Question

我的SQL返回一个用户ID列表及其角色，重点是将所有角色（组）连接成一行。目前，它返回所有内容，但我只需要为每个用户返回一个条目，基本上是每个用户的条目，该用户具有最大的rn（行号）。

SELECT 
  IT_ID, 
  SUBSTR (SYS_CONNECT_BY_PATH (grp , ','), 2) GROUPS
FROM (
  SELECT 
    U.IT_ID, 
    LAST_NAME, 
    BFIRST_NAME, 
    GRP, 
    ROW_NUMBER() OVER (partition by u.it_id ORDER BY U.IT_ID) rn, 
    COUNT(*) OVER() cnt
FROM ECG_IT_USERS U
JOIN SECUREGROUPS G ON U.IT_ID = G.IT_ID)
START WITH rn = 1
CONNECT BY rn = PRIOR rn + 1 and it_id = prior it_id
Group by it_id

它返回：

IT_ID   GROUPS
afz23   ADMIN
afz23   ADMIN, QA
klf44   USER
klf44   USER, BUSINESS

我需要返回

IT_ID   GROUPS
afz23   ADMIN, QA
klf44   USER, BUSINESS

Answer 1

添加另一个分析函数，用于计算每个IT_ID分区上行号的最大值。然后选择rn = max_rn的行。

Answer 2

select it_id, groups
from 
(
    select 
      it_id
      , groups
      , row_number () over (partition by id_id order by length(groups) desc ) rn
    from 
      (
          SELECT 
            IT_ID, 
            SUBSTR (SYS_CONNECT_BY_PATH (grp , ','), 2) GROUPS
          FROM (
            SELECT 
              U.IT_ID, 
              LAST_NAME, 
              BFIRST_NAME, 
              GRP, 
              ROW_NUMBER() OVER (partition by u.it_id ORDER BY U.IT_ID) rn, 
              COUNT(*) OVER() cnt
          FROM ECG_IT_USERS U
          JOIN SECUREGROUPS G ON U.IT_ID = G.IT_ID)
          START WITH rn = 1
          CONNECT BY rn = PRIOR rn + 1 and it_id = prior it_id
          Group by it_id
      )
 )
where rn = 1
;

Answer 3

这是一个greatest-n-per-group问题。这里有关于这个主题的非常全面的答案：SQL Select only rows with Max Value on a Column

由于现在时间有限，我无法处理您的查询。不给你鱼，但教你如何捕鱼。

如何返回行数最多的行？

3 个答案: