我从 admin.ModelAdmin 继承了一个名为 ModelAdminWithInline 的新类,并修改了方法 add_view(...) 和 change_view(...) 来调用函数 is_cross_valid(self, form, formsets) ,您可以在其中一起验证所有表单。
这两个功能都有:
#...
if all_valid(formsets) and form_validated:
#...
变成:
#...
formsets_validated = all_valid(formsets)
cross_validated = self.is_cross_valid(form, formsets)
if formsets_validated and form_validated and cross_validated:
#...
新函数 is_cross_valid(...) 定义如下:
def is_cross_valid(self, form, formsets):
return True
因此,如果您不更改 is_cross_valid(...) 函数,新类应该与 ModelAdmin 完全相同。
现在我的 admin.py 看起来像这样:
###admin.py###
class ModelAdminWithInline(admin.ModelAdmin):
def is_cross_valid(self, form, formsets):
return True
def add_view(self, request, form_url='', extra_context=None):
#modified code
def change_view(self, request, object_id, extra_context=None):
#modified code
class ModelChildInline(admin.TabularInline):
model = ModelChild
class ModelParentAdmin(ModelAdminWithInline):
inlines = [ModelChildInline]
def is_cross_valid(self, form, formsets):
#Do some cross validation on forms
#For example, here is my particular validation:
valid = True
if hasattr(form, 'cleaned_data'):
subjects_parent = form.cleaned_data.get("subjects_parent")
#You can access forms from formsets like this:
for formset in formsets:
for formset_form in formset.forms:
if hasattr(formset_form, 'cleaned_data'):
subjects_child = formset_form.cleaned_data.get("subjects_child")
delete_form = formset_form.cleaned_data.get("DELETE")
if subjects_child and (delete_form == False):
for subject in subjects_child:
if subject in subjects_parent:
valid = False
#From here you can still report errors like in regular forms:
if "subjects_child" in formset_form.cleaned_data.keys():
formset_form._errors["subjects_child"] = ErrorList([u"Subject %s is already selected in parent ModelParent" % subject])
del formset_form.cleaned_data["subjects_child"]
else:
formset_form._errors["subjects_child"] += ErrorList(u"Subject %s is already selected in parent ModelParent" % subject])
#return True on success or False otherwise.
return valid
admin.site.register(ModelParent, ModelParentAdmin)
该解决方案有点hackish,但它有效:)。这些错误的显示与常规 ModelForm 和 ModelAdmin 类相同。 Django 1.2(应该很快就会发布)应该有模型验证,所以我希望这个问题能够得到更好的解决。