我仍然对选择date_of_highest_points有问题:
$qry1 = mysql_query("INSERT INTO bbdnes_hraci (nick, sumfrags, sumpoints,
sumhours, lastdate1, highest_points, date_of_highest_points)
SELECT DISTINCT nick, SUM(frags), SUM(points), SUM(hours),
MAX(lastdate), MAX(points), ??????
FROM hraci
GROUP BY nick "); 我在这里问过这个问题,https://stackoverflow.com/questions/18817900/select-in-select,但我没有解决它,尽管议会。有人能帮我处理密码吗?
表: hraci
nick frags points hours lastdate
Gamer1 20 100 1 2013-09-17 22:16:08
Gamer1 30 150 3 2013-09-18 20:17:15
Gamer1 25 125 0.5 2013-09-18 23:16:06
Gamer2 50 250 4 2013-09-17 21:11:30
Gamer2 5 25 2 2013-09-17 23:13:59需要:
表: bbdnes_hraci
nick sumfrags sumpoints sumhours lastdate1 highest_points date_of_highest_points
Gamer1 75 375 4.5 2013-09-18 23:16:06 150 2013-09-18 20:17:15 ??
Gamer2 55 275 6 2013-09-17 23:13:59 250 2013-09-17 21:11:30 ??发布于 2013-09-18 17:30:22
你可以这样做:
select nick, SUM(frags) sumfrags, SUM(points) sumpoints, SUM([hours]) sumhours, max(lastdate) lastdate, max(points) highest_points
into #t1
from hraci
group by Nick
select t1.*, x.lastdate date_of_highest_points
from #t1 t1
outer apply
(select top 1 lastdate
from hraci t
where t.Nick = t1.Nick and t.points = t1.highest_points
order by t.lastdate desc) x发布于 2013-09-18 15:28:00
尝尝这个
select * from table_name where highest_points = (select max(highest_points) from table_name)发布于 2013-09-18 15:33:55
未经测试:
SELECT DISTINCT nick, SUM(frags), SUM(points), SUM(hours),
MAX(lastdate), MAX(highest_points),
(
SELECT lastdate As date_of_highest_points
FROM hraci
WHERE points =(SELECT MAX(points) FROM hraci)
)
(rest of query. i.e. FROM and ORDER By's)https://stackoverflow.com/questions/18876134
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