Oracle SQL 统计同月的星期几
Oracle SQL counting the days of the week of the same month
如何计算用户将提供给我的 SQL 查询的特定月份的每周天数？例如，如果用户输入 2021 年 4 月，则结果将为：
如果用户输入 2021 年 5 月，结果将为：
我认为这是您要查找的查询：
SELECT WEEK AS WEEK_NUMBER, COUNT(*) AS WEEK_DAYS
FROM (SELECT TO_CHAR(FIRST_DAY + (LEVEL-1), 'IW') AS WEEK
FROM (SELECT supplied_date AS FIRST_DAY, LAST_DAY(supplied_date) - SUPPLIED_DATE+1 AS DAYS
FROM (SELECT TO_DATE('05/2021', 'MM/YYYY') AS SUPPLIED_DATE
FROM DUAL))
CONNECT BY LEVEL <= DAYS)
GROUP BY WEEK
ORDER BY WEEK;
只需将内部 TO_DATE 替换为您的日期即可。
编辑 1：附加列（查看评论）
SELECT MONTH_NAME, WEEK AS WEEK_NUMBER, COUNT(*) AS WEEK_DAYS
FROM (SELECT TO_CHAR(FIRST_DAY, 'MONTH') AS MONTH_NAME, TO_CHAR(FIRST_DAY + (LEVEL-1), 'IW') AS WEEK
FROM (SELECT supplied_date AS FIRST_DAY, TO_CHAR(LAST_DAY(supplied_date), 'DD') AS DAYS
FROM (SELECT TO_DATE('05/2021', 'MM/YYYY') AS SUPPLIED_DATE
FROM DUAL))
CONNECT BY LEVEL <= DAYS)
GROUP BY MONTH_NAME, WEEK
ORDER BY WEEK;
注意：我还更改了计算该月天数的方式。
您可以使用以下方法生成周数（无需使用任何聚合）：
WITH input ( month ) AS (
SELECT DATE '2021-04-01' FROM DUAL
),
weeks_of_month ( month_start, week_start, end_day ) AS (
SELECT TRUNC( month, 'MM' ),
TRUNC( month, 'IW' ),
LAST_DAY( TRUNC( month ) )
FROM input
UNION ALL
SELECT month_start,
week_start + INTERVAL '7' DAY,
end_day
FROM weeks_of_month
WHERE week_start + INTERVAL '7' DAY <= end_day
)
SELECT TO_CHAR( week_start, 'IW' ) AS iso_week,
GREATEST( month_start, week_start ) AS first_day_of_week,
LEAST( end_day, week_start + INTERVAL '6' DAY ) AS last_day_of_week,
LEAST( end_day + 1, week_start + INTERVAL '7' DAY )
- GREATEST( month_start, week_start ) AS days
FROM weeks_of_month;
哪些输出（NLS_DATE_FORMAT 设置为 YYYY-MM-DD (DY)）：
ISO_WEEK
FIRST_DAY_OF_WEEK
LAST_DAY_OF_WEEK
DAYS
13
2021-04-01 (THU)
2021-04-04 (SUN)
4
14
2021-04-05 (MON)
2021-04-11 (SUN)
7
15
2021-04-12 (MON)
2021-04-18 (SUN)
7
16
2021-04-19 (MON)
2021-04-25 (SUN)
7
17
2021-04-26 (MON)
2021-04-30 (FRI)
5
db<>fiddle here
这里直接计算一下。输入以字符串形式给出，例如May 2021；您可以在其位置使用绑定变量。请记住用户可能处于非英语地区的可能性；只要他们使用当地语言将月份传递给查询，一切都应该正常。
with
inputs (mth) as (select 'May 2021' from dual)
, first_day (dt) as (select to_date(mth, 'fmMonth yyyy') from inputs)
, mondays (dt, ord, lst) as (
select trunc(dt, 'iw') + 7 * (level - 1), level, max(level) over ()
from first_day
connect by level <= 1 + (trunc(add_months(dt, 1), 'iw') - trunc(dt, 'iw')) / 7
)
select to_number(to_char(dt, 'iw')) as week_number,
case ord when 1 then dt + 7 - trunc(dt + 7, 'mm')
when lst then last_day(dt) + 1 - dt
else 7 end as week_days
from mondays
order by week_number
;
WEEK_NUMBER WEEK_DAYS
----------- ----------
17 2
18 7
19 7
20 7
21 7
22 1
如何计算用户将提供给我的 SQL 查询的特定月份的每周天数？例如，如果用户输入 2021 年 4 月，则结果将为：
如果用户输入 2021 年 5 月，结果将为：
我认为这是您要查找的查询：
SELECT WEEK AS WEEK_NUMBER, COUNT(*) AS WEEK_DAYS
FROM (SELECT TO_CHAR(FIRST_DAY + (LEVEL-1), 'IW') AS WEEK
FROM (SELECT supplied_date AS FIRST_DAY, LAST_DAY(supplied_date) - SUPPLIED_DATE+1 AS DAYS
FROM (SELECT TO_DATE('05/2021', 'MM/YYYY') AS SUPPLIED_DATE
FROM DUAL))
CONNECT BY LEVEL <= DAYS)
GROUP BY WEEK
ORDER BY WEEK;
只需将内部 TO_DATE 替换为您的日期即可。
编辑 1：附加列（查看评论）
SELECT MONTH_NAME, WEEK AS WEEK_NUMBER, COUNT(*) AS WEEK_DAYS
FROM (SELECT TO_CHAR(FIRST_DAY, 'MONTH') AS MONTH_NAME, TO_CHAR(FIRST_DAY + (LEVEL-1), 'IW') AS WEEK
FROM (SELECT supplied_date AS FIRST_DAY, TO_CHAR(LAST_DAY(supplied_date), 'DD') AS DAYS
FROM (SELECT TO_DATE('05/2021', 'MM/YYYY') AS SUPPLIED_DATE
FROM DUAL))
CONNECT BY LEVEL <= DAYS)
GROUP BY MONTH_NAME, WEEK
ORDER BY WEEK;
注意：我还更改了计算该月天数的方式。
您可以使用以下方法生成周数（无需使用任何聚合）：
WITH input ( month ) AS (
SELECT DATE '2021-04-01' FROM DUAL
),
weeks_of_month ( month_start, week_start, end_day ) AS (
SELECT TRUNC( month, 'MM' ),
TRUNC( month, 'IW' ),
LAST_DAY( TRUNC( month ) )
FROM input
UNION ALL
SELECT month_start,
week_start + INTERVAL '7' DAY,
end_day
FROM weeks_of_month
WHERE week_start + INTERVAL '7' DAY <= end_day
)
SELECT TO_CHAR( week_start, 'IW' ) AS iso_week,
GREATEST( month_start, week_start ) AS first_day_of_week,
LEAST( end_day, week_start + INTERVAL '6' DAY ) AS last_day_of_week,
LEAST( end_day + 1, week_start + INTERVAL '7' DAY )
- GREATEST( month_start, week_start ) AS days
FROM weeks_of_month;
哪些输出（NLS_DATE_FORMAT 设置为 YYYY-MM-DD (DY)）：
ISO_WEEK FIRST_DAY_OF_WEEK LAST_DAY_OF_WEEK DAYS 13 2021-04-01 (THU) 2021-04-04 (SUN) 4 14 2021-04-05 (MON) 2021-04-11 (SUN) 7 15 2021-04-12 (MON) 2021-04-18 (SUN) 7 16 2021-04-19 (MON) 2021-04-25 (SUN) 7 17 2021-04-26 (MON) 2021-04-30 (FRI) 5
db<>fiddle here
这里直接计算一下。输入以字符串形式给出，例如May 2021；您可以在其位置使用绑定变量。请记住用户可能处于非英语地区的可能性；只要他们使用当地语言将月份传递给查询，一切都应该正常。
with
inputs (mth) as (select 'May 2021' from dual)
, first_day (dt) as (select to_date(mth, 'fmMonth yyyy') from inputs)
, mondays (dt, ord, lst) as (
select trunc(dt, 'iw') + 7 * (level - 1), level, max(level) over ()
from first_day
connect by level <= 1 + (trunc(add_months(dt, 1), 'iw') - trunc(dt, 'iw')) / 7
)
select to_number(to_char(dt, 'iw')) as week_number,
case ord when 1 then dt + 7 - trunc(dt + 7, 'mm')
when lst then last_day(dt) + 1 - dt
else 7 end as week_days
from mondays
order by week_number
;
WEEK_NUMBER WEEK_DAYS
----------- ----------
17 2
18 7
19 7
20 7
21 7
22 1