Oracle:两个日期之间的天数和排除工作日如何处理负数
Oracle: Days between two date and Exclude weekdays how to handle negative numbers
我有两个日期列,并试图测量这两个日期之间的天数,不包括周末。我得到一个负数,需要帮助解决。
Table
CalendarDate DayNumber FirstAssgn FirstCnt DayNumber2 Id BusinessDays
5/21/2017 Sunday 5/21/17 5/21/17 Sunday 1 -1
查询:
TRUNC(TO_DATE(A.FIRST_CONTACT_DT, 'DD/MM/YYYY')) - TRUNC(TO_DATE(A.FIRST_ASSGN_DT, 'DD/MM/YYYY'))
- ((((TRUNC(A.FIRST_CONTACT_DT,'D'))-(TRUNC(A.FIRST_ASSGN_DT,'D')))/7)*2)
- (CASE WHEN TO_CHAR(A.FIRST_ASSGN_DT,'DY','nls_date_language=english') ='SUN' THEN 1 ELSE 0 END)
- (CASE WHEN TO_CHAR(A.FIRST_CONTACT_DT,'DY','nls_date_language=english')='SAT' THEN 1 ELSE 0 END)
- (SELECT COUNT(1) FROM HUM.CALENDAR CAL
WHERE 1=1
AND CAL.CALENDAR_DATE >= A.FIRST_ASSGN_DT
AND CAL.CALENDAR_DATE < A.FIRST_CONTACT_DT
--BETWEEN A.FIRST_ASSGN_DT AND A.FIRST_CONTACT_DT
AND CAL.GRH_HOLIDAY_IND = 'Y'
) AS Business_Days
看起来下面的文章需要编辑...
- (CASE WHEN TO_CHAR(A.FIRST_ASSGN_DT,'DY','nls_date_language=english')='SUN' THEN 1 ELSE 0 END)
改编自my answer here:
获取两个星期的星期一之间的天数(使用 TRUNC( datevalue, 'IW' )
作为查找星期几的独立方法 NLS_LANGUAGE
)然后加上星期几(星期一 = 1,星期二 = 2,依此类推,最多 5 个以忽略周末)作为结束日期并减去星期几作为开始日期。像这样:
SELECT ( TRUNC( end_date, 'IW' ) - TRUNC( start_date, 'IW' ) ) * 5 / 7
+ LEAST( end_date - TRUNC( end_date, 'IW' ) + 1, 5 )
- LEAST( start_date - TRUNC( start_date, 'IW' ) + 1, 5 )
AS WeekDaysDifference
FROM your_table
与 RANGE_TEMP 作为 (
SELECT
STARTPERIOD start_date,
ENDPERIOD end_date
FROM
TABLE_DATA -- YOUR TABLE WITH ALL DATA DATE
), DATE_TEMP AS (
SELECT
(start_date + LEVEL) DATE_ALL
FROM
RANGE_TEMP
CONNECT BY LEVEL <= (end_date - start_date)
), WORK_TMP 作为 (
SELECT
COUNT(DATE_ALL) WORK_DATE
FROM
DATE_TEMP
WHERE
TO_CHAR(DATE_ALL,'D', 'NLS_DATE_LANGUAGE=ENGLISH') NOT IN ('1','7')
), BUSINESS_TMP 作为 (
SELECT
COUNT(DATE_ALL) BUSINESS_DATE
FROM
DATE_TEMP
WHERE
TO_CHAR(DATE_ALL,'D', 'NLS_DATE_LANGUAGE=ENGLISH') IN ('1','7')
)
SELECT
L.WORK_DATE,
H.BUSINESS_DATE
从
BUSINESS_TMPH,
WORK_TMP大号
;
我有两个日期列,并试图测量这两个日期之间的天数,不包括周末。我得到一个负数,需要帮助解决。
Table
CalendarDate DayNumber FirstAssgn FirstCnt DayNumber2 Id BusinessDays
5/21/2017 Sunday 5/21/17 5/21/17 Sunday 1 -1
查询:
TRUNC(TO_DATE(A.FIRST_CONTACT_DT, 'DD/MM/YYYY')) - TRUNC(TO_DATE(A.FIRST_ASSGN_DT, 'DD/MM/YYYY'))
- ((((TRUNC(A.FIRST_CONTACT_DT,'D'))-(TRUNC(A.FIRST_ASSGN_DT,'D')))/7)*2)
- (CASE WHEN TO_CHAR(A.FIRST_ASSGN_DT,'DY','nls_date_language=english') ='SUN' THEN 1 ELSE 0 END)
- (CASE WHEN TO_CHAR(A.FIRST_CONTACT_DT,'DY','nls_date_language=english')='SAT' THEN 1 ELSE 0 END)
- (SELECT COUNT(1) FROM HUM.CALENDAR CAL
WHERE 1=1
AND CAL.CALENDAR_DATE >= A.FIRST_ASSGN_DT
AND CAL.CALENDAR_DATE < A.FIRST_CONTACT_DT
--BETWEEN A.FIRST_ASSGN_DT AND A.FIRST_CONTACT_DT
AND CAL.GRH_HOLIDAY_IND = 'Y'
) AS Business_Days
看起来下面的文章需要编辑...
- (CASE WHEN TO_CHAR(A.FIRST_ASSGN_DT,'DY','nls_date_language=english')='SUN' THEN 1 ELSE 0 END)
改编自my answer here:
获取两个星期的星期一之间的天数(使用 TRUNC( datevalue, 'IW' )
作为查找星期几的独立方法 NLS_LANGUAGE
)然后加上星期几(星期一 = 1,星期二 = 2,依此类推,最多 5 个以忽略周末)作为结束日期并减去星期几作为开始日期。像这样:
SELECT ( TRUNC( end_date, 'IW' ) - TRUNC( start_date, 'IW' ) ) * 5 / 7
+ LEAST( end_date - TRUNC( end_date, 'IW' ) + 1, 5 )
- LEAST( start_date - TRUNC( start_date, 'IW' ) + 1, 5 )
AS WeekDaysDifference
FROM your_table
与 RANGE_TEMP 作为 (
SELECT
STARTPERIOD start_date,
ENDPERIOD end_date
FROM
TABLE_DATA -- YOUR TABLE WITH ALL DATA DATE
), DATE_TEMP AS (
SELECT
(start_date + LEVEL) DATE_ALL
FROM
RANGE_TEMP
CONNECT BY LEVEL <= (end_date - start_date)
), WORK_TMP 作为 (
SELECT
COUNT(DATE_ALL) WORK_DATE
FROM
DATE_TEMP
WHERE
TO_CHAR(DATE_ALL,'D', 'NLS_DATE_LANGUAGE=ENGLISH') NOT IN ('1','7')
), BUSINESS_TMP 作为 (
SELECT
COUNT(DATE_ALL) BUSINESS_DATE
FROM
DATE_TEMP
WHERE
TO_CHAR(DATE_ALL,'D', 'NLS_DATE_LANGUAGE=ENGLISH') IN ('1','7')
) SELECT L.WORK_DATE, H.BUSINESS_DATE 从 BUSINESS_TMPH, WORK_TMP大号 ;