MySQL - SELECT query - oldest and youngest date - mysql

Question

I have to build one SQL SELECT query. Table structure is as follows:
[ID] [PHONE] [description] [DATE]
[1] [600898367] [main] [2016-01-23]
[2] [600898367] [] [2016-01-24]
[3] [600898367] [] [2016-01-26]
[4] [600898367] [] [2016-01-28]
[5] [662349093] [main] [2016-01-10]
[6] [662349093] [] [2016-01-21]
[7] [662349093] [] [2016-01-30]
[8] [662349093] [] [2016-01-31]
You have here different records grouped within the same telephone number. The first (the oldest) occurance is marked with [main] flag. There's no two identical numbers with [main] flag.
I want to select each [main] record and additionaly one youngest with the same phone number, so the result should give records 1,4,5,8.
Please help.

Answer 1

Use a WHERE clause to give you the records with the main flag. Use MAX to get the most recent record and JOIN to get the additional columns. Finally, do a UNION ALL to combine the result.
-- Get the main records first
SELECT *
FROM tbl
WHERE description = 'main'
UNION ALL
-- Get the most recent records
SELECT b.*
FROM (
SELECT
t.PHONE,
MAX(DATE) AS MaxDate
FROM tbl t
GROUP BY PHONE
) a
INNER JOIN tbl b -- Do a JOIN to get the additional columns
ON b.PHONE = a.PHONE
AND b.DATE = a.MaxDate

Answer 2

Try this;)
SQL Fiddle
MySQL 5.6 Schema:
CREATE TABLE table1
(`ID` int, `PHONE` int, `description` varchar(4), `DATE` varchar(11))
;
INSERT INTO table1
(`ID`, `PHONE`, `description`, `DATE`)
VALUES
(1, 600898367, 'main', '2016-01-23'),
(2, 600898367, NULL, '2016-01-24'),
(3, 600898367, NULL, '2016-01-26'),
(4, 600898367, NULL, '2016-01-28'),
(5, 662349093, 'main', '2016-01-10'),
(6, 662349093, NULL, '2016-01-21'),
(7, 662349093, NULL, '2016-01-30'),
(8, 662349093, NULL, '2016-01-31')
;
Query 1:
select t.*
from table1 t
inner join (
select `PHONE`, max(`DATE`) as `DATE` from table1 group by `PHONE`
) t1 on t.`PHONE` = t1.`PHONE` and (t.`DATE` = t1.`DATE` or t.`description` = 'main')
order by t.`ID`
Results:
| ID | PHONE | description | DATE |
|----|-----------|-------------|------------|
| 1 | 600898367 | main | 2016-01-23 |
| 4 | 600898367 | (null) | 2016-01-28 |
| 5 | 662349093 | main | 2016-01-10 |
| 8 | 662349093 | (null) | 2016-01-31 |

Answer 3

You can use the following query:
SELECT t1.*, t3.*
FROM mytable AS t1
LEFT JOIN (
SELECT PHONE, MAX(date) AS max_date
FROM mytable
GROUP BY PHONE
) AS t2 ON t1.PHONE = t2.PHONE
LEFT JOIN mytable AS t3 ON t1.PHONE = t3.PHONE AND t2.max_date = t3.`date`
WHERE t1.description = 'main'

Answer 4

With a GROUP BY we'll first group on both PHONE and description, thus getting 4 rows.
Next we'll create a comma separated set using GROUP_CONCAT. It can take an ORDER BY clause, to order the phone numbers by date.
Last we want to get the first item from the set, we can do this with SUBSTRING_INDEX.
SELECT
SUBSTRING_INDEX(GROUP_CONCAT(`PHONE` ORDER BY `DATE`), ',', 1) AS `PHONE`,
description
FROM table1
GROUP BY `PHONE`, `description`;
See Fiddle

Answer 5

Try the following query..
SELECT
A.*
FROM
`old_young` A
INNER JOIN
(SELECT
MIN(`DATE`) AS Res
FROM
old_young
WHERE description = 'main'
GROUP BY PHONE
UNION
ALL
SELECT
MAX(`DATE`) AS Res
FROM
old_young
WHERE description = ''
GROUP BY PHONE) B
ON A.DATE = B.Res ;
Check the FIDDLE