Question

我有以下数据，需要以下快速帮助。（提供EMP表示例，由于合规性问题我无法提供真实数据）：

SELECT EMPNO,
       JOB,
       HIREDATE,
       HIREDATE AS STARTDATE,
       LEAD(HIREDATE) OVER (ORDER BY HIREDATE) AS ENDDATE
  FROM SCOTT.EMP
 ORDER BY HIREDATE;

EMPNO   JOB        HIREDATE   STARTDATE  ENDDATE  
--------------------------------------------------
7369    CLERK      17-Dec-80  17-Dec-80  20-Feb-81  
7499    SALESMAN   20-Feb-81  20-Feb-81  22-Feb-81  
7521    SALESMAN   22-Feb-81  22-Feb-81  2-Apr-81  
7566    MANAGER    2-Apr-81   2-Apr-81   1-May-81  
7698    MANAGER    1-May-81   1-May-81   9-Jun-81  
7782    MANAGER    9-Jun-81   9-Jun-81   8-Sep-81  
7844    SALESMAN   8-Sep-81   8-Sep-81   28-Sep-81  
7654    SALESMAN   28-Sep-81  28-Sep-81  17-Nov-81  
7839    PRESIDENT  17-Nov-81  17-Nov-81  3-Dec-81  
7900    CLERK      3-Dec-81   3-Dec-81   3-Dec-81  
7902    ANALYST    3-Dec-81   3-Dec-81   23-Jan-82  
7934    CLERK      23-Jan-82  23-Jan-82  19-Apr-87  
7788    ANALYST    19-Apr-87  19-Apr-87  23-May-87  
7876    CLERK      23-May-87  23-May-87  NULL

这里我想在oracle SQL中按作业分区，这样连续的行获得相同的派生开始日期（DERSTARTDATE）和派生结束日期（DERENDDATE），如下所示。您能否在Oracle中使用SQL提供建议。

我想要这样的事情：

EMPNO   JOB       HIREDATE   DERSTARTDATE  DERENDDATE   
-----------------------------------------------------
7369    CLERK     17-Dec-80  17-Dec-80     20-Feb-81   
7499    SALESMAN  20-Feb-81  20-Feb-81     2-Apr-81    
7521    SALESMAN  22-Feb-81  20-Feb-81     2-Apr-81       
7566    MANAGER   2-Apr-81   2-Apr-81      8-Sep-81       
7698    MANAGER   1-May-81   2-Apr-81      8-Sep-81    
7782    MANAGER   9-Jun-81   2-Apr-81      8-Sep-81    
7844    SALESMAN  8-Sep-81   8-Sep-81      17-Nov-81   
7654    SALESMAN  28-Sep-81  8-Sep-81      17-Nov-81   
7839    PRESIDENT 17-Nov-81  17-Nov-81     3-Dec-81    
7900    CLERK     3-Dec-81   3-Dec-81      3-Dec-81    
7902    ANALYST   3-Dec-81   3-Dec-81      23-Jan-82   
7934    CLERK     23-Jan-82  23-Jan-82     19-Apr-87   
7788    ANALYST   19-Apr-87  19-Apr-87     23-May-87   
7876    CLERK     23-May-87  23-May-87     NULL

我的要求是如果连续重复相同的工作，例如对于Salesman，startdate和enddate应该是startdate的min和所有连续列中enddate的最大值。

示例：

两位连续员工的销售人员empno = 7499和empno = 7521应该有DERSTARTDATE = 20-Feb-81和DERENDDATE = 2-Apr-81
SALESMAN for empno = 7844 and empno = 7654应该有DERSTARTDATE = 8-Sep-81和DERENDDATE = 17-Nov-81

Answer 1

首先需要识别相同连续作业的组。 Tabibitosan是一种简单的方法。

 select empno, job, hiredate,
        lead(hiredate) over (order by hiredate,empno) startdate,
        row_number() over (order by hiredate,empno) -
        row_number() over (order by job, hiredate,empno) grp
from emp
);

识别出组后，可以使用MIN / MAX作为分析函数，并在每组中找到MIN / MAX。

with x as (
  select empno, job, hiredate,
        lead(hiredate) over (order by hiredate,empno) startdate,
        row_number() over (order by hiredate,empno) -
        row_number() over (order by job, hiredate,empno) grp
    from emp
  )
select empno, job, hiredate,
      min(hiredate) over (partition by grp) derstartdate,
      max(startdate) over (partition by grp) derenddate
  from x
order by hiredate;

Demo

ORACLE SQL Query查找组中的最小值和最大值

1 个答案: