需要使用SQL将多行连接成单个单元格？

Question

目前，我已经写了：

SELECT IT_ID, SUBSTR (SYS_CONNECT_BY_PATH (grp , ','), 2) GROUPS 
FROM (SELECT U.IT_ID, LAST_NAME, FIRST_NAME, GRP, ROW_NUMBER() OVER (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 

这将返回：

IT_ID GROUPS 

afz23 ADMIN 
afz23 ADMIN,QA 
alv77 ADMIN,QA,USER 
jaj23 ADMIN,QA,USER,USER 
klo26 ADMIN,QA,USER,USER,PROD 
klo26 ADMIN,QA,USER,USER,PROD,ADMIN 
klo26 ADMIN,QA,USER,USER,PROD,ADMIN,QA 
mav45 ADMIN,QA,USER,USER,PROD,ADMIN,QA,ADMIN 

我想不通我怎么可以让它重新遇到一个新用户后？它似乎在继续前面的组，即使用户不属于他们。

我需要看到：

IT_ID GROUPS 

afz23 ADMIN,QA 
alv77 USER 
jaj23 USER 
klo26 PROD,ADMIN,QA 
mav45 ADMIN 

Answer 1

有你需要做三件事情。

首先，您需要向row_number函数添加一个分区，以便每个IT_ID从1开始编号。您还需要将IT_ID列添加到连接方式，以便仅采用具有相同IT_ID值的行。最后，您需要按it_id列进行分组以删除重复的行。

最后的查询将

with ECG_IT_USERS as (
    select 'afz23' as it_id from dual union all 
    select 'alv77' as it_id from dual union all 
    select 'jaj23' as it_id from dual union all 
    select 'klo26' as it_id from dual union all 
    select 'mav45' as it_id from dual 
), 
securegroups as (
    select 'afz23' as it_id, 'ADMIN' as grp from dual union all 
    select 'afz23' as it_id, 'QA' as grp from dual union all 
    select 'alv77' as it_id, 'USER' as grp from dual union all 
    select 'jaj23' as it_id, 'USER' as grp from dual union all 
    select 'klo26' as it_id, 'PROD' as grp from dual union all 
    select 'klo26' as it_id, 'ADMIN' as grp from dual union all 
    select 'klo26' as it_id, 'QA' as grp from dual union all 
    select 'mav45' as it_id, 'ADMIN' as grp from dual 
) 
SELECT 
    IT_ID, 
    Max(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 

这就产生了我下面的输出：

IT_ID GROUPS 
----- -------------------- 
alv77 USER 
afz23 ADMIN,QA 
jaj23 USER 
mav45 ADMIN 
klo26 PROD,ADMIN,QA 

编辑：我添加了一个与条款与一些样本数据，这会运行我没有任何问题，尽管我注释掉了last_name和first_name列，因为它们对最终查询没有影响，我在连接条件中放了一些括号。

也许它会支付开始与我已经得到上面的查询，检查它最初适用于您，并适当修改它。