MySQL 5.5 - Select rows where record changes - mysql

Question

I have got this table and I need to produce the line with the last A and first B and then last B and first A. Expected results are here:
ID | Date | OPS
6 | 2018-12-20 | A
7 | 2018-12-20 | B
11 | 2018-12-24 | B
12 | 2018-12-25 | A
I have tried to do with the below code with nil luck. I cannot use analytic functions like ROW_NUMBER(), LEAD () or LAG ().
SELECT *
FROM T1
JOIN (
SELECT COUNT(*) cnt, t1.OPS
FROM (
SELECT DISTINCT ts1.OPS
FROM T1 ts1
JOIN T1 tx1 ON ts1.OPS = tx1.OPS
) t1
JOIN (
SELECT DISTINCT ts2.OPS
FROM T1 ts2
JOIN T1 tx2 ON ts2.OPS = tx2.OPS
) t2 on t1.OPS <= t2.OPS
GROUP BY t1.OPS
) tt ON T1.OPS = tt.OPS
My table is:
CREATE TABLE t1(
ID int,
Date1 date,
OPS CHAR
);
INSERT INTO T1 VALUES
( 1, '2018-12-17', 'A'),
( 2, '2018-12-18', 'A'),
( 3, '2018-12-19', 'A'),
( 4, '2018-12-19', 'A'),
( 5, '2018-12-19', 'A'),
( 6, '2018-12-20', 'A'),
( 7, '2018-12-20', 'B'),
( 8, '2018-12-21', 'B'),
( 9, '2018-12-22', 'B'),
(10, '2018-12-23', 'B'),
(11, '2018-12-24', 'B'),
(12, '2018-12-25', 'A'),
(13, '2018-12-26', 'A'),
(14, '2018-12-27', 'A'),
(15, '2018-12-28', 'A');

Answer 1

Without LEAD and LAG and considering that you have multiple records per-date you could use a horrendous simplified solution which assumes that the records are ordered by id instead of date. It does not require ids to be contiguous and works for more than two ops values:
SELECT *
FROM t AS ref
WHERE EXISTS (
SELECT 1
FROM t AS lag
WHERE ops <> ref.ops
AND id < ref.id
AND NOT EXISTS (
SELECT 1
FROM t AS tmp
WHERE id > lag.ID
AND id < ref.id
)
) OR EXISTS (
SELECT 1
FROM t AS led
WHERE ops <> ref.ops
AND id > ref.id
AND NOT EXISTS (
SELECT 1
FROM t AS tmp
WHERE id < led.ID
AND id > ref.id
)
)
Basically for each row, the query searches all previous rows where ids do not match then determines if there is a row in between. It searches forward likewise.

Answer 2

Try the following logic, which phrases your question using a series of unions:
(SELECT ID, Date1, OPS FROM T1 WHERE OPS = 'B' ORDER BY Date1 LIMIT 1)
UNION ALL
(SELECT ID, Date1, OPS FROM T1
WHERE OPS = 'A' AND Date1 <= (SELECT MIN(Date1) FROM T1 WHERE OPS = 'B')
ORDER BY Date1 DESC LIMIT 1)
UNION ALL
(SELECT ID, Date1, OPS FROM T1 WHERE OPS = 'B' ORDER BY Date1 DESC LIMIT 1)
UNION ALL
(SELECT ID, Date1, OPS FROM T1
WHERE OPS = 'A' AND Date1 >= (SELECT MAX(Date1) FROM T1 WHERE OPS = 'B')
ORDER BY Date1 LIMIT 1)
ORDER BY Date1, OPS;
Demo
Note generally that this is a gaps and islands problem. These are very difficult to handle without using analytic functions, which you appear to not have available. If you were using MySQL 8+, then we could phrase a much cleaner looking query.

Answer 3

if we are allowed max and min function, then it is pretty easy to calculate these 4 individually and union the individual results.
-- To select last A
select ID, Date1, OPS
from test_t1
where test_t1.OPS = 'A'
and test_t1.Date1 in
(
select max(t1_a.Date1) as Date1
from
(select ID, Date1, OPS
from test_t1
where OPS = 'A') t1_a inner join
(select min(Date1) as Date1
from test_t1
where OPS = 'B') t1_min_b
on t1_a.Date1 <= t1_min_b.Date1
)
union
-- To select first B
select ID, Date1, OPS
from test_t1
where test_t1.OPS = 'B'
and test_t1.Date1 in
(
select min(Date1) as Date1
from test_t1
where OPS = 'B'
)
union
/*
To select last B
/
select ID, Date1, OPS
from test_t1
where test_t1.OPS = 'B'
and test_t1.Date1 in
(
select max(Date1) as Date1
from test_t1
where OPS = 'B'
)
union
/
To select first A
*/
select ID, Date1, OPS
from test_t1
where test_t1.OPS = 'A'
and test_t1.Date1 in
(
select min(t1_a.Date1) as Date1
from
(select ID, Date1, OPS
from test_t1
where OPS = 'A') t1_a inner join
(select max(Date1) as Date1
from test_t1
where OPS = 'B') t1_max_b
on t1_a.Date1 >= t1_max_b.Date1
)
;

Answer 4

You could use a correlated subquery that computed the min and max ids for each ops, like
SELECT t.*
FROM T1 t
WHERE
t.id IN (
SELECT MIN(id) FROM T1 WHERE ops = t.ops
UNION ALL SELECT MAX(id) FROM T1 WHERE ops = t.ops
)
ORDER BY t.ops, t.id DESC