Python 如何动态更改Django管理模板
我有这样的Django管理模型Python 如何动态更改Django管理模板,python,django,Python,Django,我有这样的Django管理模型 class STUDENTAdmin(ModelAdmin): change_list_template = "students/student_change_list.html" def changelist_view(self, request, extra_context=None): # a lot of stuff happen here return TemplateResponse(request, self.change_
class STUDENTAdmin(ModelAdmin):
change_list_template = "students/student_change_list.html"
def changelist_view(self, request, extra_context=None):
# a lot of stuff happen here
return TemplateResponse(request, self.change_list_template or [
'admin/%s/%s/change_list.html' % (app_label, opts.object_name.lower()),
'admin/%s/change_list.html' % app_label,
'admin/change_list.html'
], context, current_app=self.admin_site.name)
现在,我想根据一些请求参数动态地改变它
差不多
if request.GET['foo']:
change_list_template = "students/student_change_list_other.html"
我该怎么做呢?我认为您需要覆盖
changelist\u视图
并对从中返回的templaterresponse()
执行操作,或者在调用之前更改保存该名称的变量
原来的函数是这样的
class STUDENTAdmin(ModelAdmin):
change_list_template = "students/student_change_list.html"
def changelist_view(self, request, extra_context=None):
# a lot of stuff happen here
return TemplateResponse(request, self.change_list_template or [
'admin/%s/%s/change_list.html' % (app_label, opts.object_name.lower()),
'admin/%s/change_list.html' % app_label,
'admin/change_list.html'
], context, current_app=self.admin_site.name)
所以我认为像这样的代码
def changelist(self, request, extra_context=None):
if request.GET['foo']:
self.change_list_template = "students/student_change_list_other.html"
return super(STUDENTAdmin, self).changelist_view(request, extra_context)