Select oldest two records from group - mysql

Question

I've found a number of examples showing how to select a single oldest/newest row from a grouped set, but am having trouble getting the oldest two rows from a data set.
Here's my sample table:
CREATE TABLE IF NOT EXISTS `orderTable` (
`customer_id` varchar(10) NOT NULL,
`order_id` varchar(4) NOT NULL,
`date_added` date NOT NULL,
PRIMARY KEY (`customer_id`,`order_id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `orderTable` (`customer_id`, `order_id`, `date_added`) VALUES
('1234', '5A', '1997-01-22'),
('1234', '88B', '1992-05-09'),
('0487', 'F9', '2002-01-23'),
('5799', 'A12F', '2007-01-23'),
('1234', '3A', '2009-01-22'),
('3333', '7FHS', '2009-01-22'),
('0487', 'Z33', '2004-06-23'),
('3333', 'FF44', '2013-09-11'),
('3333', '44f5', '2013-09-02');
This query returns more than two rows:
SELECT customer_id, order_id, date_added
FROM orderTable T1
WHERE (
select count(*) FROM orderTable T2
where T2.order_id = T1.order_id AND T2.date_added <= T1.date_added
) <= 2;
Since I am not looking for a single row, this is not a standard greatest-n-per-group type query.
What am I missing that I can get the first two orders for each customer_id?

Answer 1

The best (i.e. most performant) approach is to use a User Defined Variable in the query.
SELECT tmp.customer_id, tmp.date_added
FROM (
SELECT
customer_id, date_added,
IF (#prev <> customer_id, #rownum := 1, #rownum := #rownum+1 ) rank,
#prev := customer_id
FROM orderTable t
JOIN (SELECT #rownum := NULL, #prev := 0) r
ORDER BY t.customer_id
) tmp
WHERE tmp.rank <= 2
ORDER BY customer_id, date_added
Results:
| CUSTOMER_ID | DATE_ADDED |
|-------------|----------------------------------|
| 0487 | January, 23 2002 00:00:00+0000 |
| 0487 | June, 23 2004 00:00:00+0000 |
| 1234 | May, 09 1992 00:00:00+0000 |
| 1234 | January, 22 1997 00:00:00+0000 |
| 3333 | January, 22 2009 00:00:00+0000 |
| 3333 | September, 02 2013 00:00:00+0000 |
| 5799 | January, 23 2007 00:00:00+0000 |
Fiddle here.
Note that the join is just being used to initialise the variables.

Answer 2

Your original query should be (use customer_id in subquery)
SELECT customer_id, order_id, date_added
FROM orderTable T1
WHERE (
select count(*) FROM orderTable T2
where T2.customer_id = T1.customer_id AND T2.date_added <= T1.date_added
) <= 2;
You can also use variables:
SELECT customer_id, order_id, date_added FROM (
SELECT customer_id, order_id, date_added,
#rownum := if(#prev_cust = customer_id, #rownum + 1,1) as rn,
#prev_cust := customer_id cust_var
FROM orderTable T1,
(SELECT #rownum := 0) r,
(SELECT #prev_cust := '') c
order by customer_id, date_added
) o where o.rn < 3;
SQL DEMO

Answer 3

Here's another (deliberately incomplete) method, though others may have a point about performance...
SELECT x.*
, COUNT(*) rank
FROM ordertable x
JOIN ordertable y
ON y.customer_id = x.customer_id
AND y.date_added <= x.date_added
GROUP
BY x.customer_id
, x.date_added;

Answer 4

This should produce the results you're after, but the outer SELECT won't be the most efficient as it's filtering on a derived table.
SELECT ranked.*
FROM (
SELECT ot.* ,
#rownum := IF( ot.customer_id = #previous , #rownum +1, 1 ) rank,
#previous := ot.customer_id
FROM orderTable ot,
(SELECT #rownum :=1, #previous := NULL) init
ORDER BY customer_id, date_added
) ranked
WHERE rank <=2