Question

您需要了解的架构和数据：

SELECT * FROM 'income';  -- Returns all 309 rows.
SELECT * FROM 'income' WHERE businessday_revision = 0;  -- 308 rows
SELECT * FROM 'income' WHERE businessday_revision = 1;  -- 1 row

businessday表有：

id INTEGER,
revision INTEGER,
....
PRIMARY KEY(id, revision)

收入表有：

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

我想从每天的最新版本中选择收入。 哪一行应为308行。 但遗憾的是，我太密集了，无法弄明白。我发现我可以使用这个获得所有最新的businessday修订版：

SELECT id, MAX(revision)
FROM businessday
GROUP BY id;

我有什么办法可以用这些数据来选择我的收入吗？有点像：

-- Pseudo-code:
SELECT *
FROM income i
WHERE i.businessday_id = businessday.id THAT EXISTS IN
    (SELECT id, MAX(revision)
    FROM businessday
    GROUP BY id);

我显然在这里没有任何线索，请指出正确的方向！

Answer 1

如何使用join？

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

Answer 2

这应该有效：

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

sqlite3的。使用FK，MAX（）和GROUP BY仅选择最高版本

您需要了解的架构和数据：

问题

2 个答案: