[英]MySQL query GROUP BY two columns
我需要有关MySQL查询的帮助
我有这个场景:
表A
-page_id
-rev_id
-日期
一个page_id可以有多个rev_id
表B
-rev_id
-字
我每个修订版都有什么用
我需要为每个日期返回每个page_id中上一个rev_id中包含的单词数量
例:
table A
page_id | rev_id | date
---------------------------------
1 | 231 | 2002-01-01
2 | 345 | 2002-10-12
1 | 324 | 2002-10-13
3 | 348 | 2003-01-01
-
table B
rev_id | words
---------------
231 | 'ask'
231 | 'the'
231 | 'if'
345 | 'ask'
324 | 'ask'
324 | 'if'
348 | 'who'
此处编辑了神奇的sql,以显示其计算方式{page_id:[words]}
date | count(words)
--------------------------
2002-01-01 | 3 { 1:[ask, the, if] }
2002-10-12 | 4 { 1:[ask, the, if], 2:[ask] }
2002-10-13 | 3 { 1:[ask, if], 2:[ask] }
2003-01-01 | 4 { 1:[ask, if], 2:[ask], 3:[who] }
我做了这个查询,但是我的日期是固定的,我需要表修订中包含的所有日期:
SELECT SUM(q)
FROM (
SELECT COUNT(equation) q
FROM revision r, equation e
WHERE r.rev_id in (
SELECT max(rev_id)
FROM revision
WHERE date < '2006-01-01'
GROUP BY page_id
)
AND r.rev_id = e.rev_id
GROUP BY date
) q;
我的朋友帮助我创建查询来解决我的问题!
select s.date, count(words) from
(select d.date, r.page_id, max(r.rev_id) as rev_id
from revision r, (select distinct(date) from revision) d
where d.date >= r.date group by d.date, r.page_id) s
join words e on e.rev_id = s.rev_id
group by s.date;
我认为这是一个基本的join
和group by
:
select a.date, count(*)
from a join
b
on a.rev_id = b.rev_id
group by a.date;
编辑:
哦,我想我明白了。 这是累积的事情。 这使其变得更加复杂。
select d.date,
(select count(*)
from a join
b
on a.rev_id = b.rev_id
where a.date <= d.date and
a.rev_id = (select max(a2.rev_id) from a a2 where a2.date = a.date and a2.date <= d.date)
) as cnt
from (select date from a) d;
但是由于相关子句的嵌套,在MySQL中这是行不通的。 因此,我们可以将逻辑重构为:
select a.date, count(*)
from (select a.*,
(select max(a2.rev_id)
from a a2
where a2.date <= a.date and a2.page_id = a.page_id
) as last_rev_id
from a
) a join
b
on a.last_rev_id = b.rev_id
group by a.date;
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