101

For an example, I would like to select id with max date group by category, the result is: 7, 2, 6

id  category  date
1   a         2013-01-01
2   b         2013-01-03
3   c         2013-01-02
4   a         2013-01-02
5   b         2013-01-02
6   c         2013-01-03
7   a         2013-01-03
8   b         2013-01-01
9   c         2013-01-01

May I know how to do this in PostgreSQL?

Erwin Brandstetter
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user2412043
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3 Answers3

167

This is a perfect use-case for DISTINCT ON - a Postgres specific extension of the standard DISTINCT:

SELECT DISTINCT ON (category)
       id  -- , category, date  -- any other column (expression) from the same row
FROM   tbl
ORDER  BY category, date DESC;

Careful with descending sort order. If the column can be NULL, you may want to add NULLS LAST:

DISTINCT ON is simple and fast. Detailed explanation in this related answer:

For big tables with many rows per category consider an alternative approach:

Erwin Brandstetter
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22

Try this one:

SELECT t1.* FROM Table1 t1
JOIN 
(
   SELECT category, MAX(date) AS MAXDATE
   FROM Table1
   GROUP BY category
) t2
ON T1.category = t2.category
AND t1.date = t2.MAXDATE

See this SQLFiddle

Himanshu Jansari
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16

Another approach is to use the first_value window function: http://sqlfiddle.com/#!12/7a145/14

SELECT DISTINCT
  first_value("id") OVER (PARTITION BY "category" ORDER BY "date" DESC) 
FROM Table1
ORDER BY 1;

... though I suspect hims056's suggestion will typically perform better where appropriate indexes are present.

A third solution is:

SELECT
  id
FROM (
  SELECT
    id,
    row_number() OVER (PARTITION BY "category" ORDER BY "date" DESC) AS rownum
  FROM Table1
) x
WHERE rownum = 1;
Craig Ringer
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