如何在oracle PL中找到連續兩天的遲來員工SQL SQL

Question

我的主表包含時鐘輸入是TAD_REGISTER_VIEW。 表格很龐大，所以我只挑選一個ID爲004634的特定員工並顯示其記錄。

my table data is 

1 004634 9/1/2017 9/1/2017 9:17:00 AM 
2 004634 9/2/2017 9/2/2017 9:17:00 AM 
3 004634 9/3/2017   
4 004634 9/4/2017 9/4/2017 9:17:00 AM 
5 004634 9/5/2017 9/5/2017 9:13:00 AM 
6 004634 9/6/2017 9/6/2017 9:17:00 AM 
7 004634 9/7/2017 9/7/2017 9:17:00 AM 
8 004634 9/8/2017 9/8/2017 9:24:00 AM 
9 004634 9/9/2017 9/9/2017 9:00:00 AM 
10 004634 9/10/2017  
11 004634 9/11/2017 9/11/2017 9:04:00 AM  
12 004634 9/12/2017 9/12/2017 9:12:00 AM  
13 004634 9/13/2017 9/13/2017 10:45:00 AM 
14 004634 9/14/2017 9/14/2017 11:12:00 AM 
15 004634 9/15/2017 9/15/2017 9:12:00 AM  

我創建這個查詢之後9:15 AM找到的所有時間較晚記錄。 15分鐘是寬限期。 15分鐘後，他認爲後來者 我的查詢是

SELECT AR.EMPLOYEEID, 
      UPPER(AR.ATTENDANCETYPE) Att, 
      AR.CLOCKIN, 
      TO_CHAR(AR.ATTENDANCEDATE, 'dd-MON-yy') Day 
     FROM TAD_REGISTER_VIEW AR, 
      EMPMASTERINFO  VEI, 
      SEC_EMPLOYEES  S, 
      TAD_CALENDARDETAIL D 
    WHERE S.USERID = 'AFZAL' 
     AND S.MISFLAG = 1 
     AND AR.ATTENDANCEDATE BETWEEN '01-Sep-17' AND '30-Sep-17' 
     AND AR.EMPLOYEEID = VEI.EMPLOYEEID 
     AND AR.APPLICATIONID IS NULL 
     AND AR.CALENDARID = D.CALENDARID 
     AND S.EMPLOYEEID = VEI.EMPLOYEEID 
     AND AR.ATTENDANCETYPE = 'P' 
     AND AR.EMPLOYEEID= 004634  
     AND TRIM(TO_CHAR(AR.ATTENDANCEDATE, 'Day')) = TRIM(D.DAY) 
     AND TO_DATE(TO_CHAR(AR.CLOCKIN, 'fmHH:MI:SS AM'), 
        'fmHH:MI:SS AM') >= 
      TO_DATE(TO_CHAR(D.CLOCK_IN + 16/1440, 'fmHH:MI:SS AM'), 
        'fmHH:MI:SS AM') 
    ORDER BY AR.PATH, AR.EMPLOYEEID, AR.ATTENDANCEDATE; 

從該查詢遲到的員工我的結果是

1 004634 P 9/1/2017 9:17:00 AM  01-SEP-17 
2 004634 P 9/2/2017 9:17:00 AM  02-SEP-17 
3 004634 P 9/4/2017 9:17:00 AM  04-SEP-17 
4 004634 P 9/6/2017 9:17:00 AM  06-SEP-17 
5 004634 P 9/7/2017 9:17:00 AM  07-SEP-17 
6 004634 P 9/8/2017 9:24:00 AM  08-SEP-17 
7 004634 P 9/13/2017 10:45:00 AM 13-SEP-17 
8 004634 P 9/14/2017 11:12:00 AM 14-SEP-17 

我想1個結果連續兩晚一天 如

2 004634 P 9/2/2017 9:17:00 AM  02-SEP-17 
5 004634 P 9/7/2017 9:17:00 AM  07-SEP-17 
7 004634 P 9/13/2017 10:45:00 AM 13-SEP-17 

Answer 1

使用LAG函數獲取前一天的日期，並將其與DATE -1進行比較。但請注意，您顯示的輸出結果中連續3天（6,7,8）連續兩天都沒有提及。即使這將從這個查詢返回。

WITH late_employees as 
     (
     SELECT AR.EMPLOYEEID, 
        UPPER(AR.ATTENDANCETYPE) Att, 
        AR.CLOCKIN CLOCKIN, 
        AR.ATTENDANCEDATE ATTENDANCEDATE 
       FROM TAD_REGISTER_VIEW AR, 
        EMPMASTERINFO  VEI, 
        SEC_EMPLOYEES  S, 
        TAD_CALENDARDETAIL D 
      WHERE S.USERID = 'AFZAL' 
       AND S.MISFLAG = 1 
       AND AR.ATTENDANCEDATE BETWEEN '01-Sep-17' AND '30-Sep-17' 
       AND AR.EMPLOYEEID = VEI.EMPLOYEEID 
       AND AR.APPLICATIONID IS NULL 
       AND AR.CALENDARID = D.CALENDARID 
       AND S.EMPLOYEEID = VEI.EMPLOYEEID 
       AND AR.ATTENDANCETYPE = 'P' 
       AND AR.EMPLOYEEID= 004634  
       AND TRIM(TO_CHAR(AR.ATTENDANCEDATE, 'Day')) = TRIM(D.DAY) 
       AND TO_DATE(TO_CHAR(AR.CLOCKIN, 'fmHH:MI:SS AM'), 
          'fmHH:MI:SS AM') >= 
        TO_DATE(TO_CHAR(D.CLOCK_IN + 16/1440, 'fmHH:MI:SS AM'), 
          'fmHH:MI:SS AM') 
          ) 
     SELECT EMPLOYEE_ID, 
    att, 
    clockin, 
    DAY 
FROM 
    (SELECT l.EMPLOYEE_ID , 
    l.att, 
    l.clockin, 
    TO_CHAR(l.ATTENDANCEDATE, 'dd-MON-yy') DAY, 
    CASE 
     WHEN TRUNC(LAG(l.ATTENDANCEDATE,1) OVER (ORDER BY l.ATTENDANCEDATE)) = TRUNC(l.ATTENDANCEDATE) - 1 
     THEN 1 
    END consec_late 
    FROM late_employees l 
) 
WHERE consec_late = 1; 

輸出：

4634 P 9/2/2017 9:17:00 AM  02-09-17  
4634 P 9/7/2017 9:17:00 AM  07-09-17  
4634 P 9/8/2017 9:24:00 AM  08-09-17  
4634 P 9/14/2017 11:12:00 AM 14-09-17