My table is:
id home datetime player resource ---|-----|------------|--------|--------- 1 | 10 | 04/03/2009 | john | 399 2 | 11 | 04/03/2009 | juliet | 244 5 | 12 | 04/03/2009 | borat | 555 3 | 10 | 03/03/2009 | john | 300 4 | 11 | 03/03/2009 | juliet | 200 6 | 12 | 03/03/2009 | borat | 500 7 | 13 | 24/12/2008 | borat | 600 8 | 13 | 01/01/2009 | borat | 700
我需要选择每个home包含的最大值datetime。
home
datetime
Result would be:
id home datetime player resource ---|-----|------------|--------|--------- 1 | 10 | 04/03/2009 | john | 399 2 | 11 | 04/03/2009 | juliet | 244 5 | 12 | 04/03/2009 | borat | 555 8 | 13 | 01/01/2009 | borat | 700
I have tried:
-- 1 ..by the MySQL manual: SELECT DISTINCT home, id, datetime AS dt, player, resource FROM topten t1 WHERE datetime = (SELECT MAX(t2.datetime) FROM topten t2 GROUP BY home) GROUP BY datetime ORDER BY datetime DESC
不起作用 尽管数据库保留187个,但结果集有130行home。结果包括的一些重复项。
-- 2 ..join SELECT s1.id, s1.home, s1.datetime, s1.player, s1.resource FROM topten s1 JOIN (SELECT id, MAX(datetime) AS dt FROM topten GROUP BY id) AS s2 ON s1.id = s2.id ORDER BY datetime
不。提供所有记录。
-- 3 ..something exotic:
具有各种结果。
你好亲密!您需要做的就是选择住所及其最大日期时间,然后再加入到topten两个字段的表中:
topten
SELECT tt.* FROM topten tt INNER JOIN (SELECT home, MAX(datetime) AS MaxDateTime FROM topten GROUP BY home) groupedtt ON tt.home = groupedtt.home AND tt.datetime = groupedtt.MaxDateTime