django manage py - Page 30

Django remove list_editable on a per row basis.

- by Jason Leveille

On the back of Django 1.2RC1, I have built a great administrator for my client! It's awesome. The client has one request that, if not satisfied, could bring this house of cards crashing down. I have a list of swim meet results. A meet result is added by a superuser. A team rep can edit a meet result (which they must be able to do inline, with list_editable), but after they edit the meet result inline and save, they should no longer be able to edit that result inline. They can only perform one edit. So, my question ... can I turn off list_editable on a per row basis?

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Django - Passing arguments to models through ForeignKey attributes

- by marshall

I've got a class like this: class Image (models.Model): ... sizes = ((90,90), (300,250)) def resize_image(self): for size in sizes: ... and another class like this: class SomeClassWithAnImage (models.Model): ... an_image = models.ForeignKey(Image) what i'd like to do with that class is this: class SomeClassWithAnImage (models.Model): ... an_image = models.ForeignKey(Image, sizes=((90,90), (150, 120))) where i'm can specify the sizes that i want the Image class to use to resize itself as a argument rather than being hard coded on the class. I realise I could pass these in when calling resize_image if that was called directly but the idea is that the resize_image method is called automatically when the object is persisted to the db. if I try to pass arguments through the foreign key declaration like this i get an error straight away. is there an easy / better way to do this before I begin hacking down into django?

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Problems using User model in django unit tests

- by theycallmemorty

I have the following django test case that is giving me errors: class MyTesting(unittest.TestCase): def setUp(self): self.u1 = User.objects.create(username='user1') self.up1 = UserProfile.objects.create(user=self.u1) def testA(self): ... def testB(self): ... When I run my tests, testA will pass sucessfully but before testB starts, I get the following error: IntegrityError: column username is not unique It's clear that it is trying to create self.u1 before each test case and finding that it already exists in the Database. How do I get it to properly clean up after each test case so that subsequent cases run correctly?

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Access to field in extended flatpage in django

- by Stanislav Feldman

How to access field in extended flatpage in django? I wrote this: class ExtendedFlatPage(FlatPage): teaser = CharField(max_length=150) class ExtendedFlatPageForm(FlatpageForm): teaser = CharField(max_length=150) class Meta: model = ExtendedFlatPage class ExtendedFlatPageAdmin(FlatPageAdmin): form = ExtendedFlatPageForm fieldsets = ( (None, {'fields': ('url', 'title', 'teaser', 'content', 'sites',)}), ) admin.site.unregister(FlatPage) admin.site.register(ExtendedFlatPage, ExtendedFlatPageAdmin) And creation in admin is ok. But then in flatpages/default.html I tried this: <html> <body> <h1>{{ flatpage.title }}</h1> {{ flatpage.teaser }} {{ flatpage.content }} </body> </html> And there was no flatpage.teaser! What is wrong?

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(Django) Trim whitespaces from charField

- by zardon

How do I strip whitespaces (trim) from the end of a charField in Django? Here is my Model, as you can see I've tried putting in clean methods but these never get run. I've also tried doing name.strip(), models.charField().strip() but these do not work either. Is there a way to force the charField to trim automatically for me? Thanks. class Employee(models.Model): """(Workers, Staff, etc)""" name = models.CharField(blank=True, null=True, max_length=100) # This never gets run def clean_variable(self): data = self.cleaned_data['variable'].strip() return data def __unicode__(self): return self.name class Meta: verbose_name_plural = 'Employees' # This never gets run either class EmployeesForm(forms.ModelForm): class Meta: model = Employee def clean_description(self): #if not self.cleaned_data['description'].strip(): # raise forms.ValidationError('Your error message here') self.cleaned_data['name'].strip()

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in django am facing url problem.....

- by dpaksp

am using django.0.97 version i have model called profile in that i have few fields eg like name ,email...ects and it's backend also ready..i.e database . and all users profile is created...i have given all permission to all users. when i login ,click on profile..i able to see list of all user name thr when i click on it ,it's goin to model page where i can edit the user profile..instead of that i want to navigate to a template when i can display the user details ,i have set the URl for it so that when url of that type request comes it should call a view from view it will call my template to display user details,.....the problem is it's not calling my view.... i think my problem is brief....if any information still required ?? pls ask me....and help me

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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: Proper place to unregister ModelAdmins

- by lazerscience

Sometimes I need to UNREGISTER some ModelAdmins from the admin site, because I don't want them to be there as they are, eg. if I'm using the Sites framework, and I dont want it to appear in the admin. It's no big deal to e.g. call admin.site.unregister(Site) to do so. In most cases I put it in admin.py of some related app that I have made, but sometimes I end up putting it in a place that hasn't much to do with the original app; another possibility would be making a "dummy app" and put it there... Does anybody know a more descent place where these calls can live?

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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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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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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 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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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 - Convert python list into a javascript object

- by amcashcow

I am working on a django / python website I have a page where I want to display a table of search results The list of results is passed in to the template as normal I also want to make this list of objects accessible to the javascript code My first solution was just create another view that returned json format. But each page load required calling the query twice. So then I tried only downloading the data using the json view and printing the table using javascript. but this is also not desirable as now the presentation layer is mixed into the javascript code. is there a way to create a javascript object from the python list as the page is rendered?

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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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Use localeURL middleware with apache prefix

- by Olivier R.

Good morning everyone, I Got a question about localeURL usage. Everything works great for me with url like this : www.mysite.com/ If I type www.mysite.com/ in adress bar, it turns correctly in www.mysite.com/en/ for example. If I use the view change_locale, it's also all right (ie change www.mysite.com/en/ in www.mysite.com/fr/). But my application use apache as server, and use a prefix for the site, that gives url like this : www.mysite.com/prefix/ If I type www.mysite.com/prefix/ in the adress bar, the adress turns into www.mysite.com/en/ without prefix (so 404) I change code of view to manage our settings.SERVER_PREFIX value : def change_locale(request) : """ Redirect to a given url while changing the locale in the path The url and the locale code need to be specified in the request parameters. O. Rochaix; Taken from localeURL view, and tuned to manage : - SERVER_PREFIX from settings.py """ next = request.REQUEST.get('next', None) if not next: next = request.META.get('HTTP_REFERER', None) if not next: next = settings.SERVER_PREFIX + '/' next = urlsplit(next).path prefix = False if settings.SERVER_PREFIX!="" and next.startswith(settings.SERVER_PREFIX) : prefix = True next = "/" + next.lstrip(settings.SERVER_PREFIX) _, path = utils.strip_path (next) if request.method == 'POST': locale = request.POST.get('locale', None) if locale and check_for_language(locale): path = utils.locale_path(path, locale) if prefix : path = settings.SERVER_PREFIX + path response = http.HttpResponseRedirect(path) return response with this customized view, i'm able to correctly change language, but i'm not sure that's the right way of doing stuff. Is there any option on localeURL to manage prefix of apache ?

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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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replace only one variable in web.py templates

- by Asterisk

I am passing variable to template in web.py and have the same condition in some places. Like this: $if myvar=="string1": $passed argument1 ............ $if myvar =="striung2": $passed argument2 If say myvar is "string1" and I pass passed = "AAA" then I have AAA argument1 on my page, but the other if statements get replaced by empty string? How to avoid that? I.e. how to leave other statements intact?

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Komodo Edit 5.2 Django Template Syntax Error - Info: <head> previously mentioned

- by Lance McNearney

I am using Komodo Edit 5.2 for editing html and Django template files. It always shows a single syntax error inside the first {% block %} area on the first tag of my template. For example: {% extends "base.html" %} {% load i18n %} {% block title %}Hello{% endblock %} {% block content %} Hello <-- Syntax error on this single line Other lines have no errors {% endblock %} {% block footer %} No errors here {% endblock %} The syntax error given is: Info: <head> previously mentioned I know for a fact that the error has nothing to do with my <head> tag since it occurs in the base template and in child templates (and the IDE isn't smart enough to process the base templates when in a child, etc.) All of my html tags are closed properly and everything validates for XHTML strict. This forum post mentions a similar problem but offers no solution (and may be specific to Smarty syntax highlighting). Any ideas on how to resolve this error (or disable it from being shown)?

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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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Search found 11188 results on 448 pages for 'django manage py'.

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- by Jason Leveille

- by marshall

- by theycallmemorty

- by Stanislav Feldman

- by zardon

- by dpaksp

- by John

- by Brant

- by lazerscience

- by amr.negm

- by RexE

- by hinnye

- by aldeano

- by Tom

- by dArignac

- by equalium

- by John

- by amcashcow

- by erimar77

- by LarrikJ

- by Olivier R.

- by mfsaint

- by Asterisk

- by Lance McNearney

- by Parand

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