Reg: Find Duplicate Records and select max number of record in Table

Question

Hi Guys,
     This is Nagendra, India.
my table structure is
id  name  tempid  temptime
1  xxx     123        date
1 yyy       128       date
1 sdd     173       date
14 ree    184      date
14 fded   189     date
This is Table Structure, totally 15000+ records is there.
My Requirement is showing id and max(tempId) value.
id  name  tempid  temptime
1   sdd    173      date
14  fded   189     date
Like that, I want to show all record(after hiding duplicate values ) like that Could you please solve this issue.
With Regards,
Nagendra

Answer 1

; WITH numbering AS (
    SELECT id, name, tempid, temptime,
           rowno = row_number() OVER(PARTITION BY id ORDER BY tempid DESC)
    FROM  tbl
SELECT id, name, tempid, temptime
FROM   numbering
WHERE  rowno = 1
The WITH thing defines a Common Table Expression which is a locally defined view which only exists for this query. The row_number function numbers the rows, re-starting on 1 for every id, and they are numbering in falling tempid order. Thus, by
selecting all rows from the CTE with rono = 1, we get the row with the highest tempid for each id.
Erland Sommarskog, SQL Server MVP, [email protected]