django orm queryset - 如何执行 sql 查询 MIN/MAX FILTER答案

【问题标题】：django orm queryset - how to perform sql query MIN/MAX FILTERdjango orm queryset - 如何执行 sql 查询 MIN/MAX FILTER
【发布时间】：2021-03-08 03:49:53
【问题描述】：

我正在尝试为特定的 sql 语句创建 django ORM 查询集的帮助（您也可以在此处运行它） https://dbfiddle.uk/?rdbms=postgres_13&fiddle=31c9431d6753e2fdd8df37bbc1869e88

特别是对于这个问题，我对包含以下内容的行更感兴趣：

MIN(created_at::time) FILTER (WHERE activity_type = 1) as min_time_in,
MAX(created_at::time) FILTER (WHERE activity_type = 2) as max_time_out

如果可能的话，这里是整个 sql 以转换为 django ORM 查询集

SELECT 
    created_at::date as created_date,
    company_id,
    employee_id,
    MIN(created_at::time) FILTER (WHERE activity_type = 1) as min_time_in,
    MAX(created_at::time) FILTER (WHERE activity_type = 2) as max_time_out
FROM
   timerecord
where date(created_at) = '2020-11-18' and employee_id = 1
GROUP BY created_date, company_id, employee_id
ORDER BY created_date, employee_id

【问题讨论】：

你能添加你的模型吗？ Django 模型？

标签： sql django filter orm max

【解决方案1】：

原始查询：

SELECT 
    created_at::date as created_date,
    company_id,
    employee_id,
    MIN(created_at::time) FILTER (WHERE activity_type = 1) as min_time_in,
    MAX(created_at::time) FILTER (WHERE activity_type = 2) as max_time_out
FROM
   timerecord
where date(created_at) = '2020-11-18' and employee_id = 1
GROUP BY created_date, company_id, employee_id
ORDER BY created_date, employee_id;

是conditional/selective aggregation 的一个示例。上面的语法在 PostgreSQL/SQLite 中可用。

使用 CASE 表达式可以很容易地重写为：

SELECT 
    created_at::date as created_date,
    company_id,
    employee_id,
    MIN(CASE WHEN activity_type = 1 THEN created_at::time END)as min_time_in,
    MAX(CASE WHEN activity_type = 2 THEN created_at::time END) as max_time_out
FROM
   timerecord
where date(created_at) = '2020-11-18' and employee_id = 1
GROUP BY created_date, company_id, employee_id
ORDER BY created_date, employee_id;

db<>fiddle demo

任何现代 RDBMS 都支持，Django 也支持：

conditional-aggregation

如果我们想知道每个 account_type 有多少客户呢？我们可以在聚合函数中嵌套条件表达式来实现这一点：
Client.objects.aggregate(
...     regular=Count('pk', filter=Q(account_type=Client.REGULAR)),
...     gold=Count('pk', filter=Q(account_type=Client.GOLD)),
...     platinum=Count('pk', filter=Q(account_type=Client.PLATINUM)),
... )
此聚合在支持它的数据库上使用 SQL 2003 FILTER WHERE 语法生成查询

...

在其他数据库上，这是使用 CASE 语句模拟的

【讨论】：

【解决方案2】：

您可以使用aggregate()--(doc) 方法或（annotate()--(doc) 方法）来获取结果。为此，您必须使用 Min() 和 Max() 数据库函数以及 filter 参数。

from django.db.models import Min, Q, Max

agg_response = TimeRecord.objects.aggregate(
    min_time_in=Min("created_at", filter=Q(activity_type=1)),
    max_time_out=Max("created_at", filter=Q(activity_type=2)),
)

【讨论】：