Overriding the save method in Django ModelForm
I'm having trouble overriding a ModelForm
save method. This is the error I'm receiving:
Exception Type: TypeError
Exception Value: save() got an unexpected keyword argument 'commit'
My intentions are to have a form submit many values for 3 fields, to then create an object for each combination of those fields, and to save each of those objects. Helpful nudge in the right direction would be ace.
File models.py
class CallResultType(models.Model):
id = models.AutoField(db_column='icontact_result_code_type_id', primary_key=True)
callResult = models.ForeignKey('CallResult', db_column='icontact_result_code_id')
campaign = models.ForeignKey('Campaign', db_column='icampaign_id')
callType = models.ForeignKey('CallType', db_column='icall_type_id')
agent = models.BooleanField(db_column='bagent', default=True)
teamLeader = models.BooleanField(db_column='bTeamLeader', default=True)
active = models.BooleanField(db_column='bactive', default=True)
File forms.py
from django.forms import ModelForm, ModelMultipleChoiceField
from callresults.models import *
class CallResultTypeForm(ModelForm):
callResult = ModelMultipleChoiceField(queryset=CallResult.objects.all())
campaign = ModelMultipleChoiceField(queryset=Campaign.objects.all())
callType = ModelMultipleChoiceField(queryset=CallType.objects.all())
def save(self, force_insert=False, force_update=False):
for cr in self.callResult:
for c in self.campain:
for ct in self.callType:
m = CallResultType(self) # this line is probably wrong
m.callResult = cr
m.campaign = c
m.calltype = ct
m.save()
class Meta:
model = CallResultType
File admin.py
class CallResultTypeAdmin(admin.ModelAdmin):
form = CallResultTypeForm
In your save
you have to have the argument commit
. If anything overrides your form, or wants to modify what it's saving, it will do save(commit=False)
, modify the output, and then save it itself.
Also, your ModelForm should return the model it's saving. Usually a ModelForm's save
will look something like:
def save(self, commit=True):
m = super(CallResultTypeForm, self).save(commit=False)
# do custom stuff
if commit:
m.save()
return m
Read up on the save
method.
Finally, a lot of this ModelForm won't work just because of the way you are accessing things. Instead of self.callResult
, you need to use self.fields['callResult']
.
UPDATE: In response to your answer:
Aside: Why not just use ManyToManyField
s in the Model so you don't have to do this? Seems like you're storing redundant data and making more work for yourself (and me :P
).
from django.db.models import AutoField
def copy_model_instance(obj):
"""
Create a copy of a model instance.
M2M relationships are currently not handled, i.e. they are not copied. (Fortunately, you don't have any in this case)
See also Django #4027. From http://blog.elsdoerfer.name/2008/09/09/making-a-copy-of-a-model-instance/
"""
initial = dict([(f.name, getattr(obj, f.name)) for f in obj._meta.fields if not isinstance(f, AutoField) and not f in obj._meta.parents.values()])
return obj.__class__(**initial)
class CallResultTypeForm(ModelForm):
callResult = ModelMultipleChoiceField(queryset=CallResult.objects.all())
campaign = ModelMultipleChoiceField(queryset=Campaign.objects.all())
callType = ModelMultipleChoiceField(queryset=CallType.objects.all())
def save(self, commit=True, *args, **kwargs):
m = super(CallResultTypeForm, self).save(commit=False, *args, **kwargs)
results = []
for cr in self.callResult:
for c in self.campain:
for ct in self.callType:
m_new = copy_model_instance(m)
m_new.callResult = cr
m_new.campaign = c
m_new.calltype = ct
if commit:
m_new.save()
results.append(m_new)
return results
This allows for inheritance of CallResultTypeForm
, just in case that's ever necessary.