使用 Django 进行聚合和附加值

发布于 2024-09-08 17:36:10 字数 405 浏览 2 评论 0原文

我有一个如下所示的模型：

class MyModel(models.Model)
    value = models.DecimalField()
    date = models.DatetimeField()

我正在执行此请求：

MyModel.objects.aggregate(Min("value"))

并且我得到了预期的结果：

{"mymodel__min": the_actual_minimum_value}

但是，我无法找到一种方法来同时获取最小值和关联的日期（最小值出现的日期）。

Django ORM 是否允许这样做，还是我必须使用原始 SQL ？

原文

I have a model which looks like this:

class MyModel(models.Model)
    value = models.DecimalField()
    date = models.DatetimeField()

I'm doing this request:

MyModel.objects.aggregate(Min("value"))

and I'm getting the expected result:

{"mymodel__min": the_actual_minimum_value}

However, I can't figure out a way to get at the same time the minimum value AND the associated date (the date at which the minimum value occured).

Does the Django ORM allow this, or do I have to use raw SQL ?

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浪荡不羁 2024-09-15 17:36:10

您想要做的是对查询进行注释，以便您返回通常的结果，同时还将一些数据添加到结果中。所以：

MyModel.objects.annotate(Min("value"))

将返回正常结果，并将 mymodel__min 作为附加值

回复您的评论，我认为这就是您正在寻找的？这将返回日期及其相应的最小值。

MyModel.objects.values('date').annotate(Min("value"))

编辑：在进一步回复您的评论时，您想要最低值的条目，但也希望结果中包含附加日期字段，您可以这样做：

MyModel.objects.values('date').annotate(min_value=Min('value')).order_by('min_value')[0]

这将通过对结果进行排序来获得您要求的结果字典，然后只需采用第一个索引，该索引始终是最低值。
请参阅更多

What you want to do is annotate the query, so that you get back your usual results but also have some data added to the result. So:

MyModel.objects.annotate(Min("value"))

Will return the normal result with mymodel__min as an additional value

In reply to your comment, I think this is what you are looking for? This will return the dates with their corresponding Min values.

MyModel.objects.values('date').annotate(Min("value"))

Edit: In further reply to your comment in that you want the lowest valued entry but also want the additional date field within your result, you could do something like so:

MyModel.objects.values('date').annotate(min_value=Min('value')).order_by('min_value')[0]

This will get the resulting dict you are asking for by ordering the results and then simply taking the first index which will always be the lowest value.
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