Example table:
id computer app version build date
---|---------|------|------------|-------|---------
1 | aaaa1 | app1 | 1.0.0 | 1 | 2013-11-11 09:51:07
2 | aaaa1 | app2 | 2.0.0 | 2 | 2013-11-12 09:51:07
5 | xxxx2 | app1 | 1.0.0 | 1 | 2013-11-13 09:51:07
3 | cccc3 | app2 | 3.1.0 | 1 | 2013-11-14 09:51:07
4 | xxxx2 | app1 | 1.0.0 | 2 | 2013-11-15 09:51:07
5 | cccc3 | app2 | 3.1.1 | 3 | 2013-11-16 09:51:07
6 | xxxx2 | app1 | 1.0.2 | 1 | 2013-11-17 09:51:07
7 | aaaa1 | app1 | 1.0.2 | 3 | 2013-11-18 09:51:07
Desired output (not exact format or listing order), getting latest install for each app on each computer:
7. aaaa1 - app1 - 1.0.2 - 3 - 2013-11-18 09:51:07
2. aaaa1 - app2 - 2.0.0 - 2 - 2013-11-12 09:51:07
6. xxxx2 - app1 - 1.0.2 - 1 - 2013-11-17 09:51:07
5. cccc3 - app2 - 3.1.1 - 3 - 2013-11-16 09:51:07
My SQL statement:
SELECT
id,
computer,
app,
version,
build,
MAX(date) AS installed
FROM
data
WHERE
placement = 'xxx'
GROUP BY
app, computer
;
This gives me:
1. aaaa1 - app1 - 1.0.0 - 1 - 2013-11-11 09:51:07
and not
7. aaaa1 - app1 - 1.0.2 - 3 - 2013-11-18 09:51:07
as I expected.
MAX(date) works if I ONLY select MAX(date) and nothing else. But then I don't get any data to work with (just latest date).
SELECT
MAX(date) AS installed
I'm not an SQL ninja so I will soon go bald by scratching my head because of this.
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