sys_connect_by_path函数配合group by 进行分组拼接

最近,碰到一个需求将 approval_code值对应的多个FIRST_NAME值通过line_no的asc排序 合并为一个最长的字段  ,对应的表 如下:

对应表的sql 语句如下:

View Code 
 SELECT DISTINCT t1.FIRST_NAME,
        t2.approval_code,
        t2.line_no
      FROM K2_ACCESS_USER@k2global t1
      INNER JOIN k2_approval_path t2
      ON t1.DOMAIN_NAME=t2.USER_ID 
     right join k2_credit_limit_hist t4
      on t2.approval_code=t4.approval_code and t4.expired_date>=to_date('2012-01-01','yyyy-mm-dd')
      ORDER BY t2.  APPROVAL_CODE,t2.line_no

起初,我是打算这样获取approval_code对应的FIRST_NAME合并值(当时还不知道 可以直接配合group by 获取到分组的最大的FIRST_NAME的合并值)

View Code 
 ------------------------start to combine the approver's name-------
 SELECT  max(substr(sys_connect_by_path(FIRST_NAME,','),2))FIRST_NAME
  FROM
    (
    SELECT
       ltrim(APPROVAL_CODE,'RQP')+row_number() over(ORDER BY APPROVAL_CODE ) ROW_NUM,
            FIRST_NAME,
      APPROVAL_CODE
      
    FROM (select distinct FIRST_NAME, approval_code from
      (
      SELECT DISTINCT t1.FIRST_NAME,
        t2.approval_code,
        t2.line_no
      FROM K2_ACCESS_USER@k2global t1
      INNER JOIN k2_approval_path t2
      ON t1.DOMAIN_NAME=t2.USER_ID 
     right join k2_credit_limit_hist t4
      on t2.approval_code=t4.approval_code and t4.expired_date>=to_date('2012-01-01','yyyy-mm-dd')
      ORDER BY t2.  APPROVAL_CODE,t2.line_no
      ))
    ) t3 
    START WITH 
    t3.APPROVAL_CODE='RQP0001105'  --RQP0001105 用来作为一个测试的值
    CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM
   
  ------------------------end to combine the approver's name--------

但是很快我发现我获取到的不是我想要的:

我去掉包含sys_connect_by_path函数的max()之后,并在select 列表中增加ROW_NUM

View Code 
SELECT t3.ROW_NUM, substr(sys_connect_by_path(FIRST_NAME,','),2)FIRST_NAME
  FROM
    (
    SELECT
       ltrim(APPROVAL_CODE,'RQP')+row_number() over(ORDER BY APPROVAL_CODE ) ROW_NUM,
      --  row_number() over(partition by APPROVAL_CODE ORDER BY APPROVAL_CODE ) RANK_NUM,
      FIRST_NAME,
      APPROVAL_CODE
      
    FROM (select distinct FIRST_NAME, approval_code from
      (
      SELECT DISTINCT t1.FIRST_NAME,
        t2.approval_code,
        t2.line_no
      FROM K2_ACCESS_USER@k2global t1
      INNER JOIN k2_approval_path t2
      ON t1.DOMAIN_NAME=t2.USER_ID 
     right join k2_credit_limit_hist t4
      on t2.approval_code=t4.approval_code and t4.expired_date>=to_date('2012-01-01','yyyy-mm-dd')
      ORDER BY t2.  APPROVAL_CODE,t2.line_no
      ))
    ) t3 
    START WITH 
    t3.APPROVAL_CODE='RQP0001105'
  --t3.APPROVAL_CODE=hist.approval_code 
  --and  t3.RANK_NUM=1
    CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM


获得的结果如下:

可以看到 其实在调用sys_connect_by_path函数的过程中 已经生成了我们想要的值'Kenneth,Lawrence'  但是由于一些原因这个值最后被重写为Lawrence.

我观察了下早先的代码 sys_connect_by_path最后的条件部分:

 START WITH
    t3.APPROVAL_CODE='RQP0001105'
    CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM

我start with 用的条件是t3.APPROVAL_CODE='RQP0001105' ('RQP0001105'是代入的测试值), 而实际上在表中APPROVAL_CODE值为'RQP0001105'

有两个为别为ROW_NUM1106和1107的两条记录.于是我在执行函数sys_connect_by_path的时候其实是分为两步来执行的 ,它会分别从ROW_NUM=1106和1107两条记录开始执行一次,也就是说它是这样的

start with t3.APPROVAL_CODE='RQP0001105' and t3.ROW_NUM='1106'

 CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM

执行结果:

和

start with t3.APPROVAL_CODE='RQP0001105' and t3.ROW_NUM='1107'

CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM

执行结果:

我们可以判断出来是由于从start with t3.APPROVAL_CODE='RQP0001105' and t3.ROW_NUM='1107'的时候将上一步 调用函数生成的值'Kenneth,Lawrence' 重写为Lawrence.

于是问题就清楚了, 解决方法是在start with 的时候再加上一个条件使他只从最上面的那条记录开始执行. 我的方法是新增一个rank列,rank列的值只会和多条记录中的第一个记录的ROW_NUM相同

　　 ltrim(APPROVAL_CODE,'RQP')+row_number() over(ORDER BY APPROVAL_CODE ) ROW_NUM,
       ltrim(APPROVAL_CODE,'RQP')+RANK() over(ORDER BY APPROVAL_CODE ) RANK_NUM,

同时 下面的条件改为:

     START WITH t3.APPROVAL_CODE=hist.approval_code and  t3.ROW_NUM=t3.RANK_NUM
    CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM

小组的leader建议我的方法是在原先的代码中新增了一列

rank_num,它是由表中分块排序而来 见如下:

SELECT
       ltrim(APPROVAL_CODE,'RQP')+row_number() over(ORDER BY APPROVAL_CODE ) ROW_NUM,
        row_number() over(partition by APPROVAL_CODE ORDER BY APPROVAL_CODE ) RANK_NUM,
      FIRST_NAME,
      APPROVAL_CODE
          from (select distinct FIRST_NAME, approval_code
    FROM
      (
      SELECT DISTINCT t1.FIRST_NAME,
        t2.approval_code,
        t2.line_no
      FROM K2_ACCESS_USER@k2global t1
      INNER JOIN k2_approval_path t2
      ON t1.DOMAIN_NAME=t2.USER_ID 
     right join k2_credit_limit_hist t4
      on t2.approval_code=t4.approval_code and t4.expired_date>=to_date('2012-01-01','yyyy-mm-dd')
      ORDER BY t2.  APPROVAL_CODE,t2.line_no
      ))

执行之后可以看到获取到的数据如下:

我们将原先的  

START WITH t3.APPROVAL_CODE='RQP0001105'

CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM

修改为

  START WITH t3.APPROVAL_CODE='RQP0001105'and  t3.RANK_NUM=1
   
    CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM

即可.

View Code 
 SELECT  max(substr(sys_connect_by_path(FIRST_NAME,','),2)) FIRST_NAME
 
-- , length(FIRST_NAME),t3.ROW_NUM,t3.APPROVAL_CODE
  FROM
    (
    SELECT
       ltrim(APPROVAL_CODE,'RQP')+row_number() over(ORDER BY APPROVAL_CODE ) ROW_NUM,
      -- ltrim(APPROVAL_CODE,'RQP')+RANK() over(ORDER BY APPROVAL_CODE ) RANK_NUM,
      row_number() over(partition by APPROVAL_CODE ORDER BY APPROVAL_CODE ) RANK_NUM,
    -- row_number(),
      FIRST_NAME,
      APPROVAL_CODE
          from (select distinct FIRST_NAME, approval_code
    FROM
      (
      SELECT DISTINCT t1.FIRST_NAME,
        t2.approval_code,
        t2.line_no
      FROM K2_ACCESS_USER@k2global t1
      INNER JOIN k2_approval_path t2
      ON t1.DOMAIN_NAME=t2.USER_ID 
     right join k2_credit_limit_hist t4
      on t2.approval_code=t4.approval_code and t4.expired_date>=to_date('2012-01-01','yyyy-mm-dd')
   -- where APPROVAL_CODE='RQP0001199'
      ORDER BY t2.  APPROVAL_CODE,t2.line_no
      ))
    ) t3 
    START WITH 
  t3.APPROVAL_CODE='RQP0001105' and  t3.RANK_NUM=1
   --and t3.app
    CONNECT BY t3.ROW_NUM -1   = prior t3.ROW_NUM


执行后结果如下:

 好吧,上面写的是我之前走比较绕的路子.实际上要实现我们要的值只需要配合group by 进行分组拼接即可 代码如下:

View Code 
SELECT max(SUBSTR(SYS_CONNECT_BY_PATH(create_by, ','), 2)) create_by
  FROM
    (SELECT
       ltrim(APPROVAL_CODE,'RQP')+row_number() over(ORDER BY APPROVAL_CODE ) ROW_NUM,
      -- row_number() over(partition by APPROVAL_CODE ORDER BY APPROVAL_CODE ) RANK_NUM,
      create_by,
      approval_code
      
      from (select distinct create_by, approval_code
    FROM
      (SELECT DISTINCT t.create_by,
        t.approval_code,
        t.line_no
      FROM k2_approval_path t 
       RIGHT JOIN K2_CREDIT_LIMIT_HIST T4
      on t.approval_code=t4.approval_code and t4.expired_date>=to_date('2010-01-01','yyyy-mm-dd')
      ORDER BY t.approval_code,t.line_no
      )
    ) )T1
    START WITH  t1.approval_code= hist.approval_code 
    --t1.approval_code= hist.approval_code and t1.RANK_NUM=1
       CONNECT BY T1.ROW_NUM -1   = PRIOR T1.ROW_NUM
       group by t1.approval_code
EXEC dbo.success '坚持','细心','计划'
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原文地址：https://www.cnblogs.com/weeky/p/2582707.html