【发布时间】:2021-12-31 10:27:43
【问题描述】:
这是我的代码:
模型.py
class Members(models.Model):
first_name = models.CharField(max_length=100)
last_name = models.CharField(max_length=100)
username = models.CharField(max_length=100)
email = models.EmailField()
phone = models.CharField(max_length=14)
Views.py(无法返回任何值)
def employeeView(request):
member = User.username
members = Members.objects.filter(username= member)
return render(request,'members/user_profile.html', {'members': members})
Views.py(正确返回)
def employeeView(request):
members = Members.objects.filter(username= 'mahesh')
return render(request,'members/user_profile.html', {'members': members})
我试图通过传递登录用户的用户名过滤 Members.obejcts 来获取数据。它在硬编码时工作,但不是通过传递变量。有什么解决办法吗?
【问题讨论】:
-
你试过调试,看看
User.username到底是什么?通常登录用户是request.user,而不是User. I would think thatUser.username`是类属性,不是实例属性。 -
谢谢@9769953,成功了。
标签: python django variables filter hardcode