如何循环MySQL中的行并根据下一行的值插入数据？
这是我在MySQL中的现有表如何循环MySQL中的行并根据下一行的值插入数据？,mysql,sql,mysql-workbench,Mysql,Sql,Mysql Workbench,这是我在MySQL中的现有表 name UUID date_time org_id SK null 2017-04-19 01:36:34 2 SK null 2017-04-19 01:36:53 2 SK null 2017-04-19 01:37:23 2 TK null 2017-02-11 01:37:20 5 TK null 2017-02-11 01:37:31 5 KK nul
name UUID date_time org_id
SK null 2017-04-19 01:36:34 2
SK null 2017-04-19 01:36:53 2
SK null 2017-04-19 01:37:23 2
TK null 2017-02-11 01:37:20 5
TK null 2017-02-11 01:37:31 5
KK null 2017-01-01 11:37:33 1
name UUID date_time org_id
SK we23 2017-04-19 01:36:34 2
SK we23 2017-04-19 01:36:53 2
SK we23 2017-04-19 01:37:23 2
TK rt56 2017-02-11 01:37:20 5
TK rt56 2017-02-11 01:37:31 5
KK yu77 2017-01-01 11:37:33 1
请让我知道如何使用MySQL workbench编写SQL脚本来完成此操作。您可以使用MySQL来完成此操作，但这相当复杂，因为它需要几个子查询
SELECT
name, date_time, org_id, uuid
FROM
(
SELECT
`name`, `date_time`, `org_id`,
(SELECT
sum(step)
FROM
(
SELECT
t1.`name`, t1.`date_time`, t1.`org_id`,
case when exists
(SELECT 1
FROM t AS t2
WHERE t2.`name` = t1.`name`
AND timestampdiff(second, t1.date_time, t2.date_time) < 60
AND t2.date_time > t1.date_time
)
then 0
else 1
end AS step
FROM
t AS t1
) AS ttt
WHERE
ttt.name <= t.name AND ttt.date_time <= ttt.date_time
) AS group_number
FROM
t
) AS ttt2
JOIN
(
SELECT 1 AS group_number, UUID() AS `uuid`
UNION
SELECT 2 AS group_number, UUID() AS `uuid`
UNION
SELECT 3 AS group_number, UUID() AS `uuid`
-- and as many as needed
) u ON u.group_number = ttt2.group_number
ORDER BY
name, date_time ;
tt2ttt2ttt我还找到了这个问题的近似解决方案：
我的问题是：
update table as u,
(SELECT UUID() as UUID, name, org_id,date_time
FROM table
group by name, org_id, DATE_FORMAT(date_time, "%Y-%m-%e %H-%i")) as b
set u.UUID = b.UUID
where (u.name = b.name OR (u.name is null and u.name is null))
and (u.org_id = b.org_id OR ( u.org_id is null and b.org_id is null))
and DATE_FORMAT(u.date_time, "%Y-%m-%e %H-%i") = DATE_FORMAT(b.date_time, "%Y-%m-%e %H-%i");
您的表是否包含标识行的唯一ID列？UUID来自何处？是。。我有一个自动递增的ID列，UUID将由MySQL的UUID（）函数生成。。只是基于上述条件，它在团体中应该是独一无二的。不客气，沙维尔。如果你认为这个问题已经被问过了，那么或多或少习惯性的做法是将其标记为已被接受，这样人们就可以把注意力集中在大量尚未回答的问题上。 name | date_time | org_id | group_number :--- | :------------------ | -----: | -----------: KK | 2017-01-01 11:37:33 | 1 | 1 SK | 2017-04-19 01:36:34 | 2 | 2 SK | 2017-04-19 01:36:53 | 2 | 2 SK | 2017-04-19 01:37:23 | 2 | 2 TK | 2017-02-11 01:37:20 | 5 | 3 TK | 2017-02-11 01:37:31 | 5 | 3
SELECT
name, date_time, org_id, uuid
FROM
ttt2
JOIN
(
SELECT 1 AS group_number, UUID() AS `uuid`
UNION
SELECT 2 AS group_number, UUID() AS `uuid`
UNION
SELECT 3 AS group_number, UUID() AS `uuid`
UNION
SELECT 4 AS group_number, UUID() AS `uuid`
-- and as many as you might need
) u ON u.group_number = ttt2.group_number
ORDER BY
name, date_time ;
update table as u,
(SELECT UUID() as UUID, name, org_id,date_time
FROM table
group by name, org_id, DATE_FORMAT(date_time, "%Y-%m-%e %H-%i")) as b
set u.UUID = b.UUID
where (u.name = b.name OR (u.name is null and u.name is null))
and (u.org_id = b.org_id OR ( u.org_id is null and b.org_id is null))
and DATE_FORMAT(u.date_time, "%Y-%m-%e %H-%i") = DATE_FORMAT(b.date_time, "%Y-%m-%e %H-%i");