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.
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)
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
class CallResultTypeAdmin(admin.ModelAdmin): form = CallResultTypeForm
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
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
UPDATE: In response to your answer:
Aside: Why not just use
ManyToManyFields 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
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.