我有以下查询,它给出了上个月的第二个和第四个星期六以及上个月的所有星期日:
SELECT to_char(NEXT_DAY(NEXT_DAY(NEXT_DAY(NEXT_DAY(TRUNC((SELECT LAST_DAY(ADD_MONTHS(sysdate,-1)) FROM DUAL), 'MONTH') - 1, 'SATURDAY'), 'SATURDAY'),'SATURDAY'),'SATURDAY'),'YYYYMMDD') SECOND_SATURDAY
FROM DUAL
UNION ALL
SELECT to_char(NEXT_DAY(NEXT_DAY(TRUNC((SELECT LAST_DAY(ADD_MONTHS(sysdate,-1)) FROM DUAL), 'MONTH') - 1, 'SATURDAY'),'SATURDAY'),'YYYYMMDD') SECOND_SATURDAY
FROM DUAL
UNION ALL
select distinct day_date from
(SELECT to_char(NEXT_DAY(LEVEL + TRUNC((SELECT LAST_DAY(ADD_MONTHS(sysdate,-1)) FROM DUAL), 'MONTH') - 1,'SUNDAY'),'YYYYMMDD') day_date
FROM DUAL
CONNECT BY LEVEL <= ADD_MONTHS(TRUNC((SELECT LAST_DAY(ADD_MONTHS(sysdate,-1)) FROM DUAL), 'MONTH'), 1) - TRUNC((SELECT LAST_DAY(ADD_MONTHS(sysdate,-1)) FROM DUAL), 'MONTH'))
where substr(day_date,1,6) in (select to_char((SELECT LAST_DAY(ADD_MONTHS(sysdate,-1)) FROM DUAL),'YYYYMM') from dual)
但我觉得必须有一种更简单的方法来在oracle中获得相同的结果。欢迎提供这方面的帮助。我对日期格式的要求是'YYYYMMDD'。
编辑是因为我误解了这个问题:
我想你想要的东西如下:
SELECT TO_CHAR(my_date, 'YYYYMMDD') AS my_formatted_date
FROM (
SELECT NEXT_DAY(LAST_DAY(ADD_MONTHS(SYSDATE, -2))+7, 'SATURDAY') AS my_date
FROM dual
UNION ALL
SELECT NEXT_DAY(LAST_DAY(ADD_MONTHS(SYSDATE, -2))+21, 'SATURDAY')
FROM dual
UNION ALL
SELECT NEXT_DAY(LAST_DAY(ADD_MONTHS(SYSDATE, -2)), 'SUNDAY') + (LEVEL-1)*7
FROM dual
CONNECT BY NEXT_DAY(LAST_DAY(ADD_MONTHS(SYSDATE, -2)), 'SUNDAY') + (LEVEL-1)*7 < TRUNC(SYSDATE, 'MONTH')
);
在查询的第一部分中,我确定上个月的最后一天,然后在该日期之后的至少一周之后获得第一个星期六(该月的第二个星期六将是从8日到14日的任何地方)。然后我在前一个月的最后一天(第四个星期六将是从22日到28日的任何地方)至少三周后第一个星期六下降。最后,我从上个月的最后一天(即两个月前)之后的第一个星期日开始,循环上个月的星期日。您也可以使用NEXT_DAY(TRUNC(ADD_MONTHS(SYSDATE, -1) - 1), 'SUNDAY')
而不是NEXT_DAY(LAST_DAY(ADD_MONTHS(SYSDATE, -2)), 'SUNDAY')
MY_FORMA
--------
20190209
20190223
20190203
20190210
20190217
20190224
6 rows selected.
比如像这里:
with
t1 as (select add_months(trunc(sysdate, 'month'), -1) dt from dual),
t2 as (
select dt + level - 1 dt, to_char(dt + level - 1, 'dy', 'nls_date_language=english') dy
from t1 connect by dt + level - 1 < trunc(sysdate, 'month'))
select to_char(dt, 'yyyymmdd') dt, dy
from (
select dt, dy, sum(case when dy = 'sat' then 1 end) over (order by dt) sm from t2)
where dy = 'sun' or (dy = 'sat' and sm in (2, 4))
结果:
DT DY
-------- ---
20190203 sun
20190209 sat
20190210 sun
20190217 sun
20190223 sat
20190224 sun
我生成了上个月的所有日子,分配了英文日名称,有条件地计算了星期六,并且只显示了有趣的日期。
编辑:
这是有效的,但有没有办法绕过with子句,因为我的父查询是from:with x as(.... where date in(“required days”)and ...)。所以它提出了一个嵌套的情况。
是的,您可以轻松转换它,例如:
select to_char(dt, 'yyyymmdd') dt
from (
select dt, dn, rank() over (order by mod(dn, 6), dt) rnk
from (
select d + level - 1 dt, d + level - trunc(d + level - 1, 'iw') dn
from (select add_months(trunc(sysdate, 'month'), -1) d from dual)
connect by level <= add_months(d, 1) - d))
where dn = 7 or (dn = 6 and rnk in (2, 4))
with
条款使步骤更具可读性。现在我也使用日数和mod()
进行计数,只是为了显示不同的方法,但你可以使用更清楚的东西(日名,sum
或count
而不是rank
)。