我需要 MySQL 查询方面的帮助
我有这个百合:
表A
- page_id
- rev_id
- 日期
一个 page_id 可以有多个 rev_id
表B
- rev_id
- 单词
我有每个修订版中的单词
我需要为每个日期返回我在该日期中拥有的单词数量 每个 page_id 中的最后一个 rev_id
示例:
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'
此处神奇的 SQL 已编辑以显示其计算方式 {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] }
我执行了此查询,但我的日期是固定的,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;
<小时/>
已解决
我的 friend 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;
最佳答案
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。这是一个累积的事情。这使得事情变得更加复杂。
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,这在 MySQL 中不起作用。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;
关于MySQL 查询 GROUP BY 两列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48374195/