1.如何获取特殊对象的django admin url¶
你有显示每个英雄孩子的孩子列,你被要求将每个孩子的链接放到更改页面,你可以这样做:
@admin.register(Hero)
class HeroAdmin(admin.ModelAdmin, ExportCsvMixin):
...
def children_display(self, obj):
display_text = ", ".join([
"<a href={}>{}</a>".format(
reverse('admin:{}_{}_change'.format(obj._meta.app_label, obj._meta.model_name),
args=(child.pk,)),
child.name)
for child in obj.children.all()
])
if display_text:
return mark_safe(display_text)
return "-"
reverse('admin:{}_{}_change'.format(obj._meta.app_label, obj._meta.model_name), args=(child.pk,))
会给出对象的url
其他选项: 删除:
reverse('admin:{}_{}_delete'.format(obj._meta.app_label, obj._meta.model_name), args=(child.pk,))
历史记录:
reverse('admin:{}_{}_history'.format(obj._meta.app_label, obj._meta.model_name), args=(child.pk,))