Django 2.1 inspectdb supports Oracle ?
The [version 2.0 docs][https://docs.djangoproject.com/en/2.0/ref/django-admin/#django-admin-inspectdb] states, "inspectdb works with PostgreSQL, MySQL and SQLite. Foreign-key detection only works in PostgreSQL and with certain types of MySQL tables.". This seems to exclude Oracle. In constrast, the [version 2.1 docs][https://docs.djangoproject.com/en/2.1/ref/django-admin/#django-admin-inspectdb] doesn't state this, but I'm seeing that it doesn't produce models with Oracle; is inspectdb V2.1 expected to work with Oracle tables and views ? I've tried: python manage.py inspectdb > models.py python manage.py inspectdb --include-views > models.py In both cases, no errors are produced and the output just contains: ''' # [generated comments...] from django.db import models ''' Thanks, pk -- You received this message because you are subscribed to the Google Groups "Django users" group. To unsubscribe from this group and stop receiving emails from it, send an email to django-users+unsubscr...@googlegroups.com. To post to this group, send email to django-users@googlegroups.com. Visit this group at https://groups.google.com/group/django-users. To view this discussion on the web visit https://groups.google.com/d/msgid/django-users/a9174323-d2be-4635-b582-7152a18b0d6a%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
url template tag ; url generation
In my template, I have: Read More » I'm seeing that the url generated via the url template tag is, eg: http://mydomain.com/mysite/param0val/param1val/param2val/ 'mysite' is the project directory created via, eg, "django-admin.py startproject mysite"; I don't want it to be part of the generated url; I want the generated url to be: http://mydomain.com/param0val/param1val/param2val/ 1. Is there a way to accomplish this, preferably using the url template tag ? My urls.py contains: (r'^(?P\w+)/(?P(\w|-)+)/(?P\d+)(-(\w+))*/ $', 'myapp.views.byname'), 2. Why/how is 'mysite' built into the url ? Thanks, pk -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: automatic documentation (docutils) does not pull in class methods?
On Jun 22, 4:12 am, "euan.godd...@googlemail.com"wrote: > You could try using Spinhx if you don't think docutils gives you > enough flexibility. The autodoc extension is great. > However sphinx + autodoc does not know how to extra django model definitions, e.g. class A(models.Model): field_not_seen_by_autodoc = models.CharField(...) Whereas the django admindocs utility knows about these fields. -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: automatic documentation (docutils) does not pull in class methods?
Answering my own question -- it is django.contrib.admindocs.view that limits the admindocs view to only list methods on a model that has a single argument. The lines: for func_name, func in model.__dict__.items(): if (inspect.isfunction(func) and len(inspect.getargspec(func) [0]) == 1): There is currently an open ticket http://code.djangoproject.com/ticket/12974 that patches it to handle the case where methods have decorator on them, hence reducing the number of args to zero. My question is why are we limiting the list of methods to single argument (self) methods? I put business logic in the model and usually the models will have many more methods that take additional arguments. P.K. -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
automatic documentation (docutils) does not pull in class methods?
The automatic documentation generated, via docutils, in the admin interface for a django app does not pull in documentation for methods defined for classes, only attributes. Is this a "feature" of the django implementation or docutils? Thanks, P.K. -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
Re: select childless parents, i.e. base with no derived classes in model inheritance
The Place/Restaurant sample is from the model doc under multi table inheritance: http://docs.djangoproject.com/en/dev/topics/db/models It is a simple inherited model relationship: class Place(models.Model): name = models.CharField(max_length=50) address = models.CharField(max_length=80) class Restaurant(Place): serves_hot_dogs = models.BooleanField() serves_pizza = models.BooleanField() On Dec 30 2008, 5:09 am, Brielwrote: > The first query looks at the restaurant attribute in the place model, > and gets all that have a NULL value in that field. Now I don't know > the restaurant/place setup, the docs I read uses book/auther examples. > It looks like that the restaurant field is ForeignKey, and should be > able to work the way intended. It would do a lookup for all places > that doen't have a restaurant object associate with it. But, if the > restaurant field is a BoolianField instead, then you can't use > is_null, as both True and False isn't Null. > > Hope this helps, else be a little more explicit about the > relationships. > > --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: select childless parents, i.e. base with no derived classes in model inheritance
Alex, First I made a mistake in the original post. The first query returns *nothing*, not "all places". This is the correct "question: 1) print Place.objects.filter(restaurant__isnull=True).count() returns 0 and 2) print Place.objects.filter (restaurant__serves_hotdog__isnull=True).count() works correctly, returning a count of all Places that is not a restaurant. The first query does not even join the child table. It tries to do: SELECT COUNT(*) FROM "place" WHERE "place"."id" IS NULL Where as the second query does a LEFT OUTER JOIN between parent and child table, then added a where clause to test for the hotdog field being null, essentially filtered out the rows that have no hotdog fields at all. Seems like query 1 is broken? P.K. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
select childless parents, i.e. base with no derived classes in model inheritance
I have a need to select base classes that does not have derived class defined using the ORM. Using the Place/Restaurant example in the Django doc, I find that: 1) Place.objects.filter(restaurant__isnull=True) returns all places. and 2) Place.objects.filter(restaurant__serves_hotdog__isnull=True) works, returning all Places that is not a restaurant. I wonder though, is this the *correct* way to do the query? P.K. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Template inheritance
One simple way to think about template inheritance (which is a GREAT feature) is to think of it as "specialization": You first define the general look of a website, the logo, banner, menu bars, footers. All that goes into the base template A. In fact, you can render just the base template A and see an "empty" site. Then for each actual website page, you have a template B that defines the "insides" of the base template, e.g. the main content area. As your page specific view renders the specialized template B, which only contains the "insides", the django template engine puts B within the context of A, and renders a full page with the almost all of A with one or more blocks replaced by specific content supplied by B. B will have none of the HTML elements that make up most of the final web page. It only contains the bits that need to be inserted into a {% block %} inside A. The neat thing about the django system is that B can replaces multiple parts of A. For example, a site has a two column type display, each column is a {% block %} that can be replaced. You can write a B1 page that just replaces what is in column A, or you can write a page B2 that replaces both columns in A. P.K. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Using settings.py mechanism for application settings
To clarify the question a bit more, I am talking about an "app" in the sense of a reusable component that is developed by me but want to distribute for others to use. James is right that for any of my own app, I can just tell the user to put in some settings in the project level settings.py file. In fact that is exactly what I do now. However that does not buy me the feature where, for django's own apps, have their default settings stored in global_settings.py (or whatever is the file's name). So I need to test for and supply default if my user does not supply a settings value. Meanwhile the django built in settings mechanism has a very nice way to specify defaults -- which is what I like to "Reuse". P.K. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Using settings.py mechanism for application settings
There are always needs for application level configuration settings. I really like the way settings.py works, with a package (django) level default settings overridable by local settings. However looking at the whole LazySettings setup it is not easily used outside of django/conf. Am I missing something? Should that mechanism be refactored so that it can be used for user applications? Is there a better way to do application level configuration -- i.e. for app designed to be distributed and reused? Thanks, P.K. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Template variable in filter?
Isn't this already in settings.py? Look at the code for the filters /django/template/defaultfilters.py it reads DATE_FORMAT and TIME_FORMAT from the settings file. P.K. On Apr 29, 2:25 am, Mike Chambers <[EMAIL PROTECTED]> wrote: > Is it possible to include a template variable inside a filter? > > Specifically, I want to have a global setting for the date format string: > > Something like: > > settings.py > -- > DATE_TIME_FORMAT = "M j, F \a\t P" > > template > -- > {{ comment.date_submitted|date:{{DATE_TIME_FORMAT > > I am guessing that I could write a custom filter that handles this, was > curious if there was another way to do it. > > mike --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Should I set up Django locally or on a web server?
if you are running Leopard (IMPORTANT) it is very easy to setup Django. I wrote up what I did here: http://www.pkshiu.com/loft/archive/2008/01/django-on-leopard The key is to just use sqlite and the development server to do development. There are some issues with sqlite compare to postgresql (I use that for production). sqlite is more forgiving in some situations so you will find some subtle errors later on, but it's not a big deal. P.K. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Templating patterns
sometimes extending the template system with application specific tags solves my problem. YMMV of course. On May 2, 1:49 am, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote: > I've got a lot of templates with the same pattern, so i want to put > some code like this in a template. > > {% if iterable %} > {% for variable in iterable %} > <{{tag}}> something > {% endfor %} > > Is it posible to send the {{tag}} variable as a parameter from > another template (where i use the include tag) ? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Extending pythonpath within python code
os.path.normpath('e:/a/b/c') will give 'e:\\a\\b\\c' On Dec 22, 6:50 am, Julien <[EMAIL PROTECTED]> wrote: > Thanks for the tip! > > However, it's a bit strange because it only accepts paths with double > back slashes, instead of the forward slashes required for declaring > the template path for example. > > import sys > sys.path.append('E:\\workspace\\myproject\\trunk\\apps') > works > > but: > import sys > sys.path.append('E:/workspace/myproject/trunk/apps') > doesn't work > > Is there a way around that? As I'm storing the absolute path of my > project in a constant variable (with forward slashes), I'd like to do > something like this: > > ABSOLUTE_PROJECT_PATH = 'E:/workspace/myproject/trunk/ > sys.append(ABSOLUTE_PROJECT_PATH + 'apps/') > > TEMPLATE_DIRS = ( > ABSOLUTE_PROJECT_PATH + 'templates/', > ) > > PS: As you may have guessed, I'm running under Windows. > > On Dec 22, 8:03 pm, Matthias Kestenholz <[EMAIL PROTECTED]> wrote: > > > On Sat, 2007-12-22 at 00:58 -0800, Julien wrote: > > > Hi there, > > > > To keep my projects tidy, I'm used to having all my apps in a > > > subfolder "apps". > > > This means that I need to add that subfolder to the pythonpath. I can > > > do this in apache or by doing runserver --pythonpath=apps. > > > > However, I'd like to extend the pythonpath within the code, for > > > example in settings.py(*). That would save me from altering apache > > > config files and would therefore make deployment a little easier. > > > > Is that possible, and is it recommended practice? > > > The python path lives in sys.path (which is a simple list). This means > > that you can add new directories with sys.path.insert(0, 'your/path') or > > sys.path.append('your/path'), whichever you like. > > > I do this all the time, mainly in the FastCGI caller script, but also in > > settings.py to enable access to middleware classes and templatetags > > which I use in nearly every project. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Django documentation in chm (Dec 22, 2007)
Thanks very much for doing this. While reading on djangoproject is nice, sometimes it is faster/better to have a local version -- When I am coding on a plane, for example ! P.K. On Dec 22, 8:36 am, char101 <[EMAIL PROTECTED]> wrote: > Hi, > > I have created a chm format of the django docs taken from svn at Dec > 22, 2007. > > Link:http://charupload.wordpress.com/2007/12/02/django-documentation-chm/ > > Note: don't right click and save as on the links because it's a link > to a page on divshare, not a direct link to the file. > > Cheers, > Charles --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Database filters
Are you using the built-in Group and User classes? If what you are quoting is cut-and-pasted from your console, you maybe using your own model? Maybe you should show us the actual source code. FYI -- the built-in classes are django.contrib.auth.models.Group and ..User not Groups nor Users P>P.K. On Dec 20, 11:13 am, kbochert <[EMAIL PROTECTED]> wrote: > I have an app, and I have added 3 Users in 2 Groups. > How do I get the users that belong to a specific Group?? > > Using the shell on the built in Django models I get:>>> G = Groups.objects > >>> U = Users.objects > >>> U.all() > > [, ]>>> G.all() > > [, ] > > Ok so far? ( the double User: in U.all() concerns me a bit) > > >>>U.all().filter(groups__id = 1) > [] > >>>U.all().filter(groups__id = 2) > > [] > > Makes sense- I've used the admin to put one user in each group. > > >>>U.all().filter(groups__name = 'Admin') > > [, ] > > ??? Why both users??? > Shouldn't this give me a list of the Users in the 'Admin' group? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: Dynamic Choicefield
This is a pretty common problem. Each form object gets instantiated as you need them. So in your POST handling, you need to instantiate the form with the same product field so that it can populate the choice field setting correct. i.e. FormForOptionQuantity(product,request) P.K. On Dec 9, 7:31 am, pinco <[EMAIL PROTECTED]> wrote: > Hi, > > I'm trying to populate dynamically at runtime a choicefield and > subsequently validate the user choice, but without results. > The goal is to permit the user, in an e-commerce application, to > choose the number of products to buy through a choicefield, with a > list of integer from 1 to a maximum number set to the amount of > products in stock (i.e. if I have 5 products in stock, the choice > should be restricted to 1, 2, 3, 4, 5). > > The following snippet actually populated in a right way the > choicefiled: > > class FormForProductQuantity(forms.Form): > quantity = forms.ChoiceField() > > def __init__(self, product, *args, **kwargs): > super(FormForProductQuantity, self).__init__(*args, **kwargs) > self.fields['quantity'].choices = [(str(product.id) + "_" + > str(c), > c) for c in range (1, product.stock_quantity)] > > The problems come when I try to validate user's choice and add the > product to the shopping cart with the following snippet: > > def add_product_to_cart(request, product_id): > if request.method == 'POST': > product = Product.objects.get(pk=product_id) > form = FormForOptionQuantity(request.POST) > if form.is_valid(): > ... do something > > which raise an the error: > > Traceback: > ... > Exception Value: 'QueryDict' object has no attribute 'stock_quantity' > > I have no idea on how to correct my code. > > Thank you in advance for any help. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: How to display if a user is logged in? Do I develop a custom template tag?
There are two issues. 1) you can't change the auth_user table/class. There are different ways to "extend" the auth_user class. The "B-List" site has some article on how you may do it. http://www.b-list.org/ I have my own "user" class that is one-to-one implemented as "many-to- one" back to the auth_user, which I store all application related user data. 2) You need a logical many-to-many relationship between your products and users. However you may want to consider using an intermediate cross reference class/table as well, as you may need to store other data relating to the relationship, like when the "fav" was added, ranking information, etc etc. P.K. On Nov 26, 10:30 pm, Greg <[EMAIL PROTECTED]> wrote: > Ok...I'm looking into using django's builtin Authentication models. I > can use that to add users. However, I still need to associate a user > with their favorites once they login. So do I need to add a > ManyToManyField in my User table that points to my products table? > --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---
Re: How to display if a user is logged in? Do I develop a custom template tag?
There are a couple of things you have to do to get the "user" object automatically available to all templates. 1. Follow the authentication setup as suggested. 2. Then read the bit about request_context: http://www.djangoproject.com/documentation/templates_python/#subclassing-context-requestcontext You have to make sure to pass the additional context_instance=RequestContext(request) when calling render_to_response. I also write a little context processor that inserts extra session variables from the session table to the context so that my application session objects are always available in my templates. P.K. On Nov 26, 4:59 pm, Greg <[EMAIL PROTECTED]> wrote: > Hello, > I have a website where people can login and keep track of their > favorite products. Whenever they login I create a session variable: > 'request.session['member_id']'. If this session variable is set then > I know that a user is logged in. In order to get this to work I have > to send the session variable in the return statement of every view. > Then in my template I have an if statement that either displays > 'Login' or 'Welcome, User' depending on if > 'request.session['member_id']' is True or False. > > I've created custom tags before where I bring back data from a a > class. However, I'm not sure how I create a custom tag to send a > session variable. Because the session variable is part of a request > object, and i don't know how to have my custom tag have access to the > 'request'. > > Anybody know how I can develp this so that I don't have to send my > session variable in the return statement of every view? > > Thanks --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~--~~~~--~~--~--~---