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How can I use listagg with distinct.

Posted on 2016-08-31
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Last Modified: 2016-09-01
Dear Experts,

My requirement is to get distinct values after I use listagg.
But distinct is not allowed inside listagg,

Please help !!
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Question by:sakthikumar
  • 2
4 Comments
 
LVL 74

Expert Comment

by:sdstuber
ID: 41778145
simply wrap your source data in an inline view and make that distinct, then select listagg from that

for example...

select listagg(owner,',') within group(order by owner) from
     (select distinct owner from all_objects)
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LVL 5

Expert Comment

by:Abhimanyu Suri
ID: 41778154
Can you please quote an example

If it is one row o/p combined with listagg, below example may help

SELECT LISTAGG (usr, ',') WITHIN GROUP (ORDER BY usr)
  FROM (SELECT DISTINCT username usr FROM USERS)
0
 

Author Comment

by:sakthikumar
ID: 41778221
the column already has comma separated values, so distinct is not working.

data is like below

category  email
a                email1,email2,email3
b               email2,email3,email6

when I do a group I should get only distinct email IDs
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LVL 74

Accepted Solution

by:
sdstuber earned 2000 total points
ID: 41778241
then you need to split your data first, then combine it afterward.


SELECT LISTAGG(email, ',') WITHIN GROUP (ORDER BY email)
  FROM (SELECT DISTINCT REGEXP_SUBSTR(
                            email,
                            '[^,]+',
                            1,
                            COLUMN_VALUE
                        )
                            email
          FROM yourdata,
               TABLE(
                       SELECT COLLECT(LEVEL)
                         FROM DUAL
                   CONNECT BY LEVEL <= REGEXP_COUNT(email, '[^,]+')
               ))

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