I have a database table like this:
id version_id field1 field2 1 1 texta text1 1 2 textb text2 2 1 textc text3 2 2 textd text4 2 3 texte text5
If you didn’t work it out, it contains a number of versions of a row, and then some text data.
I want to query it and return the version with the highest number for each id. (so the second and last rows only in the above).
I’ve tried using group by whilst ordering by version_id DESC – but it seems to order after its grouped, so this doesn’t work.
Anyone got any ideas? I can’t believe it can’t be done!
Come up with this, which works, but uses a subquery:
SELECT * FROM (SELECT * FROM table ORDER BY version_id DESC) t1 GROUP BY t1.id
It’s called selecting the group-wise maximum of a column. Here are several different approaches for mysql.
Here’s how I would do it:
SELECT * FROM (SELECT id, max(version_id) as version_id FROM table GROUP BY id) t1 INNER JOIN table t2 on t2.id=t1.id and t1.version_id=t2.version_id
This will be relatively efficient, though mysql will create a temporary table in memory for the subquery. I assume you already have an index on (id, version_id) for this table.
It’s a deficiency in SQL that you more or less have to use a subquery for this type of problem (semi-joins are another example).
Subqueries are not well optimized in mysql but uncorrelated subqueries aren’t so bad as long as they aren’t so enormous that they will get written to disk rather than memory. Given that in this query only has two ints the subquery could be millions of rows long before that happened but the select * subquery in your first query could suffer from this problem much sooner.
I think this would do it, not sure if it is the best or fastest though.
SELECT * FROM table WHERE (id, version_id) IN (SELECT id, MAX(version_id) FROM table GROUP BY id)
SELECT id, version_id, field1, field2 FROM ( SELECT @prev = id AS st, (@prev := id), m.* FROM ( (SELECT @prev := NULL) p, ( SELECT * FROM mytable ORDER BY id DESC, version_id DESC ) m ) m2 WHERE NOT IFNULL(st, FALSE);
No subqueries, one pass on
UNIQUE INDEX ON MYTABLE (id, version_id) if you have one (which I think you should)
I usually do this with a subquery:
select id, version_id, field1, field2 from datatable as dt where id = (select id from datatable where id = dt.id order by version_id desc limit 1)
This is pseudo code but something like this should work just fine
select * from table inner join ( select id , max(version_id) maxVersion from table ) dvtbl ON id = dvtbl.id && versionid = dvtbl.maxVersion
This query will do the job without a group by:
SELECT * FROM table AS t LEFT JOIN table AS t2 ON t.id=t2.id AND t.version_id < t2.version_id WHERE t2.id IS NULL
It does not need any temporary tables.
I think this is what you want.
select id, max(v_id), field1, field2 from table group by id
The results I get from that are
1, 2, textb, text2
2, 3, texte, text5
I recreated the table and insert the same data with the id an version_id being a compound primary key. This gave the answer I provided earlier. It was also in MySQL.
not tested it but something like this might work:
SELECT * FROM table GROUP BY id ORDER BY MAX(version_id) DESC