Question

我有两个不同的模型引用了django User，即Recruiter和Userprofile。

我有一个函数接收User个对象作为参数。我需要检查哪些相关对象存在并相应地采取一些措施。我正在使用嵌套的try-catch来执行此操作：

def some_function(user, ....):
    ...
    try:
        profile = user.userprofile
        profile.profile_pic.save('{0}_social.jpg'.format(user.username))
        profile.save()
    except:
        try:
            recruiter = user.recruiter
            recruiter.cover_pic.save('{0}_social.jpg'.format(user.username))
            recruiter.save()
        except:
            pass

有更好/更优雅的方式吗？

编辑：考虑的模型如下

class Recruiter(models.Model):
    user = models.OneToOneField(User, unique=True, related_name='recruiter')
    ...

class UserProfile(models.Model):
    user = models.OneToOneField(User, unique=True, related_name='userprofile')
    ...

Answer 1

您可以像这样使用.exists（）QuerySet API（假设user是User模型的一个实例，Userprofile和Recruiter都是该模型的关键字，其字段也名为user ）：

def some_function(user, ...):
    if Userprofile.objects.filter(user=user).exists()
        profile = user.userprofile
        profile.profile_pic.save('{0}_social.jpg'.format(user.username))
        profile.save()
    elif Recruiter.objects.filter(user=user).exists()
        recruiter = user.recruiter
        recruiter.cover_pic.save('{0}_social.jpg'.format(user.username))
        recruiter.save()
    else:
        # User has neither Userprofile nor Recruiter associated with it!
        # Do something here to handle that case, or just get rid of the else

请注意，.exists()调用将首先执行查询以查看对象是否存在，然后访问该对象的行（例如user.userprofile）将运行另一个查询以实际访问该对象。稍微优化的方式虽然可能不太可读，但可能是：

def some_function(user, ...):
    userprofiles = Userprofile.objects.filter(user=user)
    recruiters = Recruiter.objects.filter(user=user)
    if len(userprofiles):
        profile = userprofiles[0]
        profile.profile_pic.save('{0}_social.jpg'.format(user.username))
        profile.save()
    elif len(recruiters):
        recruiter = recruiters[0]
        recruiter.cover_pic.save('{0}_social.jpg'.format(user.username))
        recruiter.save()
    else:
        # User has neither Userprofile nor Recruiter associated with it!
        # Do something here to handle that case, or just get rid of the else

Django：检查哪个相关对象存在

1 个答案: