SQLite3。使用FKs、MAX()和GROUP BY仅选择最高版本 关于架构和数据,您需要了解的是:
工作日表格包括:SQLite3。使用FKs、MAX()和GROUP BY仅选择最高版本 关于架构和数据,您需要了解的是:,sql,sqlite,Sql,Sqlite,工作日表格包括: id INTEGER, revision INTEGER, .... PRIMARY KEY(id, revision) id -- integer primary key, quite unimportant I think businessday_id -- FK businessday_revision -- FK, when a day is edited, a new revision is created 收
id INTEGER,
revision INTEGER,
....
PRIMARY KEY(id, revision)
id -- integer primary key, quite unimportant I think
businessday_id -- FK
businessday_revision -- FK, when a day is edited, a new revision is created
收入表包括:
id INTEGER,
revision INTEGER,
....
PRIMARY KEY(id, revision)
id -- integer primary key, quite unimportant I think
businessday_id -- FK
businessday_revision -- FK, when a day is edited, a new revision is created
外键如下所示:
FOREIGN KEY(businessday_id, businessday_revision) REFERENCES businessday(id, revision) ON DELETE CASCADE,
问题
我只想选择每天最新修订的收入应该是308行。
但可悲的是,我太笨了,想不出来。我发现我可以使用以下工具获得所有最新的工作日修订版:
SELECT id, MAX(revision)
FROM businessday
GROUP BY id;
有什么方法可以用这些数据来选择我的收入吗?大致如下:
-- Pseudo-code:
SELECT *
FROM income i
WHERE i.businessday_id = businessday.id THAT EXISTS IN
(SELECT id, MAX(revision)
FROM businessday
GROUP BY id);
我显然没有线索,请告诉我正确的方向 使用join怎么样
SELECT i.*
FROM income i
INNER JOIN
(
SELECT id, MAX(revision) revision
FROM businessday
GROUP BY id
) s ON i.businessday_id = s.id AND
i.businessday_revision = s.revision
这应该起作用:
SELECT i.*
FROM Income i
INNER JOIN (
SELECT id, MAX(revision) maxrevision
FROM businessDay
GROUP BY id
) t ON i.businessday_id = t.id AND i.businessday_revision = t.maxrevision
哇,我很惊讶。在短短的几秒钟内,您就可以将正在工作的SQL输入到我已经花了几个小时的东西上。谢谢。@ippi——没问题,很高兴我能帮忙!