Overriding the save method in Django ModelForm

Question

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

Solution

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.

Comments

[Josh Smeaton]

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.

[Josh Smeaton]

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.

[dqd]

By the way, you should not forget to add the *args and **kwargs arguments to the save method as well.

[Phil Gyford]

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.