数据仓库每小时报告

Question

我在我的PostgreSQL的9.1数据库下面的表格

SELECT * from hour_dimension limit 10; 
id | date | hour 
- -+------------+------ 
1 | 2013-01-01 | 5 
2 | 2013-01-01 | 6 
3 | 2013-01-01 | 7 
4 | 2013-01-01 | 8 
5 | 2013-01-01 | 9 
6 | 2013-01-01 | 10 
7 | 2013-01-01 | 11 
8 | 2013-01-01 | 12 
9 | 2013-01-01 | 13 
10 | 2013-01-01 | 14 



SELECT 

shop_id, 
trans_date_time::date as date, 
extract(hour from trans_date_time) as hour, 
round(amount_in_cents/100.1,2) as amount 

FROM transaction 
LIMIT 10; 

shop_id | date | hour | amount 
--------+------------+------+-------- 
2877 | 2013-01-02 | 9 | 3.50 
2877 | 2013-01-02 | 10 | 4.00 
2877 | 2013-01-02 | 14 | 4.00 
2877 | 2013-01-03 | 11 | 1.40 
2877 | 2013-01-03 | 11 | 4.50 
2877 | 2013-01-03 | 12 | 3.00 
2877 | 2013-01-03 | 13 | 2.00 
2877 | 2013-01-03 | 13 | 2.00 
2877 | 2013-01-03 | 14 | 1.00 
2877 | 2013-01-04 | 9 | 4.00 


SELECT id from shop limit 3; 
id 
------ 
2877 
2878 
2879 

我想写一个数据仓库类型的查询，所以我可以生成（和存储）的每日报告描述了每个本店已以小时为单位进行的，类似以下内容：

date | hour | shop_id | amount 
-----------+------+----------+-------- 
2013-01-01 | 5 |  2877 | 0.00 
2013-01-01 | 6 |  2877 | 0.00 
2013-01-01 | 7 |  2877 | 0.00 
2013-01-01 | 8 |  2877 | 0.00 
2013-01-01 | 9 |  2877 | 3.50 
2013-01-01 | 10 |  2877 | 4.00 
2013-01-01 | 11 |  2877 | 5.90 
2013-01-01 | 12 |  2877 | 3.00 
2013-01-01 | 13 |  2877 | 4.00 
2013-01-01 | 14 |  2877 | 1.00 

查询示例：

SELECT hd.date as date, hd.hour as hour, 

shop_id, 

round(sum(case when amount is null then 0 else amount end),2) as amount 

FROM (

    SELECT 

    shop_id, 
    trans_date_time::date as date, 
    extract(hour from trans_date_time) as hour, 
    amount_in_cents/100.0 as amount 
    FROM 
    transaction 

) x 

RIGHT JOIN hour_dimension hd ON (hd.date = x.date AND hd.hour = x.hour) 

AND shop_id = 2877 
where hd.date = '2013-01-10' 

GROUP BY hd.date, hd.hour, shop_id 
ORDER by hd.date, hd.hour 
LIMIT 10; 

Answer 1

如果您可以从商店的表中选择商店ID号码，您可能会获得更好的效果。我只使用了SELECT DISTINCT子查询。交叉连接为您提供日期，小时和shop_id的所有组合。

with shop_hours as (
    select hd."date", hd."hour", tr.shop_id 
    from hour_dimension hd 
    cross join (select distinct shop_id from transaction) tr 
) 
select sh."date"::date, sh."hour", sh.shop_id, coalesce(sum(tr.amount), 0) 
from shop_hours sh 
left join transaction tr 
     on tr.trans_date_time::date = sh."date" 
     and tr.hour = sh."hour" 
     and tr.shop_id = sh.shop_id 
group by sh."date", sh."hour", sh.shop_id 
order by sh.shop_id, sh."date", sh."hour" 

Answer 2

select 
    shop_id, 
    trans_date_time::date as date, 
    extract(hour from trans_date_time) as hour, 
    round(sum(coalesce(amount_in_cents, 0))/100.0, 2) as amount 
from transaction 
group by 1, 2, 3 
order by 1, 2, 3 

Answer 3

请尝试以下查询：

SELECT hd."date", hd.hour, 
     s.shop_id, 
     sum(coalesce(round(t.amount_in_cents/100.1,2),0)) amount 
    FROM hour_dimension hd 
    CROSS JOIN (SELECT DISTINCT shop_id FROM transaction) s 
    LEFT JOIN transaction t 
    ON hd."date"=t.trans_date_time::date 
    AND hd.hour=extract(hour from t.trans_date_time) 
GROUP BY 1,2,3 
ORDER BY 1,2,3; 

同样在SQL Fiddle。

请注意，使用date作为列名/别名并不好，因为它是reserved keyword。你应该总是用双引号，但最好避免它作为列名。

hour不保留给PostgreSQL，尽管SQL标准保留它。