How can I retrieve distinct data by using group by

user_18514535 0
Viewed 113

I have following table and its sample data as follows,

+------------------+-------+--------+---------------------+-------+
| PhoneCode        | SeqID | Active | Token               | CUSID |
+------------------+-------+--------+---------------------+-------+
| e29a5e1c695352b8 | 3898  | I      | JYN8CYRVzaKWL-l_K   | 50002 |
+------------------+-------+--------+---------------------+-------+
| e29a5e1c695352b8 | 3900  | I      | JYN8CYRVzaKWL-l_K   | 50002 |
+------------------+-------+--------+---------------------+-------+
| 741fb28bc72183e3 | 3899  | I      | RU3ReKEw0yin9LxZWCO | 50002 |
+------------------+-------+--------+---------------------+-------+
| 741fb28bc72183e4 | 3901  | A      | RU3ReKEw0yin9LxZWCO | 50002 |
+------------------+-------+--------+---------------------+-------+

I need to take distinct PhoneCode with latest SeqID. So I tried following query.

SELECT UD.PHONECODE, UD.SeqID, UD.ACTIVE, UD.Token
FROM   DEVICE UD
WHERE UD.CUSID = '50002' AND UD.ACTIVE = 'I'
GROUP BY PHONECODE

But its makes error. How can I retrieve following output?

+------------------+-------+--------+---------------------+-------+
| PhoneCode        | SeqID | Active | Token               | CUSID |
+------------------+-------+--------+---------------------+-------+
| e29a5e1c695352b8 | 3900  | I      | JYN8CYRVzaKWL-l_K   | 50002 |
+------------------+-------+--------+---------------------+-------+
| 741fb28bc72183e3 | 3899  | I      | RU3ReKEw0yin9LxZWCO | 50002 |
+------------------+-------+--------+---------------------+-------+
greatest-n-per-group oracle sql
5 Answers

You can do it without GROUP BY and without self-joins:

SELECT PHONECODE,
       SeqID,
       ACTIVE,
       Token
FROM   (
  SELECT d.*,
         ROW_NUMBER() OVER (PARTITION BY phonecode ORDER BY seqid DESC) AS rn
  FROM   device d
  WHERE  cusid = 50002
  AND    active = 'I'
)
WHERE  rn = 1;

Or, if you want to use GROUP BY then you can also use KEEP:

SELECT PHONECODE,
       MAX(SeqID) AS seqid,
       MAX(ACTIVE) KEEP (DENSE_RANK LAST ORDER BY SeqID) AS active,
       MAX(Token) KEEP (DENSE_RANK LAST ORDER BY SeqID) AS token
FROM   device
WHERE  cusid = 50002
AND    active = 'I'
GROUP BY phonecode;

Which, for the sample data:

CREATE TABLE device (PhoneCode, SeqID, Active, Token, CUSID ) AS
SELECT 'e29a5e1c695352b8', 3898, 'I', 'JYN8CYRVzaKWL-l_K',   50002 FROM DUAL UNION ALL
SELECT 'e29a5e1c695352b8', 3900, 'I', 'JYN8CYRVzaKWL-l_K',   50002 FROM DUAL UNION ALL
SELECT '741fb28bc72183e3', 3899, 'I', 'RU3ReKEw0yin9LxZWCO', 50002 FROM DUAL UNION ALL
SELECT '741fb28bc72183e4', 3901, 'A', 'RU3ReKEw0yin9LxZWCO', 50002 FROM DUAL;

Both output:

PHONECODE SEQID ACTIVE TOKEN
741fb28bc72183e3 3899 I RU3ReKEw0yin9LxZWCO
e29a5e1c695352b8 3900 I JYN8CYRVzaKWL-l_K

db<>fiddle here

user_15092640

You can use MAX() and a join:

SELECT UD.PHONECODE, UD.ACTIVE, UD.Token, UD.SeqID
FROM   DEVICE UD
INNER JOIN (
    SELECT PHONECODE, MAX(SeqID) as SEQID
    FROM DEVICE   
    WHERE CUSID = '50002' AND ACTIVE = 'I'
    GROUP BY PHONECODE
 ) T2 ON UD.PHONECODE = T2.PHONECODE AND UD.SEQID = T2.SEQID
user_110143100

What about a simple sub query

SELECT UD.PHONECODE, UD.SeqID, UD.ACTIVE, UD.Token 
FROM device ud
WHERE (UD.PHONECODE, UD.SeqID) IN (
  SELECT PHONECODE, MAX(SeqID) 
  FROM device
  WHERE CUSID = '50002' AND ACTIVE = 'I'
  GROUP BY PHONECODE 
)
user_31711720

The following query will work :

select phonecode,seqid,active,token,cusid
from
(
    select * , row_number() over(partition by phonecode order by seqid desc) as rownum
    from device
    where cusid = 50002 and active = 'I'
) as t1
where rownum = 1
user_149912770

If your database version is 12c+ and displaying the returning value of ROW_NUMBER() function doesn't matter, then such a query as an option without a subquery would be

SELECT d.*,
       ROW_NUMBER() OVER (PARTITION BY PhoneCode ORDER BY SeqID DESC) AS rn
  FROM device d
 WHERE cusid = 50002
   AND active = 'I'
 ORDER BY rn  
 FETCH FIRST 1 ROWS WITH TIES

where all ties(equal values) of rn are included in the result set.

Demo

user_58413060
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