) products = models.ManyToManyField(Product) In admin.py: class ProductInline(admin.StackedInline): model = Invoice.products.through class InvoiceAdmin(admin.ModelAdmin): inlines = [FilteredApartmentInline,] admin.site.register(Product, ProductAdmin) The problem is that django presents the products as a table of drop down menus (one per associated product). Each drop down contains all the products listed. So if I have 5000 products and 300 are associated with a certain invoice, django actually loads 300x5000 product names. Also the table is not aesthetic. How can I change it so that it'll just display the product's name in the inline table? Which form should I override, and how?

    Read the article

  • django cross-site reverse a url

    - by tutuca
    I have a similar question than django cross-site reverse. But i think I can't apply the same solution. I'm creating an app that lets the users create their own site. After completing the signup form the user should be redirected to his site's new post form. Something along this lines: new_post_url = 'http://%s.domain:9292/manage/new_post %site.domain' logged_user = authenticate(username=user.username, password=user.password) if logged_user is not None: login(request, logged_user) return redirect(new_product_url) Now, I know that "new_post_url" is awful and makes babies cry so I need to reverse it in some way. I thought in using django.core.urlresolvers.reverse to solve this but that only returns urls on my domain, and not in the user's newly created site, so it doesn't works for me. So, do you know a better/smarter way to solve this?

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >