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• Override Django inlineformset_factory has_changed() to always return True

  - by John

  Hi, I am using the django inlineformset_factory function. a = get_object_or_404(ModelA, pk=id) FormSet = inlineformset_factory(ModelA, ModelB) if request.method == 'POST': metaform = FormSet (instance=a, data=request.POST) if metaform.is_valid(): f = metaform.save(commit=False) for instance in f: instance.updated_by = request.user instance.save() else: metaform = FormSet(instance=a) return render_to_response('nodes/form.html', {'form':metaform}) What is happening is that if I change any of the data then everything works ok and all the data gets updated. However if I don't change any of the data then the data is not updated. i.e. only entries which are changed go through the for loop to be saved. I guess this makes sense as there is no point saving data if it has not changed. However I need to go through and save every object in the form regardless of whether it has any changes on not. So my question is how do I override this so that it goes through and saves every record whether it has any changes or not? Hope this makes sense Thanks

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• Django: query spanning multiple many-to-many relationships

  - by Brant

  I've got some models set up like this: class AppGroup(models.Model): users = models.ManyToManyField(User) class Notification(models.Model): groups_to_notify = models.ManyToManyField(AppGroup) The User objects come from django's authentication system. Now, I am trying to get all the notifications pertaining to the groups that the current user is a part of. I have tried.. notifications = Notification.objects.filter(groups_to_notify=AppGroup.objects.filter(users=request.user)) But that gives an error: more than one row returned by a subquery used as an expression Which I suppose is because the groups_to_notify is checking against several groups. How can I grab all the notifications meant for the user based on the groups he is a part of?

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• Django test client gets 301 redirection when accessing url

  - by Michal Klich

  I am writing unittests for django views. I have observed that one of my views returns redirection code 301, which is not expected. Here is my views.py mentioned earlier. def index(request): return render(request, 'index.html', {'form': QueryForm()}) def query(request): if request.is_ajax(): form = QueryForm(request.POST) return HttpResponse('valid') Below is urls.py. urlpatterns = patterns('', url(r'^$', 'core.views.index'), url(r'^query/$', 'core.views.query') ) And unittest that will fail. def so_test(self): response = self.client.post('/') self.assertEquals(response.status_code, 200) response = self.client.post('/query', {}) self.assertEquals(response.status_code, 200) My question is: why there is status 301 returned?

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• ModelName(django.contrib.auth.models.User) vs ModelName(models.Model)

  - by amr.negm

  I am developing a django project. I created some apps, some of those are related to User model, for instance, I have a feeds app that handles user feeds, and another app that deals with extra user data like age, contacts, and friends. for each of these, I created a table that should be connected to the User model, which I using for storing and authenticating users. I found two ways to deal with this issue. One, is through extending User model to be like this: ModelName(User): friends = models.ManyToMany('self') ..... Two, is through adding a foreign key to the new table like this: ModelName(models.Model): user = models.ForeignKey(User, unique=True) friends = friends = models.ManyToMany('self') ...... I can't decide which to use in which case. in other words, what are the core differences between both?

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• Django: where do I call settings.configure?

  - by RexE

  The Django docs say that I can call settings.configure instead of having a DJANGO_SETTINGS_MODULE. I would like my website's project to do this. In what file should I put the call to settings.configure so that my settings will get configured at the right time? Edit in response to Daniel Roseman's comment: The reason I want to do this is that settings.configure lets you pass in the settings variables as a kwargs dict, e.g. {'INSTALLED_APPS': ..., 'TEMPLATE_DIRS': ..., ...}. This would allow my app's users to specify their settings in a dict, then pass that dict to a function in my app that augments it with certain settings necessary to make my app work, e.g. adding entries to INSTALLED_APPS. What I envision looks like this. Let's call my app "rexe_app". In wsgi.py, my app's users would do: import rexe_app my_settings = {'INSTALLED_APPS': ('a','b'), ...} updated_settings = rexe_app.augment_settings(my_settings) # now updated_settings is {'INSTALLED_APPS': ('a','b','c'), 'SESSION_SAVE_EVERY_REQUEST': True, ...} settings.configure(**updated_settings)

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• Setting a preferred item of a many-to-one in Django

  - by Mike DeSimone

  I'm trying to create a Django model that handles the following: An Item can have several Names. One of the Names for an Item is its primary Name, i.e. the Name displayed given an Item. (The model names were changed to protect the innocent.) The models.py I've got looks like: class Item(models.Model): primaryName = models.OneToOneField("Name", verbose_name="Primary Name", related_name="_unused") def __unicode__(self): return self.primaryName.name class Name(models.Model): item = models.ForeignKey(Item) name = models.CharField(max_length=32, unique=True) def __unicode__(self): return self.name class Meta: ordering = [ 'name' ] The admin.py looks like: class NameInline(admin.TabularInline): model = Name class ItemAdmin(admin.ModelAdmin): inlines = [ NameInline ] admin.site.register(Item, ItemAdmin) It looks like the database schema is working fine, but I'm having trouble with the admin, so I'm not sure of anything at this point. My main questions are: How do I explain to the admin that primaryName needs to be one of the Names of the item being edited? Is there a way to automatically set primaryName to the first Name found, if primaryName is not set, since I'm using inline admin for the names?

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• Django admin proper urls inside listview

  - by hinnye

  Hi, My current target is to give users the chance to download CSV files from the admin site of my application. I successfully managed to create an additional column in the model's list view this way: def doc_link(self): return '<a href="files/%s">%s</a>' % (self.output, self.output) doc_link.allow_tags = True This shows the file name and creates the link, but sadly - because it's inside my 'searches' view - it has an URL: my_site/my_app/searches/files/13.csv. This is my problem, I would like to have my files stored in the admin media directory, like this: http://my_site/media/files/13.csv Does somebody know how to give url which points "outer" from the model's directory? Maybe somehow tell Django to use the ADMIN_MEDIA_PREFIX in the link? I'd really appreciate any help, thanks!

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• Is there a tool for detecting sites using the same template?

  - by KTB

  I often buy webtemplates online, and when I do so I often look at the demos to get some inspiration on how to use the components. But I would love to look at other sites which have already implemented a full website using this template. So I am looking for a tool which searches sites having a similar HTML as the demos (and therefore are probably implementing the template). I am referring to templates which do not have a static text like "Created with Template FooBar by BarFoo" in the footer.

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• Saving data in a inherited django model

  - by aldeano

  I'm building an app to save data and some calculations made with those datas, the idea is keep the data in one model and the calculations in other. So, the models are like this: class FreshData(models.Model): name = models.CharField(max_length=20) one = models.IntegerField() two = models.IntegerField() def save(self, *args, **kwargs): Calculations() Calculations.three = self.one + self.two super(FreshData, self).save(*args, **kwargs) Calculations.save() class Calculations(FreshData): three = models.IntegerField() I've got a valueerror pointing out "self.one" and "self.two" as without value. I keep the idea in witch my design is wrong and django has a simpler way to store related data.

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• Django: Site-Wide URL Prefix

  - by Tom

  I've built a Django site that will live at the root when it's live. Right now it's functioning perfectly at the IP address. For testing purposes, the client has pointed a proxy url at it, but the url has /folder/path in it, so none of the URL patterns match. I put (/folder/path)? into all the url patterns so they now respond, but all of the links are broken because I'm using the {% url %} tag and while the url patterns will match the optional path, they don't include it in that tag. Clearly I can just hard-code /folder/path into all of my urls (well, into all of the url includes) until testing is complete, but is there a better way to do this?

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• Django model class and custom property

  - by dArignac

  Howdy - today a weird problem occured to me: I have a modle class in Django and added a custom property to it that shall not be saved into the database and therefore is not represent in the models structure: class Category(models.Model): groups = models.ManyToManyField(Group) title = defaultdict() Now, when I'm within the shell or writing a test and I do the following: c1 = Category.objects.create() c1.title['de'] = 'german title' print c1.title['de'] # prints "german title" c2 = Category.objects.create() print c2.title['de'] # prints "german title" <-- WTF? It seems that 'title' is kind of global. If I change title to a simple string it works as expected, so it has to do something with the dict? I also tried setting title as a property: title = property(_title) But that did not work, too. So, how can I solve this? Thank you in advance! enter code here

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• django filefield return filename only in template

  - by John

  I've got a field in my model of type FileField. This gives me an object of type type File, which has the following method: File.name: The name of the file including the relative path from MEDIA_ROOT. What I want is something like .filename that will only give me the filename and not the path as well something like: {% for download in downloads %} <div class="download"> <div class="title">{{download.file.filename}}</div> </div> {% endfor %} which would give something like myfile.jpg thanks

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• SVN tags: How not to update/checkout them?

  - by Boldewyn

  In many projects, I check out the complete repository and have then the standard directory structure: project/ branches/ tags/ trunk/ If I do an svn up project, it's all fine with the branches and trunk folders, but, of course, the tags folder is updated, too, and filled with (mostly) lots of tagged versions that are of no value for my work and only occupy disk space. How can I except the tags folder from an svn update? Especially, how can I do this locally only, that is, without committing that back to the repository, as a solution with the svn:ignore keyword would do?

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• How can I insert a rendered form from a form application in a Django template from another applicati

  - by jul

  hi, I'm doing a form application (similar to django-contact-form) and would like to be able to insert the rendered form in a Django template of another application. What's the best way to do that? Should I create a template for the form (with for example only {{ form.as_p)? If so how can I insert it in another template? Thanks Jul

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• Stackoverflow interesting tags

  - by Tom

  So, that's how works Interesting Tags: I add into them my interested tags like php, mysql, jquery and so on. Then, if any of questions has the same tags as mine it makes background orange. I understand how to use jquery to do that (there were related questions to that), but without mysql it can't be done. Now, here is a question. How is it done? I imagine like that: There is a row in mysql for every member, let's call it "interested_tags". After I write and submit my tag through input, it is being written in a row "interested_tags". Then, the main page has a query which shows all answers and it always checks answer's with mine tags with strpos like this if(strpos($question_tags, $my_tags) === true) {        //and here will be made background orange } Am I thinking right or is there any way to do it?

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• Django modelform ForeignKey List

  - by Harry

  How do you get each item in the ForeignKey field in a list, for example: class Delegate(models.Model): excursion = models.ForeignKey(Excursion, limit_choices_to = {'is_activity': False}, related_name='excursion', null=True, blank=True) Template: {% for object in formset.excursion_set.all %} {{ object.lable }} etc {% endfor %} My reason is that I don't want the options to display as a dropdown, but in a custom way that I will style etc.

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• Django 1.2: Dates in admin forms don't work with Locales (I10N=True)

  - by equalium

  I have an application in Django 1.2. Language is selectable (I18N and Locale = True) When I select the english lang. in the site, the admin works OK. But when I change to any other language this is what happens with date inputs (spanish example): Correctly, the input accepts the spanish format %d/%m/%Y (Even selecting from the calendar, the date inserts as expected). But when I save the form and load it again, the date shows in the english form: %Y-%m-%d The real problem is that when I load the form to change any other text field and try to save it I get an error telling me to enter a valid date, so I have to write all dates again or change the language in the site to use the admin. I haven't specified anything for DATE_INPUT_FORMATS in settings nor have I overridden forms or models. Surely I am missing something but I can't find it. Can anybody give me a hint?

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• How to save links in Django

  - by xRobot

  I have a simple Blog model with a TextFields. What is the best way to save link ? 1) Saving link in the database in this form: http://www.example.com and then using some filter in the template to trasform it in form: <a rel="nofollow" href="http://www.example.com>http://www.example.com</a> 2) Saving link in the database directly in this form: <a rel="nofollow" href="http://www.example.com>http://www.example.com</a>

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• get foreign key objects in a single query - Django

  - by John

  Hi I have 2 models in my django code: class ModelA(models.Model): name = models.CharField(max_length=255) description = models.CharField(max_length=255) created_by = models.ForeignKey(User) class ModelB(models.Model): category = models.CharField(max_length=255) modela_link = models.ForeignKey(ModelA, 'modelb_link') functions = models.CharField(max_length=255) created_by = models.ForeignKey(User) Say ModelA has 100 records, all of which may or may not have links to ModelB Now say I want to get a list of every ModelA record along with the data from ModelB I would do: list_a = ModelA.objects.all() Then to get the data for ModelB I would have to do for i in list_a: i.additional_data = i.modelb_link.all() However this runs a query on every instance of i. Thus making 101 queries to run. Is there any way of running this all in just 1 query. Or at least less than the 101 queries. I've tried putting in ModelA.objects.select_related().all() but this didn't seem to have any effect. Thanks

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• Django template CSS/IMG is "off" in the URL

  - by erimar77

  I have /path/to/my/theme/static/css/frontend.css which is called by base.html <link rel="stylesheet" type="text/css" href="{{ STATIC_URL }}css/frontend.css" media="all" /> In which I've got a background for the header: #header-wrapper min-width: 960px; height: 150px; background: transparent url(img/header-bg.png) repeat-x center bottom; } The file is /path/to/my/theme/static/img I've run manage.py collectstatic to gather the files and almost everything looks correct except the link generated looks like: http://example.com/static/css/img/header-bg.png In which the image does not show, because the correct URL is: http://example.com/static/img/header-bg.png Where am I going wrong??

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• Django: Converting an entire Model into a single dictionary

  - by LarrikJ

  Is there a good way in Django to convert an entire model to a dictionary? I mean, like this: class DictModel(models.Model): key = models.CharField(20) value = models.CharField(200) DictModel.objects.all().to_dict() ... with the result being a dictionary with the key/value pairs made up of records in the Model? Has anyone else seen this as being useful for them? Thanks. Update I just wanted to add is that my ultimate goal is to be able to do a simple variable lookup inside a Template. Something like: {{ DictModel.exampleKey }} With a result of DictModel.objects.get(key__exact=exampleKey).value Overall, though, you guys have really surprised me with how helpful allof your responses are, and how different the ways to approach it can be. Thanks a lot.

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• Chef: nested data bag data to template file returns "can't convert String into Integer"

  - by Dalho Park

  I'm creating simple test recipe with a template and data bag. What I'm trying to do is creating a config file from data bag that has simple nested information, but I receive error "can't convert String into Integer" Here are my setting file 1) recipe/default.rb data1 = data_bag_item( 'mytest', 'qa' )['test'] data2 = data_bag_item( 'mytest', 'qa' ) template "/opt/env/test.cfg" do source "test.erb" action :create_if_missing mode 0664 owner "root" group "root" variables({ :pepe1 = data1['part.name'], :pepe2 = data2['transport.tcp.ip2'] }) end 2)my data bag named "mytest" $knife data bag show mytest qa id: qa test: part.name: L12 transport.tcp.ip: 111.111.111.111 transport.tcp.port: 9199 transport.tcp.ip2: 222.222.222.222 3)template file test.erb part.name=<%= @pepe1 % transport.tcp.binding=<%= @pepe2 % Error reurns when I run chef-client on my server, [2013-06-24T19:50:38+00:00] DEBUG: filtered backtrace of compile error: /var/chef/cache/cookbooks/config_test/recipes/default.rb:19:in []',/var/chef/cache/cookbooks/config_test/recipes/default.rb:19:inblock in from_file',/var/chef/cache/cookbooks/config_test/recipes/default.rb:12:in from_file' [2013-06-24T19:50:38+00:00] DEBUG: filtered backtrace of compile error: /var/chef/cache/cookbooks/config_test/recipes/default.rb:19:in[]',/var/chef/cache/cookbooks/config_test/recipes/default.rb:19:in block in from_file',/var/chef/cache/cookbooks/config_test/recipes/default.rb:12:infrom_file' [2013-06-24T19:50:38+00:00] DEBUG: backtrace entry for compile error: '/var/chef/cache/cookbooks/config_test/recipes/default.rb:19:in `[]'' [2013-06-24T19:50:38+00:00] DEBUG: Line number of compile error: '19' Recipe Compile Error in /var/chef/cache/cookbooks/config_test/recipes/default.rb TypeError can't convert String into Integer Cookbook Trace: /var/chef/cache/cookbooks/config_test/recipes/default.rb:19:in []' /var/chef/cache/cookbooks/config_test/recipes/default.rb:19:inblock in from_file' /var/chef/cache/cookbooks/config_test/recipes/default.rb:12:in `from_file' Relevant File Content: /var/chef/cache/cookbooks/config_test/recipes/default.rb: 12: template "/opt/env/test.cfg" do 13: source "test.erb" 14: action :create_if_missing 15: mode 0664 16: owner "root" 17: group "root" 18: variables({ 19 :pepe1 = data1['part.name'], 20: :pepe2 = data2['transport.tcp.ip2'] 21: }) 22: end 23: I tried many things and if I comment out "pepe1 = data1['part.name'],", then :pepe2 = data2['transport.tcp.ip2'] works fine. only nested data "part.name" cannot be set to @pepe1. Does anyone knows why I receive the errors? thanks,

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• Deploying a Django application in a virtual Ubuntu Server

  - by mfsaint

  I have a virtualbox machine running Ubuntu Server 10.04LTS. My intention is to this machine to work like a VPS, this way I can learn and prepare for when I get a VPS service. Apache+mod_wsgi for deploying the Django app seems the right choice to me. I have the domain (marianofalcon.com.ar) but nothing else, no DNS. The problem is that I'm pretty lost with all the deployment stuff. I know how to configure mod_wsgi(with the django.wsgi file) and apache(creating a VirtualHost). Something is missing and I don't know what it is. I think that I lack networking skills ant that's the big problem. Trying to host the app on a virtualbox adds some difficulty because I don't know well what IP to use. This is what I've got: file placed at: /etc/apache2/sites-available: NameVirtualHost *:80 <VirtualHost *:80> ServerAdmin [email protected] ServerName www.my-domain.com ServerAlias my-domain.com Alias /media /path/to/my/project/media DocumentRoot /path/to/my/project WSGIScriptAlias / /path/to/your/project/apache/django.wsgi ErrorLog /var/log/apache2/error.log LogLevel warn CustomLog /var/log/apache2/access.log combined </VirtualHost> django.wsgi file: import os, sys wsgi_dir = os.path.abspath(os.path.dirname(__file__)) project_dir = os.path.dirname(wsgi_dir) sys.path.append(project_dir) project_settings = os.path.join(project_dir,'settings') os.environ['DJANGO_SETTINGS_MODULE'] = 'myproject.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler()

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• django-mptt: how to successfully move nodes around

  - by Parand

  django-mptt seems determined to drive me out of my mind. I'm trying to do something relatively simple: I'm going to delete a node, and need to do something reasonable with the node's children. Namely, I'd like to move them up one level so they're children of their current parent's parent. That is, if the tree looks like: Root | Grandpa | Father | | C1 C2 I'm going to delete Father, and would like C1 and C2 to be children of Grandpa. Here's the code I'm using: class Node(models.Model): first_name = models.CharField(max_length=80, blank=True) parent = models.ForeignKey('self', null=True, blank=True, related_name='children') def reparent_children(self, parent): print "Reparenting" for child in self.get_children(): print "Working on", child.first_name, "to parent", parent.email parent = Node.objects.get(id=parent.id) child.move_to(parent, 'last-child') child.save() So I'd call: father.reparent_children(grandpa) father.parent = None father.save() This works - almost. The children report their parents as Grandpa: c1.parent == grandpa # True Grandpa counts C1 and C2 among its children c1 in grandpa.children.all() # True However, Root disowns these kids. c1.get_root() == father # c1's root is father, instead of Root c1 in root.get_descendants() # False How do I get the children to move and their root not get corrupted?

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• Limiting choices from an intermediary ManyToMany junction table in Django

  - by Matthew Rankin

  Background I've created three Django models—Inventory, SalesOrder, and Invoice—to model items in inventory, sales orders for those items, and invoices for a particular sales order. Each sales order can have multiple items, so I've used an intermediary junction table—SalesOrderItems—using the through argument for the ManyToManyField. Also, partial billing of a sales orders is allowed, so I've created a ForeignKey in the Invoice model related to the SalesOrder model, so that a particular sales order can have multiple invoices. Here's where I deviate from what I've normally seen. Instead of relating the Invoice model to the Item model via a ManyToManyField, I've related the Invoice model to the SalesOrderItem intermediary junction table through the intermediary junction table InvoiceItem. I've done this because it better models reality—our invoices are tied to sales orders and can only include items that are tied to that sales order as opposed to any item in inventory. I will admit that it does seem strange having the intermediary junction table of a ManyToManyField related to the intermediary junction table of another ManyToManyField. Question How can I limit the choices available for the invoice_items in the Invoice model to just the sales_order_items of the SalesOrder model for that particular Invoice? (I tried using limit_choices_to= {'sales_order': self.invoice.sales_order}) as part of the item = models.ForeignKey(SalesOrderItem) in the InvoiceItem model, but that didn't work. Am I correct in thinking that limiting the choices for the invoice_items should be handled in the model instead of in a form? Code class Item(models.Model): item_num = models.SlugField(unique=True) default_price = models.DecimalField(max_digits=10, decimal_places=2, blank=True, null=True) class SalesOrderItem(models.Model): item = models.ForeignKey(Item) sales_order = models.ForeignKey('SalesOrder') unit_price = models.DecimalField(max_digits=10, decimal_places=2) quantity = models.DecimalField(max_digits=10, decimal_places=4) class SalesOrder(models.Model): customer = models.ForeignKey(Party) so_num = models.SlugField(max_length=40, unique=True) sales_order_items = models.ManyToManyField(Item, through=SalesOrderItem) class InvoiceItem(models.Model): item = models.ForeignKey(SalesOrderItem) invoice = models.ForeignKey('Invoice') unit_price = models.DecimalField(max_digits=10, decimal_places=2) quantity = models.DecimalField(max_digits=10, decimal_places=4) class Invoice(models.Model): invoice_num = models.SlugField(max_length=25) sales_order = models.ForeignKey(SalesOrder) invoice_items = models.ManyToManyField(SalesOrderItem, through='InvoiceItem')

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