如何在django admin中创建链接到自定义django admin url我在django admin中有2个应用程序,我想从应用程序2中的app1链接到自定义URL
admin.py APP1
class APP1Admin(ModelAdmin):
list_display = ('xx','request_me')
def request_me(self,obj):
reverse_path = reverse("admin: APP2_TargetLink",args=(obj.pk,)) # My Problem is How to link to func APP2 target link
return '<a href="%s"> link </a>'%(reverse_path)
request_me.allow_tags =True
admin.py APP2
class APP2Admin(ModelAdmin):
def get_urls(self):
urls = super(APP2Admin, self).get_urls()
my_urls = [
url(r'(\d*)/target_link/$', self.admin_site.admin_view(self.target_link_view),name="TargetLink"),
]
return my_urls + urls
def target_link_view(self,request,id):
...
return TemplateResponse(request, template, context)
答案 0 :(得分:1)
定义网址格式时,您有name="TargetLink"
,因此可以使用
reverse("admin:TargetLink",args=(obj.pk,))
如果您希望app2
位于网址格式名称中,则必须自行添加,例如:
url(r'(\d*)/target_link/$', self.admin_site.admin_view(self.target_link_view),name="app2_TargetLink"),
然后用:
反转它reverse("admin:app2_TargetLink",args=(obj.pk,))