• 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:

    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'
  • 相关阅读:
    查看centos版本
    Internal.Cryptography.CryptoThrowHelper+WindowsCryptographicException 拒绝访问 / 出现了内部错误 c# – 当使用X509Certificate2加载p12/pfx文件时出现
    asp.net asp.net application 升级到 asp.net web 解决找不到控件 批量生成.designer文件
    netcore发布到 iis 设置 部署 环境 变量
    sqlserver 3145
    windows server 2012 远程桌面不好使
    VirtualBox 桥接模式,虚拟机ping不通宿主机
    移动端笔记
    css笔记——关于css中写上charset “utf-8”
    工作笔记——前端分页数据回显
  • 原文地址:https://www.cnblogs.com/weeky/p/2582707.html
Copyright © 2020-2023  润新知