我需要一个递归查询来获得在oracle中的计算字段

Question

我不知道什么最好的方法是获得我要找的结果。在一个oracle 10g数据库中，我试图查询所有用户，并为每个有主管人员的行生成一个计算字段，如果他们的主管类型是U.如果不是，我需要走到树上直到找到一个。这个查询可能需要递增多个级别。

所以对于一个Employee表是这样的：

Employee Table 
+-------+----------+--------------+---------------+ 
| empno | empgroup |  user  | supervisor_no | 
+-------+----------+--------------+---------------+ 
|  1 | E  | Joe Schmo |    4 | 
|  2 | E  | Mark Marin |    3 | 
|  3 | U  | Reed Patter |    7 | 
|  4 | E  | Martin Price |    7 | 
|  7 | U  | Mary Wengert |    | 
+-------+----------+--------------+---------------+ 

我希望看到这样的结果，其中manager_no将计算出的字段

Results 
+-------+----------+--------------+---------------+------------+ 
| empno | empgroup |  user  | supervisor_no | Manager_No | 
+-------+----------+--------------+---------------+------------+ 
|  1 | E  | Joe Schmo |    4 |   7 | 
|  2 | E  | Mark Marin |    3 |   3 | 
|  3 | U  | Reed Patter |    7 |   7 | 
|  4 | E  | Martin Price |    7 |   7 | 
|  7 | U  | Mary Wengert |    |   | 
+-------+----------+--------------+---------------+------------+ 

好了，所以我一直在问我试过。我并不是说这是必须这样做的方式，如果有人有更好的建议，我全都是耳朵，但这是我想要实现的。

我想象了两部分。

select * from employee em 
connect by prior supervisor_no = empno 
start with empno = 1 

[结果] [2]：即得到所有我的结果

select em.empno, em.empgroup, em.user, em.supervisor, (my subquery) as manager_no 
from employee em 

查询主查询

| EMPNO | EMPGROUP |  USERNAME | SUPERVISOR_NO | 
--------------------------------------------------- 
|  1 |  E | Joe Schmo |    4 | 
|  4 |  E | Martin Price |    7 | 
|  7 |  U | Mary Wengert |  (null) | 

好吧，我找到了一种方法来筛选对可能工作的团队，但不知道这是否是最有效的路线。

SELECT empno FROM (
SELECT empno FROM employee em 
CONNECT BY PRIOR supervisor_no = empno 
START WITH empno = 1 
order by level) d 
WHERE d.empgroup = 'U' and rownum =1 

我也创建了一个小提琴，如果有帮助。 http://www.sqlfiddle.com/#!4/c8805/4

Answer 1

那么，你的问题变得对我是个挑战，但我没有放弃，并得到这个只用SQL和内置函数（不应该慢慢地工作，而且相当可转位需要）：

select distinct 
    empno, 
    empgroup, 
    username, 
    supervisor_no, 
    manager_no 
from (
    select 
      e.*, 
      decode(
     instr(
      sys_connect_by_path(empgroup, '/'), 
      'U/', 
      -1 
     ), 
     0, 
     null, 
     substr(
      sys_connect_by_path(empno, '/'), 
      instr(
       sys_connect_by_path(empno, '/'), 
       '/', 
       1, 
       length(
        substr(
         sys_connect_by_path(empgroup, '/'), 
         1, 
         instr(
          sys_connect_by_path(empgroup, '/'), 
          'U/', 
          -1 
         ) 
        ) 
       ) - length(
        replace(
         substr(
          sys_connect_by_path(empgroup, '/'), 
          1, 
          instr(
           sys_connect_by_path(empgroup, '/'), 
           'U/', 
           -1 
          ) 
         ), 
         '/' 
        ) 
       ) 
      ) + 1, 
      instr(
       sys_connect_by_path(empno, '/'), 
       '/', 
       1, 
       length(
        substr(
         sys_connect_by_path(empgroup, '/'), 
         1, 
         instr(
          sys_connect_by_path(empgroup, '/'), 
          'U/', 
          -1 
         ) 
        ) 
       ) - length(
        replace(
         substr(
          sys_connect_by_path(empgroup, '/'), 
          1, 
          instr(
           sys_connect_by_path(empgroup, '/'), 
           'U/', 
           -1 
          ) 
              ), 
               '/' 
             ) 
           ) + 1 
         ) - instr(
           sys_connect_by_path(empno, '/'), 
           '/', 
           1, 
           length(
             substr(
               sys_connect_by_path(empgroup, '/'), 
               1, 
               instr(
                 sys_connect_by_path(empgroup, '/'), 
                 'U/', 
                 -1 
               ) 
             ) 
           ) - length(
             replace(
               substr(
                 sys_connect_by_path(empgroup, '/'), 
                 1, 
                 instr(
                   sys_connect_by_path(empgroup, '/'), 
                   'U/', 
                   -1 
                 ) 
               ), 
               '/' 
             ) 
           ) 
         ) - 1 
       ) 
     ) manager_no 
    from employee e 
    connect by prior empno = supervisor_no 
) 
where manager_no is not null or supervisor_no is null 
order by empno; 

的SQL捣鼓查询：http://www.sqlfiddle.com/#!4/c8805/27/0

UPDATE： 当我早上起床，我意识到一切都可以做更容易和查询变得更具可读性，在这里你去：

select 
    empno, 
    empgroup, 
    username, 
    supervisor_no, 
    null manager_no 
from 
    employee 
where 
    supervisor_no is null 
union all 
select 
    empno, 
    empgroup, 
    username, 
    supervisor_no, 
    substr(ep, 2, instr(ep, '/', 2)-2) manager_no 
from (
    select 
    sys_connect_by_path(empgroup, '/') gp, 
    sys_connect_by_path(empno, '/') ep, 
    e.* 
    from employee e 
    connect by prior empno = supervisor_no 
) e 
where 
    substr(gp, 1, 3) = '/U/' 
    and (length(gp) - length(replace(gp, 'U/'))) = length('U/') 
order by empno; 

它的工作原理！以及上面的查询。SQL小提琴吧：http://www.sqlfiddle.com/#!4/c8805/54/0

享受！

Answer 2

也许创建一个递归函数来获取用户的'U'型监督员？

CREATE OR REPLACE FUNCTION GET_U_SUPERVISOR(employee_no INT) 
RETURN INT IS 
    supervisor_empno INT; 
    supervisor_group VARCHAR(20); 
BEGIN 
    SELECT empno, empgroup 
    INTO supervisor_empno, supervisor_group 
    FROM employee 
    WHERE employee.empno = (SELECT supervisor_no FROM employee WHERE empno = employee_no); 

    IF 'U' = supervisor_group 
    THEN 
     RETURN supervisor_empno; 
    ELSE 
     RETURN GET_U_SUPERVISOR(supervisor_empno); 
    END IF; 
END GET_U_SUPERVISOR; 
/

然后：

SELECT empno, empgroup, username, supervisor_no, GET_U_SUPERVISOR(empno) 
FROM employee; 

一个SQL小提琴回你：http://www.sqlfiddle.com/#!4/c7540/1/0

我不能保证在任何的表现，不过，如果它是一个大桌子。它是递归的，并将在每行的基础上调用。

Answer 3

我递归，SQL Fiddle使用，

CREATE TABLE EMPLOYEE(NO INT, EMP_GROUP VARCHAR2(1), EUSER VARCHAR2(30), SUPERVISOR_NO INT); 

INSERT INTO EMPLOYEE VALUES(1, 'E', 'JOE SCHMO', 4); 
INSERT INTO EMPLOYEE VALUES(2, 'E', 'Mark Marin', 3); 
INSERT INTO EMPLOYEE VALUES(3, 'U', 'Reed Patter', 7); 
INSERT INTO EMPLOYEE VALUES(4, 'E', 'Martin Price ', 7); 
INSERT INTO EMPLOYEE(NO, EMP_GROUP, EUSER) VALUES(7, 'U', 'Mary Wengert'); 
WITH A(EMPLOYEE, MANAGER) AS 
(
    SELECT NO AS EMPLOYEE, SUPERVISOR_NO AS MANAGER FROM EMPLOYEE 
    UNION ALL 
    SELECT A.EMPLOYEE, E.SUPERVISOR_NO AS MANAGER 
    FROM A, EMPLOYEE E 
    WHERE A.MANAGER = E.NO 
    AND E.SUPERVISOR_NO IS NOT NULL 
) 

SELECT * FROM A 
ORDER BY 1