Oracle SQL 創建匯總總數

Question

我有以下測試數據，我可以使用一些幫助生成以下輸出

    YEAR 2021 
       Week 30
         1  Smith,Jane HRS 11  MIN 45  SEC 08
         2   Doe,John HRS 07  MIN 01  SEC 11

     Week Total HRS 18  MIN 46  SEC 19

     Week 31
         1   Smith,Jane HRS 13 MIN 36 SEC 01
         2   Doe,John HRS 03 MIN 27 SEC 00

    Week Total  HRS 17 MIN 03 SEC 01

    Year Total   HRS 35 MIN 49 SEC 20

我想從 emp_attendance 表中提取每個唯一的 YYYY，然后創建 YYYY（年總數）的總數。

在每個 YYYY 中（樣本只有 2 個周數 30,31，但每年最多可以有 52 個）我想創建周數總計。

在每個星期內，我想要每個員工 ID 的總數。

我已經使用以下查詢來獲取小時、分鍾、秒，但我不確定如何匯總數據以在上面的示例輸出中生成我想要的幾個總數。 我更喜歡 ALL sql 解決方案。

    select  
     e.employee_id       
     e.first_name,
     e.last_name,
     trunc(sum(a.end_date - a.start_date) * 24) hours,
   
    trunc(mod(sum(a.end_date - a.start_date) * 24 * 60,60)) minutes,

    round(mod(sum(a.end_date - a.start_date) * 24 * 60 * 60,60)) seconds

   from  employees e,
    emp_attendance a

    where a.employee_id = e.employee_id 
    group by e.employee_id, e.first_name, e.last_name

   order by e.employee_id, e.first_name,
       e.last_name;

    Emp Fname Lname  hrs  min sec
    1      Jane     Smith    25   21   9
    2      John     Doe      10   28  11  

我的測試用例。

    ALTER SESSION SET NLS_DATE_FORMAT = 'MMDDYYYY HH24:MI:SS';

    Create table employees(
     employee_id NUMBER(6), 
     first_name VARCHAR2(20),
     last_name VARCHAR2(20) 
   );

    Insert into employees (employee_id, 
     first_name,
      last_name) values
       (1, 'Jane', 'Smith');

    Insert into employees (employee_id, 
    first_name,last_name)
    values(2, 'John', 'Doe');

    CREATE TABLE  emp_attendance(    
    
    seq_num integer  GENERATED BY DEFAULT AS IDENTITY (START WITH 1) NOT NULL,
   employee_id NUMBER(6),
   start_date DATE,
   end_date DATE,
   week_number NUMBER(2),
   create_date DATE DEFAULT SYSDATE
   );


    INSERT INTO emp_attendance (employee_id, start_date, end_date,  week_number)VALUES (1, to_date('20210714 18:35:40','YYYYMMDD HH24:MI:SS'), to_date('20210715 02:34:56','YYYYMMDD HH24:MI:SS'),30);


    INSERT INTO emp_attendance (employee_id, start_date, end_date,  week_number)        VALUES (1, to_date('20210715 19:37:10','YYYYMMDD HH24:MI:SS'), to_date('20210715 23:23:02','YYYYMMDD HH24:MI:SS'),30);


    INSERT INTO emp_attendance (employee_id, c start_date, end_date,  week_number)VALUES (1, to_date('20210722 09:17:11','YYYYMMDD HH24:MI:SS'), to_date('20210722 22:53:12','YYYYMMDD HH24:MI:SS'),31);

    INSERT INTO emp_attendance (employee_id, start_date, end_date,  week_number)
     VALUES (2, to_date('20210716 18:24:11','YYYYMMDD HH24:MI:SS'), to_date('20210717 01:25:22','YYYYMMDD HH24:MI:SS'),30);

    INSERT INTO emp_attendance (employee_id, start_date, end_date,  week_number)
    VALUES (2, to_date('20210722 09:17:11','YYYYMMDD HH24:MI:SS'),      to_date('20210722 12:44:11','YYYYMMDD HH24:MI:SS'),31);

Answer 1

GROUPING SETS子句可用於創建多個聚合，而GROUPING_ID函數可用於識別每行屬於哪個聚合。

例如：

select
    to_char(a.start_date, 'YYYY') the_year,
    week_number,
    e.employee_id,
    e.first_name,
    e.last_name,
    trunc(sum(a.end_date - a.start_date) * 24) hours,
    trunc(mod(sum(a.end_date - a.start_date) * 24 * 60,60)) minutes,
    round(mod(sum(a.end_date - a.start_date) * 24 * 60 * 60,60)) seconds,
    grouping_id(to_char(a.start_date, 'YYYY'), week_number, e.employee_id, e.first_name, e.last_name) the_grouping_id
from employees e
join emp_attendance a
    on a.employee_id = e.employee_id
group by grouping sets
(
    (to_char(a.start_date, 'YYYY'), week_number, e.employee_id, e.first_name, e.last_name),
    (to_char(a.start_date, 'YYYY'), week_number),
    (to_char(a.start_date, 'YYYY')),
    ()
)
order by week_number, employee_id;

THE_YEAR   WEEK_NUMBER   EMPLOYEE_ID   FIRST_NAME   LAST_NAME   HOURS   MINUTES   SECONDS   THE_GROUPING_ID
--------   -----------   -----------   ----------   ---------   -----   -------   -------   ---------------
    2021            30             1         Jane       Smith      11        45         8                 0
    2021            30             2         John         Doe       7         1        11                 0
    2021            30                                             18        46        19                 7
    2021            31             1         Jane       Smith      13        36         1                 0
    2021            31             2         John         Doe       3        27         0                 0
    2021            31                                             17         3         1                 7
    2021                                                           35        49        20                15
                                                                   35        49        20                31

這些結果包含您想要的數字，但不是您想要的確切格式。 我假設您將使用報告工具進行格式化，但如果您想純粹在 SQL 中執行此操作，您可以使用復雜的CASE(WHEN THE_GROUPING_ID = 'X' THEN ...表達式來構建您想要的確切字符串。