当两个不同的用户在 Django 中向数据库插入新记录时的竞争条件

发布于 2025-01-15 00:28:14 字数 961 浏览 2 评论 0原文

当我想创建模型 Order 的新实例时,会出现竞争条件情况。
有一个 daily_id 字段,任何类别的每一天都从 1 开始。这意味着每个类别都有自己的每日 ID。

class Order(models.Model):
    daily_id = models.SmallIntegerField(default=0)
    category = models.ForeignKey(Categoty, on_delete=models.PROTECT, related_name="orders")
    declare_time = models.DateField()
    ...
}

使用这种方法计算新记录的 daily_id 字段:

def get_daily_id(category, declare_time):
    try:
        last_order = Order.objects.filter(declare_time=declare_time,
                                          category=category).latest('daily_id')
        return last_order.daily_id + 1
    except Order.DoesNotExist:
        # If no order has been registered in declare_time date.
        return 1

问题是,当两个不同的用户同时注册同一类别的订单时,订单很可能存在重复的 <代码>daily_id 值。

我已经尝试过 DRF APIView 的 post 方法的 @transaction.atomic 装饰器,但它不起作用!

There is a race condition situation, when I want to create a new instance of model Order.
There is a daily_id field that everyday for any category starts from one. It means every category has its own daily id.

class Order(models.Model):
    daily_id = models.SmallIntegerField(default=0)
    category = models.ForeignKey(Categoty, on_delete=models.PROTECT, related_name="orders")
    declare_time = models.DateField()
    ...
}

daily_id field of new record is being calculated using this method:

def get_daily_id(category, declare_time):
    try:
        last_order = Order.objects.filter(declare_time=declare_time,
                                          category=category).latest('daily_id')
        return last_order.daily_id + 1
    except Order.DoesNotExist:
        # If no order has been registered in declare_time date.
        return 1

The problem is that when two different users are registering orders in the same category at the same time, it is highly likely that the orders have the repetitive daily_id values.

I have tried @transaction.atomic decorator for post method of DRF APIView and it didn't work!

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不知所踪 2025-01-22 00:28:14

您必须使用自动增量并添加一个计算语义顺序的视图,例如:

SELECT *, ROW_NUMBER() OVER(PARTITION BY MyDayDate ORDER BY id_autoinc) AS daily_id

You must use an auto increment and add a view that computes your semantic order like :

SELECT *, ROW_NUMBER() OVER(PARTITION BY MyDayDate ORDER BY id_autoinc) AS daily_id
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