使用工作时间表的动态周数计算工作时间

问题描述 投票：0回答：1

我想大致了解当前和未来几周每位员工的可用时间。 Wk1（第 1 周）是当前一周。结果会是这样的：

enter image description here

我有一张表 1，其中有可用的缺勤时间，我有一张表 2，其中包含员工的工作计划（大部分每周 40 小时，但有时周末也有工作时间）：

table 1 table 2

我现在有以下查询：

WITH WEEK AS ( 
    SELECT  TO_CHAR(TRUNC(SYSDATE, 'IW') + (level - 1) * 7, 'IW') AS week_number
    , TRUNC(SYSDATE, 'IW') + (level - 1) * 7 AS week_start
    , TRUNC(SYSDATE, 'IW') + level * 7 - 1 AS week_end
    FROM DUAL
    WHERE level <= 7
    CONNECT BY TRUNC(SYSDATE, 'IW') + (level - 1) * 7 <= SYSDATE + 49
    ) 
    
, ABSENCE AS (
    SELECT  EMP_P.EMPLOYEE_NUMBER
    , EMP_P.START_DATE AS START_DATE_ABSENCE
    , EMP_P.END_DATE AS END_DATE_ABSENCE
    , sum(TOTAL_ABSENCE_HOURS_PER_WEEK) AS ABSENCE_HOURS
    , WEEK_NUMBER
    FROM XXAS.XXAS_FHT_EMP_PERIODS_R EMP_P
    JOIN XXAS.XXAS_FHT_EMPLOYEES_ALL_MV EMP_A
      ON EMP_A.EMPLOYEE_NUMBER = EMP_P.EMPLOYEE_NUMBER
    CROSS APPLY (
        SELECT  TO_CHAR((EMP_P.START_DATE + LEVEL - 1), 'IW') AS WEEK_NUMBER
        ,(
              CASE to_number(to_char((EMP_P.START_DATE + LEVEL - 1),'D'))
              WHEN 1 THEN EMP_A.MONDAY
              WHEN 2 THEN EMP_A.TUESDAY
              WHEN 3 THEN EMP_A.WEDNESDAY
              WHEN 4 THEN EMP_A.THURSDAY
              WHEN 5 THEN EMP_A.FRIDAY
              WHEN 6 THEN EMP_A.SATURDAY
              WHEN 7 THEN EMP_A.SUNDAY
              END
            ) AS TOTAL_ABSENCE_HOURS_PER_WEEK
        FROM DUAL
        CONNECT BY EMP_P.START_DATE + LEVEL - 1 <= EMP_P.END_DATE
        )
    WHERE EMP_A.EMPLOYEE_TYPE = 'Factory'
    AND EMP_A.FUNCTION = 'Fitter'
    AND (EMP_A.EFFECTIVE_END_DATE >= SYSDATE
        OR EMP_A.EFFECTIVE_END_DATE IS NULL)
    AND EMP_P.START_DATE >= SYSDATE
        
    GROUP BY EMP_P.EMPLOYEE_NUMBER
    , WEEK_NUMBER
    , EMP_P.START_DATE
    , EMP_P.END_DATE
    
    
)

SELECT EMP_A.FULL_NAME
, EMP_A.EMPLOYEE_NUMBER
, WK.week_number
, WK.week_start
, WK.week_end 
, SUM(EMP_A.monday + EMP_A.tuesday + EMP_A.wednesday + EMP_A.thursday + EMP_A.friday + EMP_A.saturday + EMP_A.sunday) AS WORK_HOURS
, A.ABSENCE_HOURS
, NVL((SUM(EMP_A.monday + EMP_A.tuesday + EMP_A.wednesday + EMP_A.thursday + EMP_A.friday + EMP_A.saturday + EMP_A.sunday) - A.ABSENCE_HOURS)
       ,SUM(EMP_A.monday + EMP_A.tuesday + EMP_A.wednesday + EMP_A.thursday + EMP_A.friday + EMP_A.saturday + EMP_A.sunday)) AS AVAILABLE_HOURS
,
case
    when (
        NVL((SUM(EMP_A.monday + EMP_A.tuesday + EMP_A.wednesday + EMP_A.thursday + EMP_A.friday + EMP_A.saturday + EMP_A.sunday) - A.ABSENCE_HOURS)
       ,SUM(EMP_A.monday + EMP_A.tuesday + EMP_A.wednesday + EMP_A.thursday + EMP_A.friday + EMP_A.saturday + EMP_A.sunday))
        ) 
        <
        (
        SUM(EMP_A.monday + EMP_A.tuesday + EMP_A.wednesday + EMP_A.thursday + EMP_A.friday + EMP_A.saturday + EMP_A.sunday)
        ) then 'red'
    else 'green'
end as field_color
FROM xxas.XXAS_FHT_EMPLOYEES_ALL_MV EMP_A

LEFT OUTER JOIN XXAS.XXAS_FHT_EMP_PERIODS_R EMP_P
ON EMP_P.EMPLOYEE_NUMBER = EMP_A.EMPLOYEE_NUMBER
AND EMP_P.WORK_ORDER_NAME = 'Leave or absence'
AND EMP_P.END_DATE >= TRUNC(SYSDATE, 'IW')

CROSS JOIN WEEK WK

LEFT OUTER JOIN ABSENCE A
  ON A.EMPLOYEE_NUMBER = EMP_A.EMPLOYEE_NUMBER
 AND A.WEEK_NUMBER = WK.WEEK_NUMBER

WHERE EMP_A.EMPLOYEE_TYPE = 'Factory'
 AND EMP_A.FUNCTION = 'Fitter'
 AND (EMP_A.EFFECTIVE_END_DATE >= SYSDATE
      OR EMP_A.EFFECTIVE_END_DATE IS NULL
      )
      
 AND EMP_A.EMPLOYEE_NUMBER = '1000599'

GROUP BY EMP_A.EMPLOYEE_NUMBER 
, WK.WEEK_NUMBER   
, WK.week_start
, WK.week_end
, EMP_A.EMPLOYEE_NUMBER
, EMP_A.FULL_NAME
, EMP_P.START_DATE
, EMP_P.END_DATE
, A.ABSENCE_HOURS

ORDER BY WK.week_number
;

这导致：

enter image description here

我需要一个好的解决方案的帮助。我最好的猜测是为不同的周创建 7 个带有 as 的子句，并在缺勤周加入它们。但在我投入数小时的工作之前，我想知道我的思考方向是否正确。

我试图改变我已经拥有的结果。但我偶然发现你需要数据透视表中的静态数据，这样是行不通的。

要用数据创建表 1 和 2：

    CREATE TABLE employee_schedule (
  employee_number VARCHAR2(50),
  person_id NUMBER,
  first_name VARCHAR2(50),
  last_name VARCHAR2(50),
  function VARCHAR2(50),
  employee_type VARCHAR2(50),
  employment_start_date DATE,
  monday NUMBER,
  tuesday NUMBER,
  wednesday NUMBER,
  thursday NUMBER,
  friday NUMBER,
  saturday NUMBER,
  sunday NUMBER
);
INSERT INTO employee_schedule (
  employee_number, person_id, first_name, last_name, function, employee_type, employment_start_date, monday, tuesday, wednesday, thursday, friday, saturday, sunday
) VALUES (
  '1000599', 43010, 'Sead', 'Babahmetovic', 'Fitter', 'Factory', TO_DATE('01-01-2021 00:00:00', 'MM-DD-YYYY HH24:MI:SS'), 8, 8, 8, 8, 8, 0, 0
);

CREATE TABLE work_orders (
  employee_number VARCHAR2(50),
  employee_type VARCHAR2(50),
  first_name VARCHAR2(50),
  last_name VARCHAR2(50),
  work_order_name VARCHAR2(100),
  start_date DATE,
  end_date DATE
);

INSERT INTO work_orders (
  employee_number, employee_type, first_name, last_name, work_order_name, start_date, end_date
) VALUES (
  '43010', '1000599', 'Sead', 'Babahmetovic', 'Leave or absence', TO_DATE('26-04-2023 00:00:00', 'DD-MM-YYYY HH24:MI:SS'), TO_DATE('03-05-2023 00:00:00', 'DD-MM-YYYY HH24:MI:SS')
);
sql oracle date-range week-number
1个回答
0
投票

您可以使用 

LATERAL
连接和条件聚合来生成每周的小时数，然后 
PIVOT
将值作为列：

WITH weeks AS ( 
  SELECT LEVEL AS week_number,
         TRUNC(SYSDATE, 'IW') + (level - 1) * 7 AS week_start,
         TRUNC(SYSDATE, 'IW') + level * 7 - 1 AS week_end
  FROM   DUAL
  CONNECT BY level <= 7
),
worked_hours (week_number, employee_number, first_name, last_name, hours) AS (
  SELECT w.week_number,
         s.employee_number,
         s.first_name,
         s.last_name,
         s.monday    * (1 - a.monday)
         + s.tuesday   * (1 - a.tuesday)
         + s.wednesday * (1 - a.wednesday)
         + s.thursday  * (1 - a.thursday)
         + s.friday    * (1 - a.friday)
         + s.saturday  * (1 - a.saturday)
         + s.sunday    * (1 - a.sunday)
  FROM   weeks w
         CROSS JOIN employee_schedule s
         LEFT OUTER JOIN LATERAL (
           SELECT LEAST(COUNT(CASE WHEN w.week_start + 0 BETWEEN o.start_date AND o.end_date THEN 1 END), 1) AS monday,
                  LEAST(COUNT(CASE WHEN w.week_start + 1 BETWEEN o.start_date AND o.end_date THEN 1 END), 1) AS tuesday,
                  LEAST(COUNT(CASE WHEN w.week_start + 2 BETWEEN o.start_date AND o.end_date THEN 1 END), 1) AS wednesday,
                  LEAST(COUNT(CASE WHEN w.week_start + 3 BETWEEN o.start_date AND o.end_date THEN 1 END), 1) AS thursday,
                  LEAST(COUNT(CASE WHEN w.week_start + 4 BETWEEN o.start_date AND o.end_date THEN 1 END), 1) AS friday,
                  LEAST(COUNT(CASE WHEN w.week_start + 5 BETWEEN o.start_date AND o.end_date THEN 1 END), 1) AS saturday,
                  LEAST(COUNT(CASE WHEN w.week_start + 6 BETWEEN o.start_date AND o.end_date THEN 1 END), 1) AS sunday
           FROM   work_orders o
           WHERE  o.employee_number = s.employee_number
           AND    o.start_date <= w.week_end
           AND    o.end_date   >= w.week_start
        ) a
        ON (1 = 1)
)
SELECT *
FROM   worked_hours
PIVOT (
  SUM(hours) FOR week_number IN (
    1 AS wk1,
    2 AS wk2,
    3 AS wk3,
    4 AS wk4,
    5 AS wk5,
    6 AS wk6,
    7 AS wk7
  )
);

其中，对于您的样本数据：

CREATE TABLE employee_schedule (
  employee_number VARCHAR2(50),
  person_id NUMBER,
  first_name VARCHAR2(50),
  last_name VARCHAR2(50),
  function VARCHAR2(50),
  employee_type VARCHAR2(50),
  employment_start_date DATE,
  monday NUMBER,
  tuesday NUMBER,
  wednesday NUMBER,
  thursday NUMBER,
  friday NUMBER,
  saturday NUMBER,
  sunday NUMBER
);

CREATE TABLE work_orders (
  employee_number VARCHAR2(50),
  employee_type VARCHAR2(50),
  first_name VARCHAR2(50),
  last_name VARCHAR2(50),
  work_order_name VARCHAR2(100),
  start_date DATE,
  end_date DATE
);

INSERT INTO employee_schedule (
  employee_number, person_id, first_name, last_name, function, employee_type, employment_start_date, monday, tuesday, wednesday, thursday, friday, saturday, sunday
) VALUES (
  '1000599', 43010, 'Sead', 'Babahmetovic', 'Fitter', 'Factory', TO_DATE('01-01-2021 00:00:00', 'MM-DD-YYYY HH24:MI:SS'), 8, 8, 8, 8, 8, 0, 0
);

INSERT INTO work_orders (
  employee_number, employee_type, first_name, last_name, work_order_name, start_date, end_date
) VALUES (
  '1000599', '43010', 'Sead', 'Babahmetovic', 'Leave or absence', TO_DATE('26-04-2023 00:00:00', 'DD-MM-YYYY HH24:MI:SS'), TO_DATE('03-05-2023 00:00:00', 'DD-MM-YYYY HH24:MI:SS')
);

注意：您在最终

employee_number
中似乎有
employee_type
和
INSERT
的顺序错误。

输出：

EMPLOYEE_NUMBER 名字 LAST_NAME WK1 WK2 WK3 WK4 WK5 WK6 WK7
1000599 西德 巴巴梅托维奇 40 40 16 16 40 40 40

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