函数计算postgresql中的总计数

Question

是否有一个函数计算像下面这样的完整月的总计数？我不确定是否postgres。我正在寻找总价值。

2012-08=# select date_trunc('day', time), count(distinct column) from table_name group by 1 order by 1; 

    date_trunc  | count 
---------------------+------- 
2012-08-01 00:00:00 | 22 
2012-08-02 00:00:00 | 34 
2012-08-03 00:00:00 | 25 
2012-08-04 00:00:00 | 30 
2012-08-05 00:00:00 | 27 
2012-08-06 00:00:00 | 31 
2012-08-07 00:00:00 | 23 
2012-08-08 00:00:00 | 28 
2012-08-09 00:00:00 | 28 
2012-08-10 00:00:00 | 28 
2012-08-11 00:00:00 | 24 
2012-08-12 00:00:00 | 36 
2012-08-13 00:00:00 | 28 
2012-08-14 00:00:00 | 23 
2012-08-15 00:00:00 | 23 
2012-08-16 00:00:00 | 30 
2012-08-17 00:00:00 | 20 
2012-08-18 00:00:00 | 30 
2012-08-19 00:00:00 | 20 
2012-08-20 00:00:00 | 24 
2012-08-21 00:00:00 | 20 
2012-08-22 00:00:00 | 17 
2012-08-23 00:00:00 | 23 
2012-08-24 00:00:00 | 25 
2012-08-25 00:00:00 | 35 
2012-08-26 00:00:00 | 18 
2012-08-27 00:00:00 | 16 
2012-08-28 00:00:00 | 11 
2012-08-29 00:00:00 | 22 
2012-08-30 00:00:00 | 26 
2012-08-31 00:00:00 | 17 
(31 rows) 
-------------------------------- 
     Total   | 12345 

Answer 1

尽我所能从您的问题和意见中猜出，您希望按月份计算不同计数的小计。你不能这样做，group by date_trunc('month',time)，因为它会做count(distinct column)这是不同的所有天。

为此，您需要一个子查询或CTE：

WITH day_counts(day,day_col_count) AS (
    select date_trunc('day', time), count(distinct column) 
    from table_name group by 1 
) 
SELECT 'Day', day, day_col_count 
FROM day_counts 
UNION ALL 
SELECT 'Month', date_trunc('month', day), sum(day_col_count) 
FROM day_counts 
GROUP BY 2 
ORDER BY 2; 

---

我的评论之前更早的猜测是：集团通过一个月？

select date_trunc('month', time), count(distinct column) 
from table_name 
group by date_trunc('month', time) 
order by time 

或者你是否试图包含运行总计或小计行？对于运行总计，您需要使用sum作为窗口函数。小计只是一种痛苦，因为SQL并不真正把自己借给他们;您需要UNION两个查询，然后将它们包装在外部ORDER BY。

Answer 2

select 
    date_trunc('day', time)::text as "date", 
    count(distinct column) as count 
from table_name 
group by 1 
union 
select 
    'Total', 
    count(distinct column) 
from table_name 
group by 1, date_trunc('month', time) 
order by "date" = 'Total', 1