Re: modelformset_factory error: (Hidden field id) with this None already exists.
Thanks for the idea Karen,
What's the right mechanism for passing the queryset back into the
form? I don't see an example in the docs.
Thanks!
-Josh
On Dec 4, 4:10 pm, "Karen Tracey" <[EMAIL PROTECTED]> wrote:
> On Thu, Dec 4, 2008 at 2:12 AM, cyberjack <[EMAIL PROTECTED]> wrote:
>
> > Hi,
>
> > I haven't been able to solve this problem, but have made some
> > progress:
>
> > It's definitely *not* bug 9039. I've download the tests from that
> > ticket and confirmed the fix was included in 1.0.2 final.
>
> > The problem is related to using querysets with formset_factory. Ie,
> > if I include a queryset filter, then I see this error. If I don't
> > include a queryset, and thus edit all status reports, I don't see the
> > error.
>
> > Lastly, I've simplified my code producing the problem:
>
> > my model:
>
> > class StatusReport(models.Model):
> > vehicle = models.IntegerField()
> > report_date = models.DateField()
> > status = models.CharField(max_length=10)
>
> > and my view:
>
> > def detail(request, vehicle_id):
> > limited_reports = StatusReport.objects.filter(vehicle=vehicle_id)
>
> > StatusFormSet = modelformset_factory(StatusReport, extra=0)
> > if request.method == 'POST':
> > formset = StatusFormSet(request.POST)
> > if formset.is_valid():
> > formset.save()
> > else:
> > # with error
> > #formset = StatusFormSet(queryset=limited_reports)
> > # without error
> > formset = StatusFormSet()
> > return render_to_response('manage_articles.html',
> > {'formset': formset,
> > 'vehicle_id': vehicle_id})
>
> > Is this a bug in Django? If so, what should I include in the bug
> > report?
>
> I think the bug is that you need to pass the queryset parameter when you are
> dealing with a POST (if you've specified it on the GET) so that the posted
> data can be matched up to the existing model instances. Otherwise I think
> it will be trying to validate and save as all new instances, and I think the
> error message is saying you wind up with duplicate primary key ids when you
> try that (though I will say that error message might could use some
> improvement).
>
> Karen
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Re: modelformset_factory error: (Hidden field id) with this None already exists.
Hi,
I haven't been able to solve this problem, but have made some
progress:
It's definitely *not* bug 9039. I've download the tests from that
ticket and confirmed the fix was included in 1.0.2 final.
The problem is related to using querysets with formset_factory. Ie,
if I include a queryset filter, then I see this error. If I don't
include a queryset, and thus edit all status reports, I don't see the
error.
Lastly, I've simplified my code producing the problem:
my model:
class StatusReport(models.Model):
vehicle = models.IntegerField()
report_date = models.DateField()
status= models.CharField(max_length=10)
and my view:
def detail(request, vehicle_id):
limited_reports = StatusReport.objects.filter(vehicle=vehicle_id)
StatusFormSet = modelformset_factory(StatusReport, extra=0)
if request.method == 'POST':
formset = StatusFormSet(request.POST)
if formset.is_valid():
formset.save()
else:
# with error
#formset = StatusFormSet(queryset=limited_reports)
# without error
formset = StatusFormSet()
return render_to_response('manage_articles.html',
{'formset': formset,
'vehicle_id': vehicle_id})
Is this a bug in Django? If so, what should I include in the bug
report?
Thanks,
-Josh
On Dec 2, 3:49 pm, cyberjack <[EMAIL PROTECTED]> wrote:
> Hi all, I've been stumped by this error for the past few days and
> haven't made any progress. Could someone please help out?
>
> I've created a formset of seven entries, one for each day of the
> previous week. The first submission works fine; all seven entries are
> successfully submitted. Subsequent submissions, however, generate the
> following error for each of the seven entries:
>
> (Hidden field id) Status report with this None already exists.
>
> I found an django issue from a few months ago which might be related
> [1]. However, this bug was resolved back in Oct. and I'm running
> 1.0.2, so I assume the CL is included. Is there any way to be
> certain?
>
> Here are my models, simplified for space:
> class Vehicle(models.Model):
> number = models.IntegerField(unique=True)
>
> class StatusReport(models.Model):
> date = models.DateField()
> vehicle = models.ForeignKey(Vehicle, to_field='number')
> status = models.CharField
> ('status',max_length=8,choices=STATUS_REPORT_CHOICES)
>
> class Meta:
> unique_together = ("date", "vehicle")
>
> And here's my view:
>
> def GenerateReportDate(request, vehicle_id, date):
> # removed verification code for clarity
> # week_start = monday before date.
> # week_end = week_start + datetime.timedelta(7)
> for i in range(7):
> check_date = week_start + datetime.timedelta(i)
> obj, created = StatusReport.objects.get_or_create(vehicle = int
> (vehicle_id),
> date = check_date,
> defaults={'date':
> check_date,
>
> 'vehicle':vehicle[0]})
> # Pull 7 reports for this week
> reports = StatusReport.objects.filter(
> vehicle__exact=vehicle_id
> ).filter(
> date__gte = week_start
> ).filter(
> date__lte = week_end
> ).order_by("date")
>
> ReportFormSet = modelformset_factory(StatusReport, extra=0)
>
> if request.method == 'POST':
> formset = ReportFormSet(request.POST, request.FILES)
> if formset.is_valid():
> formset.save()
> return HttpResponseRedirect('../done/') # Redirect after POST
> else:
> formset = ReportFormSet(queryset=reports)
> return render_to_response('ops/report.html', {'formset': formset,
> 'number': vehicle_id,
> 'weekof': week_start})
>
> Thanks for any help,
>
> -Josh
>
> [1]http://code.djangoproject.com/ticket/9039
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modelformset_factory error: (Hidden field id) with this None already exists.
Hi all, I've been stumped by this error for the past few days and
haven't made any progress. Could someone please help out?
I've created a formset of seven entries, one for each day of the
previous week. The first submission works fine; all seven entries are
successfully submitted. Subsequent submissions, however, generate the
following error for each of the seven entries:
(Hidden field id) Status report with this None already exists.
I found an django issue from a few months ago which might be related
[1]. However, this bug was resolved back in Oct. and I'm running
1.0.2, so I assume the CL is included. Is there any way to be
certain?
Here are my models, simplified for space:
class Vehicle(models.Model):
number = models.IntegerField(unique=True)
class StatusReport(models.Model):
date = models.DateField()
vehicle = models.ForeignKey(Vehicle, to_field='number')
status= models.CharField
('status',max_length=8,choices=STATUS_REPORT_CHOICES)
class Meta:
unique_together = ("date", "vehicle")
And here's my view:
def GenerateReportDate(request, vehicle_id, date):
# removed verification code for clarity
# week_start = monday before date.
# week_end = week_start + datetime.timedelta(7)
for i in range(7):
check_date = week_start + datetime.timedelta(i)
obj, created = StatusReport.objects.get_or_create(vehicle = int
(vehicle_id),
date = check_date,
defaults={'date':
check_date,
'vehicle':vehicle[0]})
# Pull 7 reports for this week
reports = StatusReport.objects.filter(
vehicle__exact=vehicle_id
).filter(
date__gte = week_start
).filter(
date__lte = week_end
).order_by("date")
ReportFormSet = modelformset_factory(StatusReport, extra=0)
if request.method == 'POST':
formset = ReportFormSet(request.POST, request.FILES)
if formset.is_valid():
formset.save()
return HttpResponseRedirect('../done/') # Redirect after POST
else:
formset = ReportFormSet(queryset=reports)
return render_to_response('ops/report.html', {'formset': formset,
'number': vehicle_id,
'weekof': week_start})
Thanks for any help,
-Josh
[1] http://code.djangoproject.com/ticket/9039
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Re: How to return Admin to previous filters after save?
Thanks, that's good to know. -Josh --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
How to return Admin to previous filters after save?
I have an admin interface managing a few thousand records. Finding a record requires heavy use of multiple list_filter selections. However, once a user edits a record, the admin interface redirects to the root admin URL and loses the filter state. Is there an easy way to have the admin editor return to the calling URL after a form submit? Ie, I'd like the Admin tool to redirect to /admin/app/?is_active__exact=0&type__exact=A&model__exact=B instead of /admin/app/? As a work-around, I have my users bookmark common filter configurations, but it's still pretty frustrating. I feel like there must be some way to customize the Admin save function, but I don't know where to start. Thanks, -Josh --~--~-~--~~~---~--~~ 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 return Admin to previous filters after save?
> Not a solution but a workaround. I had exactly the same problem as a user, I > solved it by bookmarking the right url. I've recommended this to our users, but it's a real pain. Each user is pretty active, so there is a lot of (load-bookmark, make edit, load- bookmark, repeat) annoyance. Does anyone know how to have the admin page redirect to HTTP_REFERER? Thanks, -Josh --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
How to return Admin to previous filters after save?
I have an admin interface managing a few thousand records. Finding a record requires heavy use of multiple list_filter selections. However, once a user edits a record, the admin interface redirects to the root admin URL and loses the filter state. Is there an easy way to have the admin editor return to the calling URL after a form submit? Ie, I'd like the Admin tool to redirect to /admin/foo/?is_active__exact=0&type__exact=A&model__exact=B instead of /admin/foo/? Thanks, -Josh --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
