Re: Image Model, Form and View -> Easily change "upload_to" path from form.save()?

[Kurtis]

I'm not sure if there's a better solution, but here's one I kind of
came up with after reading into the documentation and some other
Discussions. Let me know if there's a better solution out there.
Hopefully this will help someone else out as well.

from django.db import models
from django.contrib.auth.models import User

# Provides the Path for User Images
def user_image_folder(instance = None, filename = None):
return 'uploaded/' + instance.user.username + '/images/' +
filename

class Image(models.Model):

date_created = models.DateField(auto_now_add = True)
user = models.ForeignKey(User)
# Note: upload_to is referencing the above function in the same
file.
data = models.ImageField(upload_to = user_image_folder)

On Nov 1, 10:55 pm, Kurtis  wrote:
> Hey,
>
> I've got a very simple setup for a sample application. Basically, I
> have a Model w/ an ImageField. I have a ModelForm using that Model.
> Finally, I have a FormView for using that Form.
>
> In my View, I call my form's .save() method and pass along the request
> user. I'd like to manipulate where this file is stored.
>
> Here's what I tried so far but it did *not* work -- the thing just
> uses the preset upload_to variable.
>
> ImageForm.save(self, user):
>
>     # Grab Model Instance
>     m = super(ImageForm, self).save(commit = False)
>
>     # Try to Change "upload_to" Value
>     # Note: "data" is the name of my ImageField
>     m.data.upload_to = 'some_other_folder'
>
>     # Save the Model
>     m.save()
>
>     # Done
>     return m

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Image Model, Form and View -> Easily change "upload_to" path from form.save()?

[Kurtis]

Hey,

I've got a very simple setup for a sample application. Basically, I
have a Model w/ an ImageField. I have a ModelForm using that Model.
Finally, I have a FormView for using that Form.

In my View, I call my form's .save() method and pass along the request
user. I'd like to manipulate where this file is stored.

Here's what I tried so far but it did *not* work -- the thing just
uses the preset upload_to variable.

ImageForm.save(self, user):

# Grab Model Instance
m = super(ImageForm, self).save(commit = False)

# Try to Change "upload_to" Value
# Note: "data" is the name of my ImageField
m.data.upload_to = 'some_other_folder'

# Save the Model
m.save()

# Done
return m

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Re: Suggestions on "Loading" for possible long load-times?

[Kurtis Mullins]

Thanks a lot Russel! It sounds like a pretty reasonable approach to me.

On Tue, Nov 1, 2011 at 7:54 PM, Russell Keith-Magee  wrote:

> On Wed, Nov 2, 2011 at 7:37 AM, Kurtis  wrote:
> > Hey,
> >
> > I've got a part of my site that is just prototyped at the moment. It's
> > not particularly optimized since it's still in the works.
> >
> > I don't think it's going to take a considerable amount of time to load
> > while we have few users. But, when we start migrating our old users to
> > the new system, I can imagine there will be a delay before the
> > calculations are finished and the data is ready to display.
> >
> > Unfortunately, this is real-time calculations so it's not a simple
> > matter of caching it to death. I'm not ready to do any major
> > optimization at this point anyways since it's still a work-in-
> > progress.
> >
> > Anyways, I want to display a "loading" page while the data is
> > prepared. I don't mind if it's a matter of a redirect, done using
> > AJAX, or just using some sort of a Javascript technique that will
> > display the "Loading" dialog while the rest of the data is sent to the
> > browser. What are some suggestions on doing this with Django? I prefer
> > a simple, straight-forward method that won't cause a lot of
> > complication with the existing code base if at all possible.
>
> There's a very common pattern for this kind of activity:
>
>  1) User requests /start page to get some data that is expensive to
> calculate
>
>  2) View for /start creates a Task object that represents the work to
> be executed in the background.
>
>  3) An external process is running, waiting for new Task objects to be
> created. Exactly how this is serviced is up to you, but the important
> part is that the expensive calculation isn't performed in the view
> code -- it's handled by a process external to the web server. Celery
> is a common way to handle this sort of task; you can also knock
> something together quickly with a script running under a cron (run
> once per minute, find all new tasks, process each, then quit. Not
> ideal for production, but it works as a proof of concept).
>
>  4) Once the task has been *created* (Not finished -- just created),
> /start redirects to a /loading page.
>
>  5) /loading contains an ajax lookup that polls the completion status
> of the task
>
>  6) The task executes and completes; on completion, the task is marked
> as "complete".
>
>  7) Once the task is complete, /loading reports the completion, and
> redirects to /finished to display the results of the calculation.
>
> At every step, the web server's interaction is short lived -- create a
> task; check if the task is complete; display the result of the task --
> so it scales well. If you're careful about the design of your task
> processor, you can make sure that this processor scales too; e.g., add
> a second Celery processor for the task, and you've just doubled your
> capacity.
>
> Hope this helps!
>
> Yours,
> Russ Magee %-)
>
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>
>

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Re: Can't locate Django source code path

[Kurtis Mullins]

To grab the entire source package, follow the instructions on this page:

https://code.djangoproject.com/#Browsingthecodeonline

To grab just that single HTML file, it can be found here:

https://code.djangoproject.com/browser/django/trunk/django/contrib/admin/templates/admin

Note, that there's different versions of Django so if you are using version
1.3, then you will have to go to the 1.3 release instead of using the
"trunk". Good luck!

On Tue, Nov 1, 2011 at 10:43 PM, BillB1951  wrote:

> FRom part 2 of the tutorial I am instructed to copy a file ---
>
>
> Now copy the template admin/base_site.html from within the default
> Django admin template directory in the source code of Django itself
> (django/contrib/admin/templates) i
>
> I can't find the path.  How do I get to it?
>
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>
>

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Re: Can't locate Django source code path

[Nikolas Stevenson-Molnar]

It will be in the "site-packages" directory of your Python installation.
For Windows, this is typically C:\Python27\Lib\site-packages\django. The
location varies based on operating system. You can always open the
python interpreter and run

>>> import sys
>>> sys.path

to get a clue where your PYthon installation is (look for a path which
includes 'site-packages').

On 11/1/2011 7:43 PM, BillB1951 wrote:
> FRom part 2 of the tutorial I am instructed to copy a file ---
>
>
> Now copy the template admin/base_site.html from within the default
> Django admin template directory in the source code of Django itself
> (django/contrib/admin/templates) i
>
> I can't find the path.  How do I get to it?
>

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Can't locate Django source code path

[BillB1951]

FRom part 2 of the tutorial I am instructed to copy a file ---


Now copy the template admin/base_site.html from within the default
Django admin template directory in the source code of Django itself
(django/contrib/admin/templates) i

I can't find the path.  How do I get to it?

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Re: Django aggregation does excessive GROUP BY clauses

[Karen Tracey]

On Tue, Nov 1, 2011 at 6:19 PM, christian.oudard  wrote:

> I am doing a very simple aggregation using the Django ORM, and it is
> producing a GROUP BY clause that includes the data field, which is
> very large, and is slowing down the query by over 100-fold.
>
> Here is a simplified version of the model:
>
> class Document(models.Model):
>data = models.TextField()
>
> class Attachment(models.Model):
>document = models.ForeignKey(Document)
>
> And the query I am running:
>
> Document.objects.annotate(num_attachments=Count('attachment'))
>
>
The SQL generated by the ORM for this query changed between Django version
1.2 and 1.3. The 1.2 SQL did a group by only on the id field. With 1.3
we're getting id twice and then all other fields in the model. Bisection
shows the change was made with r14715:

https://code.djangoproject.com/changeset/14715

It certainly looks to me like the old SQL was correct and preferable for
this particular case. In a brief search I did not find a ticket reporting
this issue -- could you open one?

Karen
-- 
http://tracey.org/kmt/

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Re:Field available via Django shell but not via web application (crossposted from StackOverFlow)

[Andy McKay]

Is this using the Django built in runserver or some other way of serving
pages? If not try using runserver.

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Field available via Django shell but not via web application (crossposted from StackOverFlow)

[Jordan]

On the web page, I get the following error:

FieldError at /foo/bar/
Cannot resolve keyword 'foos' into field. Choices are: __unused__,
[snip]

The problem code is

User.objects.filter(foos__name='bar')

When I run this in the shell, it works and I get a recordset:

>>> User.objects.filter(foos__name='bar')
[]

But on the webpage I get the exception above.

Already checked:
 - INSTALLED_APPS are identical for both settings
 - runserver version also works (as would be expected)
 - User used is identical in both cases and is
`django.contrib.auth.models.User`
 - related names for the shell's User and the web app's User are
*definitely* different. User._meta.get_all_related_objects() in the
shell displays around 7 more related fields than the if I dump that
from the web app.

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Re: Suggestions on "Loading" for possible long load-times?

[Russell Keith-Magee]

On Wed, Nov 2, 2011 at 7:37 AM, Kurtis  wrote:
> Hey,
>
> I've got a part of my site that is just prototyped at the moment. It's
> not particularly optimized since it's still in the works.
>
> I don't think it's going to take a considerable amount of time to load
> while we have few users. But, when we start migrating our old users to
> the new system, I can imagine there will be a delay before the
> calculations are finished and the data is ready to display.
>
> Unfortunately, this is real-time calculations so it's not a simple
> matter of caching it to death. I'm not ready to do any major
> optimization at this point anyways since it's still a work-in-
> progress.
>
> Anyways, I want to display a "loading" page while the data is
> prepared. I don't mind if it's a matter of a redirect, done using
> AJAX, or just using some sort of a Javascript technique that will
> display the "Loading" dialog while the rest of the data is sent to the
> browser. What are some suggestions on doing this with Django? I prefer
> a simple, straight-forward method that won't cause a lot of
> complication with the existing code base if at all possible.

There's a very common pattern for this kind of activity:

 1) User requests /start page to get some data that is expensive to calculate

 2) View for /start creates a Task object that represents the work to
be executed in the background.

 3) An external process is running, waiting for new Task objects to be
created. Exactly how this is serviced is up to you, but the important
part is that the expensive calculation isn't performed in the view
code -- it's handled by a process external to the web server. Celery
is a common way to handle this sort of task; you can also knock
something together quickly with a script running under a cron (run
once per minute, find all new tasks, process each, then quit. Not
ideal for production, but it works as a proof of concept).

 4) Once the task has been *created* (Not finished -- just created),
/start redirects to a /loading page.

 5) /loading contains an ajax lookup that polls the completion status
of the task

 6) The task executes and completes; on completion, the task is marked
as "complete".

 7) Once the task is complete, /loading reports the completion, and
redirects to /finished to display the results of the calculation.

At every step, the web server's interaction is short lived -- create a
task; check if the task is complete; display the result of the task --
so it scales well. If you're careful about the design of your task
processor, you can make sure that this processor scales too; e.g., add
a second Celery processor for the task, and you've just doubled your
capacity.

Hope this helps!

Yours,
Russ Magee %-)

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Re: User data being exposed with mod_wsgi/apache

[Karen Tracey]

On Tue, Nov 1, 2011 at 4:40 PM, Jennifer Bell wrote:

>def
> __init__(self,data=None,files=None,initial={},first_update=False,user=None,
> report=None):
>   if user and user.is_authenticated() and
> UserProfile.objects.filter(user=user).exists():
>   initial[ 'author' ] = user.first_name + " " + user.last_name
>   initial[ 'phone' ] = user.get_profile().phone
>   initial[ 'email' ] = user.email
>
> super(ReportUpdateForm,self).__init__(data,files=files,initial=initial)
>
> >
>
> ... I'm guessing because the 'initial' declaration in the form
> constructor prototype is not on the stack, like I would have thought.
> Changing the view to construct the ReportUpdateForm like so:
>
>  "update_form": ReportUpdateForm(user=request.user,
> initial={}),
>

This is a classic learning Python gotcha, see for example:
http://www.ferg.org/projects/python_gotchas.html#contents_item_6

which describes the why in a fair amount of detail.

You should really replace initial={} in the __init__ definition with
initial=None and replace None with an empty dictionary inside the method
body...otherwise you are relying on all ReportUpdateForm callers to do the
right thing (that is, always specify initial, even though it's not
"required") to prevent that default initial dictionary from getting
modified.

Karen
-- 
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Suggestions on "Loading" for possible long load-times?

[Kurtis]

Hey,

I've got a part of my site that is just prototyped at the moment. It's
not particularly optimized since it's still in the works.

I don't think it's going to take a considerable amount of time to load
while we have few users. But, when we start migrating our old users to
the new system, I can imagine there will be a delay before the
calculations are finished and the data is ready to display.

Unfortunately, this is real-time calculations so it's not a simple
matter of caching it to death. I'm not ready to do any major
optimization at this point anyways since it's still a work-in-
progress.

Anyways, I want to display a "loading" page while the data is
prepared. I don't mind if it's a matter of a redirect, done using
AJAX, or just using some sort of a Javascript technique that will
display the "Loading" dialog while the rest of the data is sent to the
browser. What are some suggestions on doing this with Django? I prefer
a simple, straight-forward method that won't cause a lot of
complication with the existing code base if at all possible.

Thanks!

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Django aggregation does excessive GROUP BY clauses

[christian.oudard]

I am doing a very simple aggregation using the Django ORM, and it is
producing a GROUP BY clause that includes the data field, which is
very large, and is slowing down the query by over 100-fold.

Here is a simplified version of the model:

class Document(models.Model):
data = models.TextField()

class Attachment(models.Model):
document = models.ForeignKey(Document)

And the query I am running:

Document.objects.annotate(num_attachments=Count('attachment'))

And the SQL output:

SELECT
  `document_document`.`id`,
  `document_document`.`data`,
  COUNT(`document_attachment`.`id`) AS `num_attachments`
FROM `document_document`
  LEFT OUTER JOIN `document_attachment`
ON (`document_document`.`id` =
`document_attachment`.`document_id`)
GROUP BY
  `document_document`.`id`,
  `document_document`.`id`,
  `document_document`.`data`
ORDER BY NULL

Doing GROUP BY on the data field is unnecessary and ridiculous. I can
stop this by doing a values query:

Document.objects.values('pk').annotate(num_attachments=Count('attachment'))

But then how do I get a real, annotated Document query as the result?

Note: I cross-posted this to StackOverflow:
http://stackoverflow.com/questions/7973461/django-aggregation-does-excessive-group-by-clauses

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Re: User data being exposed with mod_wsgi/apache

[Jennifer Bell]

Well... you were right.  The problem was with my code.  As a public
service, the code below will expose the data of a logged in user for
anyone viewing the site:

BAD code >

view.py:

def show( request, report_id ):
report = get_object_or_404(Report, id=report_id)
return render_to_response("reports/show.html",
{ "report": report,
  "update_form": ReportUpdateForm(user=request.user),
context_instance=RequestContext(request))

form.py:

class ReportUpdateForm(forms.ModelForm):

class Meta:
model = ReportUpdate
fields = ('desc','author','email','phone','is_fixed')

def
__init__(self,data=None,files=None,initial={},first_update=False,user
= None, report=None):
   if user and user.is_authenticated() and
UserProfile.objects.filter(user=user).exists():
   initial[ 'author' ] = user.first_name + " " +
user.last_name
   initial[ 'phone' ] = user.get_profile().phone
   initial[ 'email' ] = user.email
   super(ReportUpdateForm,self).__init__(data,files=files,
initial=initial)


>

... I'm guessing because the 'initial' declaration in the form
constructor prototype is not on the stack, like I would have thought.
Changing the view to construct the ReportUpdateForm like so:

  "update_form": ReportUpdateForm(user=request.user,
initial={}),

puts the values on the stack, instead of in the apparently persistent
dict declared in the constructor prototype.  This was confirmed with a
unit test:

>

def test_update_form(self):
# check that default values are already filled in.
c = Client()
r = c.login(username='user1',password='user1')
url = '/reports/4'
r = c.get( url )
self.assertEquals( r.status_code, 200 )
self.assertContains(r,"Clark Kent")
self.assertContains(r,"us...@test.com")
self.assertContains(r,"555-111-")

# check that default values are NOT already filled in
# for a second anonymous client (problem in the field)

c2 = Client()
r = c2.get( url )
self.assertEquals( r.status_code, 200 )
self.assertNotContains(r,"Clark Kent")
self.assertNotContains(r,"us...@test.com")
self.assertNotContains(r,"555-111-")

--->

Which passes or fails according to the change above.   Thank you for
your advice.

Jennifer


On Oct 25, 2:36 pm, Daniel Roseman  wrote:
> On Monday, 24 October 2011 23:14:40 UTC+1, Jennifer Bell wrote:
>
> > On my site, some user data is automatically filled in to a form if a
> > user is logged in by accessing request.user in the view code.
>
> > On deployment, it seems that if *any* user is logged in, forms
> > requested via another browser will be filled in with their data.  The
> > data is not filled in if no user is logged in.
>
> > I'm mystified.  Where is this coming from?  I'm using django 1.3, and
> > caching is not enabled in my settings (though I have set
> > CACHE_MIDDLEWARE_ANONYMOUS_ONLY=True just in case).
>
> > The WSGIDeamonProcess is set up like this:
> > WSGIDaemonProcess lalala user=lalala group=lalala threads=1
> > processes=3
>
> > Is this apache?  mod_wsgi?
>
> > Jennifer
>
> No, it's your code. You've got something somewhere that's providing default
> arguments to your form, but is doing so at the module or class level rather
> than per-request. You'd better show your form and view code.
> --
> DR.

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Re: Django app, part 1, error with the database

[Nicolas]

Hi,
Thanks for your messages. The problem was indeed some missing commas
after 'ENGINE': ' ' and 'NAME': ' '
Best
-Nicolas-

On Nov 1, 6:32 pm, Sandro Dutra  wrote:
>  'NAME': 'C:/Users/songbird/Desktop/bdd/' # Or path to database file
> if using sqlite3.
>
> You're using a SQLite database, so you've to provide the full path to
> SQLite database file, for example:  'NAME':
> 'C:/Users/songbird/Desktop/bdd/mydb.db3' instead only the path to
> directory.
>
> 2011/11/1 Nikolas Stevenson-Molnar :
>
>
>
>
>
>
>
> > Hi Nicolas,
>
> > It looks like you have a syntax error in your settings file. It's
> > possible you left of a comma after one of your dictionary values, or
> > didn't close a brace. I can't tell for sure without more context: could
> > you please provide the complete "DATABASES" section of your settings file?
>
> > On 11/1/2011 9:46 AM, Nicolas wrote:
> >> Dear all,
> >> I'm completly new to the world of django and python.
> >> I'm trying to go through the tuto "Writing your first Django app, part
> >> 1" and of course I have an error.
> >> This error happens in the step "database setup". I edited the settings
> >> file, wrote 'django.db.backends.sqlite3' for engine and a path to a db
> >> file (as it's said in the tuto that "If the file doesn't exist, it
> >> will automatically be created when you synchronize the database for
> >> the first time" I just set a path to an empty folder).
> >> And then, I run manage.py syncdb and I get this :
>
> >> C:\Python27\Lib\site-packages\django\bin\mysite>manage.py syncdb
> >> Traceback (most recent call last):
> >>   File "C:\Python27\Lib\site-packages\django\bin\mysite\manage.py",
> >> line 9, in <
> >> module>
> >>     execute_from_command_line(sys.argv)
> >>   File "C:\Python27\lib\site-packages\django\core\management
> >> \__init__.py", line
> >> 422, in execute_from_command_line
> >>     utility.execute()
> >>   File "C:\Python27\lib\site-packages\django\core\management
> >> \__init__.py", line
> >> 361, in execute
> >>     self.fetch_command(subcommand).run_from_argv(self.argv)
> >>   File "C:\Python27\lib\site-packages\django\core\management
> >> \__init__.py", line
> >> 234, in fetch_command
> >>     app_name = get_commands()[subcommand]
> >>   File "C:\Python27\lib\site-packages\django\core\management
> >> \__init__.py", line
> >> 102, in get_commands
> >>     apps = settings.INSTALLED_APPS
> >>   File "C:\Python27\lib\site-packages\django\utils\functional.py",
> >> line 185, in
> >> inner
> >>     self._setup()
> >>   File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
> >> 42, in _set
> >> up
> >>     self._wrapped = Settings(settings_module)
> >>   File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
> >> 90, in __in
> >> it__
> >>     mod = importlib.import_module(self.SETTINGS_MODULE)
> >>   File "C:\Python27\lib\site-packages\django\utils\importlib.py", line
> >> 35, in im
> >> port_module
> >>     __import__(name)
> >>   File "C:\Python27\Lib\site-packages\django\bin\mysite\mysite
> >> \settings.py", lin
> >> e 15
> >>     'NAME': 'C:/Users/songbird/Desktop/bdd/' # Or path to database
> >> file if using sqlite3.
> >>           ^
> >> SyntaxError: invalid syntax
>
> >> I'm on windows 7.
> >> So, my question is simple: what am I doing wrong?
> >> Thanks a lot in advance.
> >> Best
> >> -Nicolas-
>
> > --
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Re: django model forms validation and error display

[Kurtis Mullins]

I recommend using Class Based Views if you're running Django 1.3. I tried
to go over your app and re-create a prototype using CBVs. I didn't spend a
lot of time on the actual logic or field types. I also didn't even try
running this. You'll need to fill in a couple of blanks but hopefully you
get the idea. I hope it helps :)

http://dpaste.com/hold/645547/

On Tue, Nov 1, 2011 at 11:39 AM, luke lukes  wrote:

> i have this template: http://dpaste.com/645468/ that send data to this
> view: http://dpaste.com/645469/. the template display a form made by
> modelforms, with a custom layout. i have to validate the fields value
> send to view, and then redirect to the same template if the field
> validation fails. i've tried to do it by creating an instance of the
> related model, then gathering all fields one by one (checking before
> if they are present in request.POST) and assigning it to the model
> field, and then assigning that model instance to a new modelform
> instance. then i've prefromed a .is_valid() check, but it fails since
> it require the model instance pc, so that model instance need to exist
> on db. so how can i validate those data, and if validation fails,
> redirect to another template and display the errors?
>
> --
> You received this message because you are subscribed to the Google Groups
> "Django users" group.
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>
>

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Re: Django app, part 1, error with the database

[Sandro Dutra]

 'NAME': 'C:/Users/songbird/Desktop/bdd/' # Or path to database file
if using sqlite3.

You're using a SQLite database, so you've to provide the full path to
SQLite database file, for example:  'NAME':
'C:/Users/songbird/Desktop/bdd/mydb.db3' instead only the path to
directory.

2011/11/1 Nikolas Stevenson-Molnar :
> Hi Nicolas,
>
> It looks like you have a syntax error in your settings file. It's
> possible you left of a comma after one of your dictionary values, or
> didn't close a brace. I can't tell for sure without more context: could
> you please provide the complete "DATABASES" section of your settings file?
>
> On 11/1/2011 9:46 AM, Nicolas wrote:
>> Dear all,
>> I'm completly new to the world of django and python.
>> I'm trying to go through the tuto "Writing your first Django app, part
>> 1" and of course I have an error.
>> This error happens in the step "database setup". I edited the settings
>> file, wrote 'django.db.backends.sqlite3' for engine and a path to a db
>> file (as it's said in the tuto that "If the file doesn't exist, it
>> will automatically be created when you synchronize the database for
>> the first time" I just set a path to an empty folder).
>> And then, I run manage.py syncdb and I get this :
>>
>> C:\Python27\Lib\site-packages\django\bin\mysite>manage.py syncdb
>> Traceback (most recent call last):
>>   File "C:\Python27\Lib\site-packages\django\bin\mysite\manage.py",
>> line 9, in <
>> module>
>>     execute_from_command_line(sys.argv)
>>   File "C:\Python27\lib\site-packages\django\core\management
>> \__init__.py", line
>> 422, in execute_from_command_line
>>     utility.execute()
>>   File "C:\Python27\lib\site-packages\django\core\management
>> \__init__.py", line
>> 361, in execute
>>     self.fetch_command(subcommand).run_from_argv(self.argv)
>>   File "C:\Python27\lib\site-packages\django\core\management
>> \__init__.py", line
>> 234, in fetch_command
>>     app_name = get_commands()[subcommand]
>>   File "C:\Python27\lib\site-packages\django\core\management
>> \__init__.py", line
>> 102, in get_commands
>>     apps = settings.INSTALLED_APPS
>>   File "C:\Python27\lib\site-packages\django\utils\functional.py",
>> line 185, in
>> inner
>>     self._setup()
>>   File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
>> 42, in _set
>> up
>>     self._wrapped = Settings(settings_module)
>>   File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
>> 90, in __in
>> it__
>>     mod = importlib.import_module(self.SETTINGS_MODULE)
>>   File "C:\Python27\lib\site-packages\django\utils\importlib.py", line
>> 35, in im
>> port_module
>>     __import__(name)
>>   File "C:\Python27\Lib\site-packages\django\bin\mysite\mysite
>> \settings.py", lin
>> e 15
>>     'NAME': 'C:/Users/songbird/Desktop/bdd/' # Or path to database
>> file if using sqlite3.
>>           ^
>> SyntaxError: invalid syntax
>>
>> I'm on windows 7.
>> So, my question is simple: what am I doing wrong?
>> Thanks a lot in advance.
>> Best
>> -Nicolas-
>>
>
> --
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>

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Re: Django app, part 1, error with the database

[Nikolas Stevenson-Molnar]

Hi Nicolas,

It looks like you have a syntax error in your settings file. It's
possible you left of a comma after one of your dictionary values, or
didn't close a brace. I can't tell for sure without more context: could
you please provide the complete "DATABASES" section of your settings file?

On 11/1/2011 9:46 AM, Nicolas wrote:
> Dear all,
> I'm completly new to the world of django and python.
> I'm trying to go through the tuto "Writing your first Django app, part
> 1" and of course I have an error.
> This error happens in the step "database setup". I edited the settings
> file, wrote 'django.db.backends.sqlite3' for engine and a path to a db
> file (as it's said in the tuto that "If the file doesn't exist, it
> will automatically be created when you synchronize the database for
> the first time" I just set a path to an empty folder).
> And then, I run manage.py syncdb and I get this :
>
> C:\Python27\Lib\site-packages\django\bin\mysite>manage.py syncdb
> Traceback (most recent call last):
>   File "C:\Python27\Lib\site-packages\django\bin\mysite\manage.py",
> line 9, in <
> module>
> execute_from_command_line(sys.argv)
>   File "C:\Python27\lib\site-packages\django\core\management
> \__init__.py", line
> 422, in execute_from_command_line
> utility.execute()
>   File "C:\Python27\lib\site-packages\django\core\management
> \__init__.py", line
> 361, in execute
> self.fetch_command(subcommand).run_from_argv(self.argv)
>   File "C:\Python27\lib\site-packages\django\core\management
> \__init__.py", line
> 234, in fetch_command
> app_name = get_commands()[subcommand]
>   File "C:\Python27\lib\site-packages\django\core\management
> \__init__.py", line
> 102, in get_commands
> apps = settings.INSTALLED_APPS
>   File "C:\Python27\lib\site-packages\django\utils\functional.py",
> line 185, in
> inner
> self._setup()
>   File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
> 42, in _set
> up
> self._wrapped = Settings(settings_module)
>   File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
> 90, in __in
> it__
> mod = importlib.import_module(self.SETTINGS_MODULE)
>   File "C:\Python27\lib\site-packages\django\utils\importlib.py", line
> 35, in im
> port_module
> __import__(name)
>   File "C:\Python27\Lib\site-packages\django\bin\mysite\mysite
> \settings.py", lin
> e 15
> 'NAME': 'C:/Users/songbird/Desktop/bdd/' # Or path to database
> file if using sqlite3.
>   ^
> SyntaxError: invalid syntax
>
> I'm on windows 7.
> So, my question is simple: what am I doing wrong?
> Thanks a lot in advance.
> Best
> -Nicolas-
>

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Django app, part 1, error with the database

[Nicolas]

Dear all,
I'm completly new to the world of django and python.
I'm trying to go through the tuto "Writing your first Django app, part
1" and of course I have an error.
This error happens in the step "database setup". I edited the settings
file, wrote 'django.db.backends.sqlite3' for engine and a path to a db
file (as it's said in the tuto that "If the file doesn't exist, it
will automatically be created when you synchronize the database for
the first time" I just set a path to an empty folder).
And then, I run manage.py syncdb and I get this :

C:\Python27\Lib\site-packages\django\bin\mysite>manage.py syncdb
Traceback (most recent call last):
  File "C:\Python27\Lib\site-packages\django\bin\mysite\manage.py",
line 9, in <
module>
execute_from_command_line(sys.argv)
  File "C:\Python27\lib\site-packages\django\core\management
\__init__.py", line
422, in execute_from_command_line
utility.execute()
  File "C:\Python27\lib\site-packages\django\core\management
\__init__.py", line
361, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
  File "C:\Python27\lib\site-packages\django\core\management
\__init__.py", line
234, in fetch_command
app_name = get_commands()[subcommand]
  File "C:\Python27\lib\site-packages\django\core\management
\__init__.py", line
102, in get_commands
apps = settings.INSTALLED_APPS
  File "C:\Python27\lib\site-packages\django\utils\functional.py",
line 185, in
inner
self._setup()
  File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
42, in _set
up
self._wrapped = Settings(settings_module)
  File "C:\Python27\lib\site-packages\django\conf\__init__.py", line
90, in __in
it__
mod = importlib.import_module(self.SETTINGS_MODULE)
  File "C:\Python27\lib\site-packages\django\utils\importlib.py", line
35, in im
port_module
__import__(name)
  File "C:\Python27\Lib\site-packages\django\bin\mysite\mysite
\settings.py", lin
e 15
'NAME': 'C:/Users/songbird/Desktop/bdd/' # Or path to database
file if using sqlite3.
  ^
SyntaxError: invalid syntax

I'm on windows 7.
So, my question is simple: what am I doing wrong?
Thanks a lot in advance.
Best
-Nicolas-

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django model forms validation and error display

[luke lukes]

i have this template: http://dpaste.com/645468/ that send data to this
view: http://dpaste.com/645469/. the template display a form made by
modelforms, with a custom layout. i have to validate the fields value
send to view, and then redirect to the same template if the field
validation fails. i've tried to do it by creating an instance of the
related model, then gathering all fields one by one (checking before
if they are present in request.POST) and assigning it to the model
field, and then assigning that model instance to a new modelform
instance. then i've prefromed a .is_valid() check, but it fails since
it require the model instance pc, so that model instance need to exist
on db. so how can i validate those data, and if validation fails,
redirect to another template and display the errors?

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Re: django request.POST data caching

[luke lukes]

thanks kurtis for the reply.
Anyway: solved -> fixed many error in the view (a lot of lines were
written by copy and paste)

On 30 Ott, 08:48, Kurtis Mullins  wrote:
> I apologize if this answer doesn't help much -- I'm confused by your
> question.
>
> request.POST doesn't cache anything. It's simply filled with data submitted
> by the browser on that specific request.
>
> If you click "Back" on your browser, many times the browsers will simply
> re-fill those form fields in with what you had typed in before. This is
> nothing specific to Django.
>
> I'm not aware of any HTML methods to inhibit this behavior. Keep in mind
> that it is good practise to have a redirect after every POST. For example,
> when the user submits the form, process that form, then redirect them to
> display whatever you want them to see, even if it is just additional form
> fields (within the same, or another, view).
>
> I'm not sure what you're trying to achieve by clicking on the "back" button
> -- but if you give some more information on what you're actually trying to
> achieve, I'd be happy to try give you some better help.
>
> Good luck!
>
>
>
>
>
>
>
> On Sat, Oct 29, 2011 at 10:06 AM, luke lukes  wrote:
> > hi have a template with a form and many inputs that pass some data
> > trough a POST request to a view, that process them and send the result
> > to another template. in the final template, if i use the browser back
> > button to jump to the first view, i can see again old data. i refresh
> > the page and i insert new data, i submit again but some old data
> > remain when i see the final view. the problem remain even if i restart
> > the debug server. it seems that there's some data-caching that i can
> > solve (and somethimes not) only flushing browser cache. this is the
> > view code:http://dpaste.com/640956/and the first template code:
> >http://dpaste.com/640960/. someone on stackoverflow.com that is cache
> > navigator and suggest me to use a custom middleware to disable it:
> >http://stackoverflow.com/questions/7921457/django-post-request-data-c...
> > ,
> > so i followed the advice. i also added the pragma no cache meta tag in
> > the html head section, but none of these helped. the problem remains.
> > any suggestion?
>
> > --
> > You received this message because you are subscribed to the Google Groups
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Re: Set of choices of which several can be chosen without using a ManyToManyField

[J. Cliff Dyer]

On 11/01/2011 09:05 AM, Jaroslav Dobrek wrote:

You are confusing model fields with form fields. MultipleChoiceField
is a form field, not a model field.

I wasn't aware of the existence of MultipleChoiceFields. The idea of
the above code was to express that I wanted to use this code

class Candidate(models.Model):

 programming_languages = models.CharField(max_length=50, choices=(

 (u'Python)', u'Python'),
 (u'C++', u'C++'),
 (u'Java', u'Java'),
 # ...
 ), blank=True)

with the only exception that, in the admin interface, several choices
are possible when one creates a new candidate object. I.e. I want
admins to be able to create a candidate that knows, say Python *and* C+
+ by choosing both of these languages during the creation of the
object. I used the string "MultipleChoiceField" as a dummy for
whatever should be used instead.

Jaroslav







If you want a field that will be represented by a MultipleChoiceField
in model, you simply need to define 'choices' on a field class.

https://docs.djangoproject.com/en/1.3/ref/models/fields/#choices

Cheers

Tom
Still, you want to control the input at the form level, not the model 
level.  Create a ModelForm for your Candidate, but override the language 
field to take a MultipleChoiceField, and then override the __init__() 
and save() methods to handle them properly.  "Properly," of course, is 
determined by your application, and how you want to store the 
information in the database.  You could choose to store it in a 
CharField as a comma separated list of language names, or in an 
IntegerField as a bit field (0x1 = 'Python', 0x2 = 'Perl', 0x4 = PHP, so 
0x5 means the user knows Python and PHP, but not Perl).  The latter is a 
more efficient way to store the data, but the former is arguably more 
human-friendly.


Ultimately, though, it seems that you are adding complexity rather than 
removing it.  This is exactly what many to many relationships are 
designed to handle.  If you want to use another method, you have to 
figure out the details for yourself.


Cheers,
Cliff

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Re: Set of choices of which several can be chosen without using a ManyToManyField

[Tom Evans]

On Tue, Nov 1, 2011 at 1:05 PM, Jaroslav Dobrek
 wrote:
>> You are confusing model fields with form fields. MultipleChoiceField
>> is a form field, not a model field.
>
> I wasn't aware of the existence of MultipleChoiceFields. The idea of
> the above code was to express that I wanted to use this code
>
> class Candidate(models.Model):
>
>    programming_languages = models.CharField(max_length=50, choices=(
>
>            (u'Python)', u'Python'),
>            (u'C++', u'C++'),
>            (u'Java', u'Java'),
>            # ...
>            ), blank=True)
>
> with the only exception that, in the admin interface, several choices
> are possible when one creates a new candidate object. I.e. I want
> admins to be able to create a candidate that knows, say Python *and* C+
> + by choosing both of these languages during the creation of the
> object. I used the string "MultipleChoiceField" as a dummy for
> whatever should be used instead.
>
> Jaroslav
>

That isn't possible with a single field using Django's default fields.
You want a ManyToManyField to a separate LanguageChoices model.

As an alternative, you could define a new field type that would use
MultipleChoiceField as the form field, and compresses and decompresses
from a single value. This would be a lot more work than simply
defining the additional model.

Cheers

Tom

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Re: Set of choices of which several can be chosen without using a ManyToManyField

[Jaroslav Dobrek]

> You are confusing model fields with form fields. MultipleChoiceField
> is a form field, not a model field.

I wasn't aware of the existence of MultipleChoiceFields. The idea of
the above code was to express that I wanted to use this code

class Candidate(models.Model):

programming_languages = models.CharField(max_length=50, choices=(

(u'Python)', u'Python'),
(u'C++', u'C++'),
(u'Java', u'Java'),
# ...
), blank=True)

with the only exception that, in the admin interface, several choices
are possible when one creates a new candidate object. I.e. I want
admins to be able to create a candidate that knows, say Python *and* C+
+ by choosing both of these languages during the creation of the
object. I used the string "MultipleChoiceField" as a dummy for
whatever should be used instead.

Jaroslav






> If you want a field that will be represented by a MultipleChoiceField
> in model, you simply need to define 'choices' on a field class.
>
> https://docs.djangoproject.com/en/1.3/ref/models/fields/#choices
>
> Cheers
>
> Tom

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Closing all connections to DB.

[Thomas Guettler]

Hi,

I want to close all connections to the database, because I develop a long 
running daemon process:

for connection in connections.all():
connection.close()

But the above code does not work.

select * from pg_stat_activity ; --> There Script is still "idle in transaction"


I found out, that deleting the connection objects solves the problem:

for connection in connections.all():
connection.close() # Der Server braucht keine DB-Verbindung. Ansonsten 
bleibt die ewig offen.
del(connection.connection)

Has someone seen this before?

Postgres Version 9.0.3
python-psycopg2 2.3.2

-- 
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E-Mail: guettli (*) thomas-guettler + de

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Re: Set of choices of which several can be chosen without using a ManyToManyField

[Tom Evans]

On Tue, Nov 1, 2011 at 11:07 AM, Jaroslav Dobrek
 wrote:
> Hello,
>
> what is the most straightforward way to use a set of choices of which
> several can be chosen without using a ManyToManyField?
>
> Using a ManyToManyField would make my program unnecessarily
> complicated.
>
> Example:
>
> class Candidate(models.Model):
>
>    programming_languages = models.MultipleChoiceField(max_length=50,
> choices=( # replace MultipleChoiceField with existing possibility
>
>            (u'Python)', u'Python'),
>            (u'C++', u'C++'),
>            (u'Java', u'Java'),
>            # ...
>            ), blank=True)
>
> Jaroslav
>

You are confusing model fields with form fields. MultipleChoiceField
is a form field, not a model field.

If you want a field that will be represented by a MultipleChoiceField
in model, you simply need to define 'choices' on a field class.

https://docs.djangoproject.com/en/1.3/ref/models/fields/#choices

Cheers

Tom

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Re: Disk I/O Error

[Jani Tiainen]

On 1.11.2011 8:23, Nicole Button wrote:

I'm getting a DatabaseError, looking like this:

Django Version: 1.3
Exception Type: DatabaseError
Exception Value:disk I/O error
Exception Location: /usr/local/lib/python2.6/dist-packages/django/db/
backends/sqlite3/base.py in execute, line 234

whenever I try to read or write to any of the databases associated
with any of the websites on one of our servers. The fact that all
these websites work locally, but not in their live versions seems to
suggest to me that something has gone wrong on the server, rather than
with django, but I thought I'd better ask and see if any one has had
any similar issues. I'm running python 2.6.6, django 1.3 and they're
on a Linux server.



My wild guess is that you don't have permissions to write in db 
directory... SQLite (if you're using it) requires read and write rights 
to directory where actual database resides.


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Disk I/O Error

[Nicole Button]

I'm getting a DatabaseError, looking like this:

Django Version: 1.3
Exception Type: DatabaseError
Exception Value:disk I/O error
Exception Location: /usr/local/lib/python2.6/dist-packages/django/db/
backends/sqlite3/base.py in execute, line 234

whenever I try to read or write to any of the databases associated
with any of the websites on one of our servers. The fact that all
these websites work locally, but not in their live versions seems to
suggest to me that something has gone wrong on the server, rather than
with django, but I thought I'd better ask and see if any one has had
any similar issues. I'm running python 2.6.6, django 1.3 and they're
on a Linux server.

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Set of choices of which several can be chosen without using a ManyToManyField

[Jaroslav Dobrek]

Hello,

what is the most straightforward way to use a set of choices of which
several can be chosen without using a ManyToManyField?

Using a ManyToManyField would make my program unnecessarily
complicated.

Example:

class Candidate(models.Model):

programming_languages = models.MultipleChoiceField(max_length=50,
choices=( # replace MultipleChoiceField with existing possibility

(u'Python)', u'Python'),
(u'C++', u'C++'),
(u'Java', u'Java'),
# ...
), blank=True)

Jaroslav

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Re: new formwizard - how to pass data between forms

[Kurtis Mullins]

Awesome! Let me know how it works out for you. I'll probably be giving it a
try shortly as well. Thanks for sharing the good information.

On Mon, Oct 31, 2011 at 6:43 PM, andreas  wrote:

> Hey Kurtis,
>
> thanks for your answer.
>
> I think hidden fields were used in the old implementation.
> At least the release notes for 1.4 say:
> "It features a pluggable storage API and doesn’t require the wizard to
> pass around hidden fields for every previous step."
> But checking the example apps (https://github.com/stephrdev/django-
> formwizard/blob/master/test_project/testapp2/views.py#L20) i found
> what i was looking for:
> get_cleaned_data_for_step()
>
> It does exactly what i need :-)
>
> On 31 Okt., 03:07, Kurtis Mullins  wrote:
> > I haven't read that particular back ports docs but when I looked at the
> dev docs for the new form wizard a little while back, I believe that it
> stores all previous data as hidden fields throughout the process. So I
> would think form.cleaned_data should contain everything. Hopefully that
> helps a little :)
> >
> > Sent from my iPad
> >
> > On Oct 29, 2011, at 11:19 AM, andreas  wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Hi all,
> >
> > > i am using the backport of django's new formwizard (
> https://github.com/
> > > stephrdev/django-formwizard).
> > > and i am looking for the recommended way to pass data between the
> > > different forms/steps.
> >
> > > Basically i am using the formwizard to process a series of forms that
> > > let the user (further) specify filter options from step to step.
> > > For example in the first step you would choose from a list of
> > > countries, the next form would give you the states form these
> > > countries and the final result of the wizard (done()) renders a list
> > > of the major cities in these states.
> >
> > > If i use get_form_kwargs (https://docs.djangoproject.com/en/dev/ref/
> > > contrib/formtools/form-wizard/
> > > #django.contrib.formtools.wizard.views.WizardView.get_form_kwargs) i
> > > can access the data from the previous step via self.request.POST and
> > > pass   the selected choices to the form constructor of the next form
> > > where i then filter the choices of the relevant field.
> >
> > > But is it also possible to access the selected data fron step 1 in
> > > step4?
> > > Something like get_all_cleaned_data() but just for the processed
> > > steps?
> >
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>

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