[英]Django - Query by foreign key in views
我正在尝试在帖子上制作一个按钮,当用户点击它时,请求将其作为出席添加到帖子中,然后帖子的作者必须批准该请求。
模型.py
class Attending(models.Model):
is_approved = models.BooleanField(default=False)
attending = models.ManyToManyField(User, related_name='user_event_attending')
class Post(models.Model):
title = models.CharField(max_length=100)
content = models.TextField(blank=True)
date_posted = models.DateTimeField(default=timezone.now)
author = models.ForeignKey(User, on_delete=models.CASCADE)
attending = models.ForeignKey(Attending, on_delete=models.CASCADE, verbose_name='atending', null=True)
我的问题是,每次我为按钮写一个查询时都会给我错误,我无法弄清楚如何获得外键的反向。
这是我在views.py上的代码
def request_event(request, pk):
previous = request.META.get('HTTP_REFERER')
try:
query = Attending.objects.get(pk=pk)
request_attending = query.post_set.add(request.user)
messages.success(request, f'Request sent!')
return redirect(previous)
except query.DoesNotExist:
return redirect('/')
非常感谢您的提前帮助!
这: query.post_set
只是关系。 您不能像那样调用方法add
。 您可以添加到ManyToMany
关系,我相信您想将用户添加到Attending.attending
字段,而不是直接添加到Post
object。将其更改为:
...
query = Attending.objects.get(pk=pk)
query.attending.add(request.user)
messages.success(request, f'Request sent!')
....
| 更新 |
我认为你应该考虑重新安排你的关系。 如果我了解你的计划,你应该这样拨打 go:
class Attending(models.Model):
...
attendant = models.ForeignKey(User, related_name='events_attending', on_delete=models.CASCADE)
post = models.ForeignKey('Post', on_delete=models.CASCADE)
class Post(models.Model):
...
author = models.ForeignKey(User, on_delete=models.CASCADE)
对于一个Post
object 可以有多个Attending
对象,那么你可以使用这样的关系:
att = Attending.objects.first()
att.post # get related Post object from ForeignKey | there is only one
post = Post.objects.first()
post.attending_set.all() # get all related Attending objects
Post.objects.get(attending=att) # get Post object that the Attending object have in ForeignKey field
user = User.objects.first()
user.post_set.all() # get all Post objects that User is author in
user.events_attending.all() # get all related Attending objects
有关更多信息,请查看Django 文档。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.