Groupwise maximum - mysql

Question

I have a table from which I am trying to retrieve the latest position for each security:
The Table:
My query to create the table: SELECT id, security, buy_date FROM positions WHERE client_id = 4
+-------+----------+------------+
| id | security | buy_date |
+-------+----------+------------+
| 26 | PCS | 2012-02-08 |
| 27 | PCS | 2013-01-19 |
| 28 | RDN | 2012-04-17 |
| 29 | RDN | 2012-05-19 |
| 30 | RDN | 2012-08-18 |
| 31 | RDN | 2012-09-19 |
| 32 | HK | 2012-09-25 |
| 33 | HK | 2012-11-13 |
| 34 | HK | 2013-01-19 |
| 35 | SGI | 2013-01-17 |
| 36 | SGI | 2013-02-16 |
| 18084 | KERX | 2013-02-20 |
| 18249 | KERX | 0000-00-00 |
+-------+----------+------------+
I have been messing with versions of queries based on this page, but I cannot seem to get the result I'm looking for.
Here is what I've been trying:
SELECT t1.id, t1.security, t1.buy_date
FROM positions t1
WHERE buy_date = (SELECT MAX(t2.buy_date)
FROM positions t2
WHERE t1.security = t2.security)
But this just returns me:
+-------+----------+------------+
| id | security | buy_date |
+-------+----------+------------+
| 27 | PCS | 2013-01-19 |
+-------+----------+------------+
I'm trying to get the maximum/latest buy date for each security, so the results would have one row for each security with the most recent buy date. Any help is greatly appreciated.
EDIT: The position's id must be returned with the max buy date.

Answer 1

You can use this query. You can achieve results in 75% less time. I checked with more data set. Sub-Queries takes more time.
SELECT p1.id,
p1.security,
p1.buy_date
FROM positions p1
left join
positions p2
on p1.security = p2.security
and p1.buy_date < p2.buy_date
where
p2.id is null;
SQL-Fiddle link

Answer 2

You can use a subquery to get the result:
SELECT p1.id,
p1.security,
p1.buy_date
FROM positions p1
inner join
(
SELECT MAX(buy_date) MaxDate, security
FROM positions
group by security
) p2
on p1.buy_date = p2.MaxDate
and p1.security = p2.security
See SQL Fiddle with Demo
Or you can use the following in with a WHERE clause:
SELECT t1.id, t1.security, t1.buy_date
FROM positions t1
WHERE buy_date = (SELECT MAX(t2.buy_date)
FROM positions t2
WHERE t1.security = t2.security
group by t2.security)
See SQL Fiddle with Demo

Answer 3

This is done with a simple group by. You want to group by the securities and get the max of buy_date. The SQL:
SELECT security, max(buy_date)
from positions
group by security
Note, this is faster than bluefeet's answer but does not display the ID.

Answer 4

The answer by #bluefeet has two more ways to get the results you want - and the first will probably be more efficient than your query.
What I don't understand is why you say that your query doesn't work. It seems pretty fine and returns the expected result. Tested at SQL-Fiddle
SELECT t1.id, t1.security, t1.buy_date
FROM positions t1
WHERE buy_date = ( SELECT MAX(t2.buy_date)
FROM positions t2
WHERE t1.security = t2.security ) ;
If the problems appears when you add the client_id = 4 condition, then it's because you add it only in one WHERE clause while you have to add it in both:
SELECT t1.id, t1.security, t1.buy_date
FROM positions t1
WHERE client_id = 4
AND buy_date = ( SELECT MAX(t2.buy_date)
FROM positions t2
WHERE client_id = 4
AND t1.security = t2.security ) ;

Answer 5

select security, max(buy_date) group by security from positions;
is all you need to get max buy date for each security (when you say out loud what you want from a query and you include the phrase "for each x", you probably want a group by on x)
When you use a group by, all columns in your select must either be columns that have been grouped by or aggregates, so if, for example, you wanted to include id, you'd probably have to use a subquery similar to what you had before, since there doesn't seem to be any aggregate you can reasonably use on the ids, and another group by would give you too many rows.