在 Django admin 中,我有一个打印按钮,可以将信息发送到模板。我想添加到 Action 下 zipper 接中,我在其中复选框选择所需的条目并在 Action 下拉列表中选择 打印。 但是,当我向 Aсtion 下拉列表添加链接时,我收到有关获取第三个参数的错误。我无法理解这是第三个参数。
admin.py
@admin.register(Salary)
class SalaryAdmin (admin.ModelAdmin):
list_display = ('worker', 'salary_uah', 'dates', 'button')
search_fields = ('worker', 'salary_uah', 'dates')
list_filter = ('worker', 'date')
actions = ['button']
def button(self, obj):
return '<a class="button" href="{}">Print</a>'.format(reverse('act', args=[obj.pk]))
button.short_description = 'Actions'
button.allow_tags = True
urls.py
urlpatterns = [
url(r'^$',home, name='home'),
url(r'^add/$',add_worker, name='add'),
url(r'^act/(?P<obj>[\w-]+)$',acts, name='act')
]
模型.py
class Salary (models.Model):
worker = models.ForeignKey(Worker)
salary_uah = models.IntegerField ('Salary')
date = models.DateTimeField('Date', default=datetime.datetime.utcnow())
views.py
def acts (request, obj):
if not request.user.is_authenticated():
return redirect('admin:login')
salary = Salary.objects.get(id=obj)
workers = Worker.objects.filter(id=salary.worker.pk).values()
salary = Salary.objects.filter(id=obj).values()
return render(request, 'zpapp/act.html', {'workers':workers, 'salary':salary })
错误信息:
TypeError at /admin/zpapp/salary/
button() takes 2 positional arguments but 3 were given
你能帮我在 Action 中添加一个链接吗?
请您参考如下方法:
我认为你走错了路。操作应该只是将由 django 自动添加到下拉列表中的文本。此处不支持 HTML 链接 AFAIK。进一步:
First, we’ll need to write a function that gets called when the action is triggered from the admin. Action functions are just regular functions that take three arguments:
The current ModelAdmin An HttpRequest representing the current request, A QuerySet containing the set of objects selected by the user. Our publish-these-articles function won’t need the ModelAdmin or the request object, but we will use the queryset:
来源:
如果要让你的 action 可用,就必须是
def button(self, request, queryset):
另一方面,如果您想要将 HTML 按钮添加到 ListView 或编辑 View ,则需要修改管理模板。
