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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
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  • Just to clarify, you want to save a CallResultType for each callType, for each callResult, for each campaign? Commented May 3, 2009 at 16:03

1 Answer 1

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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 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 :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.

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4 Comments

I'm not using Many To Many fields because I need extra data on the mapping table, so I'd need to go through='' a table anyways and create the data on the intermediate table.
I don't think I can create the behaviour I want in the admin application. When I try to view the object, an error is thrown because the model fields are foreign keys not many to many relationships.
By the way, you should not forget to add the *args and **kwargs arguments to the save method as well.
For anyone coming to this now (currently Django 1.10), the ModelClass.save() method only has the commit=True argument now -- it doesn't have the force_insert or force_update arguments.

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