I have a table that contains too many records and each bunch of records belong to someone:
---------------------
id | data | username
---------------------
1 | 10 | ali
2 | 11 | ali
3 | 12 | ali
4 | 20 | omid
5 | 21 | omid
6 | 30 | reza
now I want to create a query to result me like this:
1-10-ali 4-20-omid 6-30-reza 2-11-ali 5-21-omid 3-12-ali
Is there anyway to create a query to result me one record per each username and then one from another, and another to the end?
Unfortunately MySQL doesn't have a ranking system so you can use UDV (user defined variables) to rank your records like so.
SELECT id, `data`, name
FROM
( SELECT
id, `data`, name,
@rank := if(@name = name, @rank + 1, 1) as rank,
@name := name
FROM test
CROSS JOIN (SELECT @rank := 1, @name := '') temp
ORDER BY name, `data`
) t
ORDER BY t.rank, t.name, t.data
Output:
+---------------------+
| id | data | name |
+-----+------+--------+
| 1 | 10 | ali |
+---------------------+
| 4 | 20 | omid |
+---------------------+
| 6 | 30 | reza |
+---------------------+
| 2 | 11 | ali |
+---------------------+
| 5 | 21 | omid |
+---------------------+
| 3 | 12 | ali |
+---------------------+
The classic SQL approach is a self join and grouping that lets you determine a row's ranking position by counting the number of rows that come before it. As this is probably slower I doubt I could talk you out of the proprietary method but I mention it to give you an alternative.
select t.id, min(t.`data`), min(t.username)
from test t inner join test t2
on t2.username = t.username and t2.id <= t.id
group by t.id
order by count(*), min(t.username)
Your example would work with
SELECT id, `data`, name
FROM tbl
ORDER BY `data` % 10,
username
`data`;
If data and username do not have the desired pattern, then improve on the example.