我正在编写一个看起来像这样的查询:
select parent.id,
parent.date,
sum(child.amount) filter (where child.is_ok) as child_sum,
sum(sum(child.amount) filter (where child.is_ok)) over (order by parent.date)
from parent
left join child on parent.id = child.parent_id
group by parent.id, parent.date
order by parent.date desc
如您所见,我正在使用一个窗口函数来获取超过 child.amount
的运行总计。
问题是,是否有任何标准或非标准的方法来引用 child_sum
而无需在窗口函数 sum
中复制其表达式?
我正在使用 Postgres 10。
最佳答案
您可以使用子查询:
SELECT id, date, child_sum,
sum(child_sum) over (order by date)
FROM (SELECT parent.id,
parent.date,
sum(child.amount) FILTER (WHERE child.is_ok) AS child_sum
FROM parent
LEFT JOIN child ON parent.id = child.parent_id
GROUP BY parent.id, parent.date
) AS subq
ORDER BY date DESC;
关于sql - 计算列上的窗口函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55413113/