覆盖Django ModelForm中的save方法

小编典典

覆盖Django ModelForm中的save方法

django

我无法覆盖ModelForm保存方法。这是我收到的错误：

Exception Type:     TypeError  
Exception Value:    save() got an unexpected keyword argument 'commit'

我的意图是让表单为3个字段提交许多值，然后为这些字段的每个组合创建一个对象，并保存每个对象。朝正确的方向发展将是王牌。

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

阅读 940

2020-03-26

共1个答案

小编典典

在你里面save你必须有论点commit。如果有任何内容覆盖你的表单，或者想要修改其保存的内容，它将执行操作save(commit=False)，修改输出，然后将其保存。

另外，你的ModelForm应该返回它保存的模型。通常，ModelForm的save外观类似于：

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

最后，由于你访问事物的方式，许多ModelForm都无法使用。相反self.callResult，你需要使用self.fields['callResult']。

更新：针对你的回答：

旁：为什么不在ManyToManyField Model中使用s而不用这样做呢？似乎你正在存储冗余数据并为自己（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

允许继承CallResultTypeForm，以防万一。

2020-03-26