覆盖 Django ModelForm 中的保存方法

在你的save分中，你必须有commit分的论点.如果有任何内容覆盖了您的表单，或者想要修改它正在保存的内容，它将执行save(commit=False)，修改输出，然后保存它自己.

此外，您的ModelForm应该返回它正在保存的模型.通常，ModelForm的save如下所示:

def save(self, commit=True):
    m = super(CallResultTypeForm, self).save(commit=False)
    # do custom stuff
    if commit:
        m.save()
    return m

读the save method.

最后，这种模型形式中的很多都不起作用，这仅仅是因为您访问事物的方式.你需要使用self.fields['callResult']而不是self.callResult.

UPDATE:针对您的回答:

Aside:为什么不在模型中使用ManyToManyField，这样就不必这样做呢？看起来您在存储冗余数据，并为您自己(和我:P)做更多的工作.

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

这允许继承CallResultTypeForm英镑，以防万一.