我有两个日期，我必须找出这两个日期之间的星期日和节假日的数量。我可以使用 BETWEEN 做到这一点吗？如果有，怎么做？
SELECT date1, date2, trunc(deposit_date - transaction_date) TOTAL
FROM Table_Name FULL OUTER JOIN Holidays ON date2 = hdate
WHERE hdate IN (date1, date2)
使用它，我绝对可以检查这两天是否有假期，即 date1 或 date2，但我无法确定这两个日期之间是否有假期或星期日。救命！
【问题讨论】：
标签： oracle date datetime oracle-sqldeveloper
您发布的解决方案效率极低；您可以在一条 SQL 语句中完成所有这些操作：
首先生成两者之间所有可能的日期：
select trunc(:min_date) + level - 1
from dual
connect by level <= trunc(:min_date) - trunc(:max_date)
然后使用您的 HOLIDAY 表来限制您想要的内容：
with all_dates as (
select trunc(:min_date) + level - 1 as the_date
from dual
connect by level <= trunc(:min_date) - trunc(:max_date)
)
select count(*)
from all_dates a
left outer join holiday b
on a.the_date = b.hdate
where b.hdate is null
and to_char(a.the_date, 'DY') <> 'SUN'
【讨论】：
如果你想检查hdate是否在两个日期之间，你可以使用
where hdate between date1 and date2
如果你想检查hdate是否与日期1或日期2在同一天，你可以这样查询
where trunc(hdate) in (trunc(date1) ,trunc(date2))
trunc 函数删除了时间。
【讨论】：
您应该创建一个包含假期的表格并自行维护它。
CREATE TABLE holidays
(
holiday VARCHAR2(100)
, d_date DATE
);
INSERT INTO holidays VALUES ('National Developer Day', DATE'2013-06-01');
SELECT *
FROM holidays;
-- National Developer Day 2013-06-01 00:00:00
剩下的只是一条 SQL 语句的问题
场景 1：存在
SELECT COUNT
(
CASE
WHEN TRIM(TO_CHAR(d.start_date_level, 'DAY')) = 'SUNDAY'
OR CASE
WHEN EXISTS (SELECT 1 FROM holidays h WHERE d.start_date_level = h.d_date)
THEN 1
ELSE NULL
END = 1
THEN 1
ELSE NULL
END
) AS holiday_check
FROM
(
SELECT start_date + (LEVEL - 1) AS start_date_level
FROM
(
SELECT start_date, end_date, end_date - start_date AS diff_date
FROM
(
SELECT TRUNC(ADD_MONTHS(SYSDATE, -2)) AS start_date
, TRUNC(SYSDATE) AS end_date
FROM DUAL
)
)
CONNECT BY
LEVEL <= (diff_date + 1)
) d
场景 2：左连接
SELECT COUNT
(
CASE
WHEN TRIM(TO_CHAR(d.start_date_level, 'DAY')) = 'SUNDAY'
OR h.d_date IS NOT NULL
THEN 1
ELSE NULL
END
) AS holiday_check
FROM
(
SELECT start_date + (LEVEL - 1) AS start_date_level
FROM
(
SELECT start_date, end_date, end_date - start_date AS diff_date
FROM
(
SELECT TRUNC(ADD_MONTHS(SYSDATE, -2)) AS start_date
, TRUNC(SYSDATE) AS end_date
FROM DUAL
)
)
CONNECT BY
LEVEL <= (diff_date + 1)
) d
LEFT JOIN holidays h
ON d.start_date_level = h.d_date
9 个星期日 + 1 个“全国开发者日”=10
【讨论】：
CREATE OR REPLACE FUNCTION workdays (dt1 DATE, dt2 DATE) RETURN NUMBER IS
weekday_count NUMBER := 0;
date1 DATE := dt1;
date2 DATE := dt2;
cur_dt date;
holiday_count number;
begin
if date1 = date2 then
return 0;
end if;
cur_dt := transaction_date;
while cur_dt <= date2 loop
if cur_dt = date2 then
null;
else
SELECT count(*) INTO holiday_count
FROM holiday
WHERE hdate = cur_dt;
IF holiday_count = 0 THEN
IF to_char(cur_dt,'DY') NOT IN ('SUN') THEN
weekday_count := weekday_count + 1;
END IF;
END IF;
END IF;
cur_dt := cur_dt +1;
END LOOP;
RETURN weekday_count;
END;
然后我查询了我的数据库并得到了正确的结果。如果您对此有最佳解决方案，请发布。
【讨论】：
这里有一个更好更有效的解决方案，
SELECT A.ID,
COUNT(A.ID) AS COUNTED
FROM tableA A
LEFT JOIN TableB B
ON A.tableB_id=B.id
LEFT JOIN holiday C
ON TRUNC(C.hdate) BETWEEN (TRUNC(a.date1) +1) AND TRUNC(B.date2)
WHERE c.hdate IS NOT NULL
GROUP BY A.ID;
其中 TableA 包含 date1，tableB 包含 date2。假期包含假期和周日的列表。并且此查询从计数中排除了“date1”。
结果逻辑
trunc(date2) - trunc(date1) = x
x - result of the query
【讨论】：
用您的假期（HDATE 列）制作一张 T$HOLIDAYS 表。这些日期将被排除在给定期间内的工作日计算之外（sdate 是开始日期，edate 是期间的结束日期）。以下是计算给定时间段内的工作日（不包括节假日、周六和周日）的函数：
CREATE OR REPLACE FUNCTION WorkingDays(sdate IN DATE,edate IN DATE) RETURN NUMBER IS
days NUMBER;
BEGIN
WITH dates AS (SELECT sdate+LEVEL-1 AS d FROM DUAL CONNECT BY LEVEL<=edate-sdate+1)
SELECT COUNT(*) INTO days
FROM dates
WHERE d NOT IN (SELECT hdate FROM t$holidays) --exclude holidays
AND TO_CHAR(d,'D') NOT IN (6,7); --exclude saturdays + sundays
RETURN days;
END WorkingDays;
/
【讨论】：
select sum(qq) from (
select case when to_number(to_char((trunc(sysdate-10) + level - 1),'D'))<=5 then 1 else 0 end as qq
from dual
connect by level <= trunc(sysdate) - trunc(sysdate-10))
【讨论】：