single sql query to perform some group by - mysql

Question

CREATE TABLE IF NOT EXISTS `projects` (
`idproject` int(10) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`date` datetime NOT NULL,
`status` enum('new','active','closed') NOT NULL,
`priority` enum('low','medium','high') NOT NULL,
PRIMARY KEY (`idproject`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=20
Here are some data:
INSERT INTO `projects` (`idproject`, `name`, `date`, `status`, `priority`) VALUES
(1, 'iCompany', '2011-03-23 11:41:44', 'new', 'medium'),
(2, 'John Doe & Co.', '2011-04-09 14:38:04', 'closed', 'low'),
(3, 'ACME, Inc.', '2011-05-21 11:43:11', 'active', 'high'),
(4, 'John Doe & Co.', '2011-03-28 15:19:45', 'active', 'low'),
(5, 'John Doe & Co.', '2011-03-08 15:16:32', 'new', 'low'),
(6, 'ACME, Inc.', '2011-04-05 20:58:42', 'active', 'low'),
(7, 'Mega Corp', '2011-04-21 08:08:53', 'new', 'low'),
(8, 'iCompany', '2011-04-17 08:40:36', 'active', 'medium'),
(9, 'iCompany', '2011-05-18 14:36:48', 'active', 'low'),
(10, 'John Doe & Co.', '2011-04-18 19:08:25', 'new', 'medium'),
(11, 'ACME, Inc.', '2011-05-19 13:11:04', 'active', 'low'),
(12, 'Foo Bars', '2011-03-03 17:19:29', 'new', 'high'),
(13, 'ACME, Inc.', '2011-04-23 20:42:33', 'active', 'medium'),
(14, 'Foo Bars', '2011-05-13 09:18:15', 'active', 'medium'),
(15, 'ACME, Inc.', '2011-03-20 14:37:18', 'new', 'low'),
(16, 'Foo Bars', '2011-04-18 13:46:23', 'active', 'high'),
(17, 'iCompany', '2011-05-31 07:13:32', 'closed', 'low'),
(18, 'Foo Bars', '2011-05-31 15:43:39', 'active', 'low'),
(19, 'John Doe & Co.', '2011-05-28 11:28:32', 'active', 'medium')
I'd like to have the list of all projects:
- with their latest (chronologically) status and priority,
- with the number of days between the first and latest entry (0 if there is
only one entry), you may ignore the hours,
- sorted by by priority ('high' first), then by name,
- without the projects where the latest status is 'closed' (omitted from
result).
Output should be:
+---------------+-----------+---------------+----------------+
¦name ¦total_days ¦latest_status ¦latest_priority ¦
+---------------+-----------+---------------+----------------+
¦ACME, Inc. ¦62 ¦active ¦high ¦
¦John Doe & Co. ¦81 ¦active ¦medium ¦
¦Foo Bars ¦89 ¦active ¦low ¦
¦Mega Corp ¦0 ¦new ¦low ¦
+---------------+-----------+---------------+----------------+
So far i got to write this:
SELECT name,status FROM projects group by name order by priority desc,name
please help?

Answer 1

SELECT *
FROM (
SELECT name,
DATEDIFF(MAX(date), MIN(date)) total_days,
(SELECT tt.status FROM projects tt
WHERE t.name = tt.name AND tt.date = MAX(t.DATE)) latest_status,
(SELECT tt.priority FROM projects tt
WHERE t.name = tt.name AND tt.date = MAX(t.DATE)) latest_priority
FROM projects t
GROUP BY name
) t
WHERE latest_status != 'closed'
ORDER BY (CASE latest_priority
WHEN 'high' THEN 0
WHEN 'medium' THEN 1
WHEN 'low' THEN 2
END), name;
total days: take the DATEDIFF of the MAX and MIN date, which will give you the number of days in between;
latest status: fetch the status for which the row's date is equal to the MAX date;
latest priority: fetch the priorityfor which the row's date is equal to the MAX date;
order by: translate each priority string to a numerical value and order by it.
Here's an sqlfiddle.

Answer 2

Try this: http://www.sqlfiddle.com/#!2/b3962/1
select p.name, r.total_days, p.status as latest_status, p.priority as latest_priority
from projects p
join
(
select name, max(date) as recent_date, datediff(max(date),min(date)) as total_days
from projects
group by name
)
-- recent
r on(r.name,r.recent_date) = (p.name,date)
where p.status not in ('closed')
order by
(case latest_priority
when 'high' then 0
when 'medium' then 1
when 'low' THEN 2
end), p.name
Output:
| NAME | TOTAL_DAYS | LATEST_STATUS | LATEST_PRIORITY |
-----------------------------------------------------------------
| ACME, Inc. | 62 | active | high |
| John Doe & Co. | 81 | active | medium |
| Foo Bars | 89 | active | low |
| Mega Corp | 0 | new | low |
If you want to make it more shorter(by using USING), make the alias of recent_date to date: http://www.sqlfiddle.com/#!2/b3962/2
select
p.name, r.total_days, p.status as latest_status, p.priority as latest_priority
from projects p
join
(
select name, max(date) as date, datediff(max(date),min(date)) as total_days
from projects
group by name
) r using(name,date)
where p.status not in ('closed')
order by
(case latest_priority
when 'high' then 0
when 'medium' then 1
when 'low' then 2
end), p.name
How it works
Work it from inside out. First step, find the latest:
select name, max(date) as recent_date, datediff(max(date),min(date)) as total_days
from projects
group by name
Output:
| NAME | DATE | TOTAL_DAYS |
--------------------------------------------------------------
| ACME, Inc. | May, 21 2011 11:43:11-0700 | 62 |
| Foo Bars | May, 31 2011 15:43:39-0700 | 89 |
| iCompany | May, 31 2011 07:13:32-0700 | 69 |
| John Doe & Co. | May, 28 2011 11:28:32-0700 | 81 |
| Mega Corp | April, 21 2011 08:08:53-0700 | 0 |
Final step, join the above results to main table:
select
p.name, r.total_days, p.status as latest_status, p.priority as latest_priority
from projects p
join
(
select name, max(date) as date, datediff(max(date),min(date)) as total_days
from projects
group by name
) r using(name,date)
where p.status not in ('closed')
order by
(case latest_priority
when 'high' then 0
when 'medium' then 1
when 'low' then 2
end), p.name
For overriding the ordering, use CASE statement on ORDER BY clause.
Output:
| NAME | TOTAL_DAYS | LATEST_STATUS | LATEST_PRIORITY |
-----------------------------------------------------------------
| ACME, Inc. | 62 | active | high |
| John Doe & Co. | 81 | active | medium |
| Foo Bars | 89 | active | low |
| Mega Corp | 0 | new | low |