SQL 查询以查找员工的相同缺勤次数
SQL query to find the count of number of same absences for an employee
我有一个 table:
PERSON_NUMBER ABS_DATE ABS_TYPE_NAME ABS_DAYS
-----------------------------------------------------------------------------
1010 01-01-2022 PTO 1
1010 01-01-2022 PTO 1
1010 06-01-2022 PTO 0.52
1020 02-02-2022 VACATION 1
1020 02-02-2022 VACATION 0.2
1030 01-12-2021 PTO 1
1030 01-12-2021 PTO 1
1040 02-12-2021 sick 1
1040 30-12-2021 sick 1
1050 30-01-2022 SICK 1
我想在输出中添加另一列 COUNT，它告诉我一个人在同一日期有相同 ABS_TYPE 的重复数据的实例。
PERSON_NUMBER ABS_DATE ABS_TYPE_NAME ABS_DAYS COUNT
------------------------------------------------------------------------------
1010 01-01-2022 PTO 1 2
1010 01-01-2022 PTO 1
1010 06-01-2022 PTO 0.52 1
1020 02-02-2022 VACATION 1 2
1020 02-02-2022 VACATION 1
1030 01-12-2021 PTO 1 2
1030 01-12-2021 PTO 1
1040 02-12-2021 sick 1 1
1040 30-12-2021 sick 1 1
1050 30-01-2022 SICK 1 1
我正在使用-
COUNT(ABS_DATE) OVER (PARTITION BY ABS_DATE, PERSON_NUMBER, ABS_TYPE_NAME
ORDER BY PERSON_NUMBER, ABS_TYPE_NAME)
但是这将返回第一行的输出 4。它还为所有行返回。我想要这些记录的一个值。
例如-如果 2 出现在第一行，它不应该出现在第二行
当您使用分区子句执行 COUNT 时，不需要 ORDER BY，因为记录的计数顺序无关紧要。
您还可以使用第二个分析函数来确定将包含“计数”的第一行，并仅在结果中显示该行。
查询
WITH
absences (PERSON_NUMBER,
ABS_DATE,
ABS_TYPE_NAME,
ABS_DAYS)
AS
(SELECT 1010, DATE '2022-01-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1010, DATE '2022-01-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1010, DATE '2022-01-06', 'PTO', 0.52 FROM DUAL
UNION ALL
SELECT 1020, DATE '2022-02-02', 'VACATION', 1 FROM DUAL
UNION ALL
SELECT 1020, DATE '2022-02-02', 'VACATION', 0.2 FROM DUAL
UNION ALL
SELECT 1030, DATE '2021-12-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1030, DATE '2021-12-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1040, DATE '2021-12-02', 'sick', 1 FROM DUAL
UNION ALL
SELECT 1040, DATE '2021-12-30', 'sick', 1 FROM DUAL
UNION ALL
SELECT 1050, DATE '2022-01-30', 'SICK', 1 FROM DUAL)
SELECT person_number,
abs_date,
abs_type_name,
abs_days,
CASE rn WHEN 1 THEN cnt END AS COUNT
FROM (SELECT a.*,
COUNT (ABS_DATE) OVER (PARTITION BY ABS_DATE, PERSON_NUMBER, ABS_TYPE_NAME)
AS cnt,
ROW_NUMBER () OVER (PARTITION BY ABS_DATE, PERSON_NUMBER, ABS_TYPE_NAME ORDER BY 1)
AS rn
FROM absences a)
ORDER BY person_number, abs_type_name, abs_date, rn;
结果
PERSON_NUMBER ABS_DATE ABS_TYPE_NAME ABS_DAYS COUNT
________________ ____________ ________________ ___________ ________
1010 01-JAN-22 PTO 1 2
1010 01-JAN-22 PTO 1
1010 06-JAN-22 PTO 0.52 1
1020 02-FEB-22 VACATION 1 2
1020 02-FEB-22 VACATION 0.2
1030 01-DEC-21 PTO 1 2
1030 01-DEC-21 PTO 1
1040 02-DEC-21 sick 1 1
1040 30-DEC-21 sick 1 1
1050 30-JAN-22 SICK 1 1
我有一个 table:
PERSON_NUMBER ABS_DATE ABS_TYPE_NAME ABS_DAYS
-----------------------------------------------------------------------------
1010 01-01-2022 PTO 1
1010 01-01-2022 PTO 1
1010 06-01-2022 PTO 0.52
1020 02-02-2022 VACATION 1
1020 02-02-2022 VACATION 0.2
1030 01-12-2021 PTO 1
1030 01-12-2021 PTO 1
1040 02-12-2021 sick 1
1040 30-12-2021 sick 1
1050 30-01-2022 SICK 1
我想在输出中添加另一列 COUNT，它告诉我一个人在同一日期有相同 ABS_TYPE 的重复数据的实例。
PERSON_NUMBER ABS_DATE ABS_TYPE_NAME ABS_DAYS COUNT
------------------------------------------------------------------------------
1010 01-01-2022 PTO 1 2
1010 01-01-2022 PTO 1
1010 06-01-2022 PTO 0.52 1
1020 02-02-2022 VACATION 1 2
1020 02-02-2022 VACATION 1
1030 01-12-2021 PTO 1 2
1030 01-12-2021 PTO 1
1040 02-12-2021 sick 1 1
1040 30-12-2021 sick 1 1
1050 30-01-2022 SICK 1 1
我正在使用-
COUNT(ABS_DATE) OVER (PARTITION BY ABS_DATE, PERSON_NUMBER, ABS_TYPE_NAME
ORDER BY PERSON_NUMBER, ABS_TYPE_NAME)
但是这将返回第一行的输出 4。它还为所有行返回。我想要这些记录的一个值。
例如-如果 2 出现在第一行，它不应该出现在第二行
当您使用分区子句执行 COUNT 时，不需要 ORDER BY，因为记录的计数顺序无关紧要。
您还可以使用第二个分析函数来确定将包含“计数”的第一行，并仅在结果中显示该行。
查询
WITH
absences (PERSON_NUMBER,
ABS_DATE,
ABS_TYPE_NAME,
ABS_DAYS)
AS
(SELECT 1010, DATE '2022-01-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1010, DATE '2022-01-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1010, DATE '2022-01-06', 'PTO', 0.52 FROM DUAL
UNION ALL
SELECT 1020, DATE '2022-02-02', 'VACATION', 1 FROM DUAL
UNION ALL
SELECT 1020, DATE '2022-02-02', 'VACATION', 0.2 FROM DUAL
UNION ALL
SELECT 1030, DATE '2021-12-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1030, DATE '2021-12-01', 'PTO', 1 FROM DUAL
UNION ALL
SELECT 1040, DATE '2021-12-02', 'sick', 1 FROM DUAL
UNION ALL
SELECT 1040, DATE '2021-12-30', 'sick', 1 FROM DUAL
UNION ALL
SELECT 1050, DATE '2022-01-30', 'SICK', 1 FROM DUAL)
SELECT person_number,
abs_date,
abs_type_name,
abs_days,
CASE rn WHEN 1 THEN cnt END AS COUNT
FROM (SELECT a.*,
COUNT (ABS_DATE) OVER (PARTITION BY ABS_DATE, PERSON_NUMBER, ABS_TYPE_NAME)
AS cnt,
ROW_NUMBER () OVER (PARTITION BY ABS_DATE, PERSON_NUMBER, ABS_TYPE_NAME ORDER BY 1)
AS rn
FROM absences a)
ORDER BY person_number, abs_type_name, abs_date, rn;
结果
PERSON_NUMBER ABS_DATE ABS_TYPE_NAME ABS_DAYS COUNT
________________ ____________ ________________ ___________ ________
1010 01-JAN-22 PTO 1 2
1010 01-JAN-22 PTO 1
1010 06-JAN-22 PTO 0.52 1
1020 02-FEB-22 VACATION 1 2
1020 02-FEB-22 VACATION 0.2
1030 01-DEC-21 PTO 1 2
1030 01-DEC-21 PTO 1
1040 02-DEC-21 sick 1 1
1040 30-DEC-21 sick 1 1
1050 30-JAN-22 SICK 1 1