我正在尝试对表中的记录进行计数,并按小时对它们进行分组,我在查询中获得结果,但是我希望它每小时返回一次,即使没有记录也是如此。
我当前的查询是
SELECT nvl(count(*),0) AS transactioncount, trunc(date_modified, 'HH') as TRANSACTIONDATE
FROM TABLE
WHERE date_modified between to_date('23-JAN-19 07:00:00','dd-MON-yy hh24:mi:ss') and to_date('24-Jan-19 06:59:59','dd-MON-yy hh24:mi:ss')
group by trunc(date_modified, 'HH');
这将返回这样的结果
TRANSACTIONCOUNT | TRANSACTIONDATE
43 | 23-Jan-19 07:00:00
47 | 23-Jan-19 08:00:00
156 | 23-Jan-19 14:00:00
558 | 23-Jan-19 15:00:00
我想要的是让它在我的两个约会之间每小时返回一次,
TRANSACTIONCOUNT | TRANSACTIONDATE
43 | 23-Jan-19 07:00:00
47 | 23-Jan-19 08:00:00
0 | 23-Jan-19 09:00:00
0 | 23-Jan-19 10:00:00
0 | 23-Jan-19 11:00:00
0 | 23-Jan-19 12:00:00
0 | 23-Jan-19 13:00:00
156 | 23-Jan-19 14:00:00
558 | 23-Jan-19 15:00:00
--......
0 | 24-Jan-19 00:00:00
0 | 24-Jan-19 01:00:00
0 | 24-Jan-19 02:00:00
--and so on
答案 0 :(得分:2)
要填补交易时间内的漏洞,您首先要创建一个完整的小时表。
您可以使用Recursive Subquery Factoring来完成
WITH hour_table(TRANSACTIONDATE) AS (
SELECT to_date('23-JAN-19 07:00:00','dd-MON-yy hh24:mi:ss') /* init hour here */
FROM DUAL
UNION ALL
SELECT TRANSACTIONDATE + 1/24
FROM hour_table
WHERE TRANSACTIONDATE + 1/24 < to_date('24-JAN-19 06:59:59','dd-MON-yy hh24:mi:ss') /* limit here */
)
select * from hour_table;
TRANSACTIONDATE
-------------------
23.01.2019 07:00:00
23.01.2019 08:00:00
...
24.01.2019 05:00:00
24.01.2019 06:00:00
请注意,在此查询中使用的是开始日期和结束日期,开始日期必须精确到一个小时。
下一步很简单,就是将此小时表加入外部到您的汇总表,并使用NVL
设置缺失小时数的默认值。
with hour_table(TRANSACTIONDATE) AS (
SELECT to_date('23-JAN-19 07:00:00','dd-MON-yy hh24:mi:ss') /* init hour here */
FROM DUAL
UNION ALL
SELECT TRANSACTIONDATE + 1/24
FROM hour_table
WHERE TRANSACTIONDATE + 1/24 < to_date('24-JAN-19 06:59:59','dd-MON-yy hh24:mi:ss') /* limit */
),
agg as (
SELECT nvl(count(*),0) AS transactioncount, trunc(date_modified, 'HH') as TRANSACTIONDATE
FROM "TABLE"
WHERE date_modified between to_date('23-JAN-19 07:00:00','dd-MON-yy hh24:mi:ss') and to_date('24-Jan-19 06:59:59','dd-MON-yy hh24:mi:ss')
group by trunc(date_modified, 'HH')
)
select t.TRANSACTIONDATE, nvl(transactioncount,0) transactioncount
from hour_table t
left outer join agg a
on t.TRANSACTIONDATE = a.TRANSACTIONDATE
order by 1;
答案 1 :(得分:1)
您可以考虑将以下内容与CONNECT BY level
逻辑结合使用:
SELECT sum(transactioncount) as transactioncount, transactiondate
FROM
(
with "TABLE"(date_modified) as
(
SELECT timestamp'2019-01-23 08:00:00' FROM dual union all
SELECT timestamp'2019-01-23 08:30:00' FROM dual union all
SELECT timestamp'2019-01-23 09:00:00' FROM dual union all
SELECT timestamp'2019-01-24 05:01:00' FROM dual
)
SELECT nvl(count(*),0) AS transactioncount, trunc(date_modified, 'hh24') as transactiondate
FROM "TABLE" t
GROUP BY trunc(date_modified, 'HH24')
UNION ALL
SELECT 0, timestamp'2019-01-23 07:00:00' + ( level - 1 )/24
FROM dual
CONNECT BY level <= 24 * extract( day from
timestamp'2019-01-24 06:59:59'-
timestamp'2019-01-23 07:00:00') +
extract( hour from
timestamp'2019-01-24 06:59:59'-
timestamp'2019-01-23 07:00:00') + 1
)
GROUP BY transactiondate
ORDER BY transactiondate