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mysql - 如何将mysql行转换为没有标签的列

转载 作者:行者123 更新时间:2023-11-30 21:31:48 24 4
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我有一个带有行的 mysql 表,它基本上是用户的进出时间,但它没有任何行是进出行的标记。

找出答案的唯一方法是为每个用户的每个日期计算最短和最长时间,然后在两列中选择它们。但是我很难做到这一点..请帮忙

My current data looks like this:

+-----------+--------------+-----------------------+
| "User Id" | Date | "Time" |
+-----------+--------------+-----------------------+
| userA | "2019-04-22" | "2019-04-22 08:48:23" |
+-----------+--------------+-----------------------+
| userA | "2019-04-22" | "2019-04-22 17:47:51" |
+-----------+--------------+-----------------------+
| userB | "2019-04-22" | "2019-04-22 08:19:52" |
+-----------+--------------+-----------------------+
| userB | "2019-04-22" | "2019-04-22 18:06:27" |
+-----------+--------------+-----------------------+
| userC | "2019-04-22" | "2019-04-22 08:27:28" |
+-----------+--------------+-----------------------+
| userc | "2019-04-22" | "2019-04-22 17:19:25" |
+-----------+--------------+-----------------------+

Desired Output is This:
+--------+--------------+-----------------------+-----------------------+
| Userid | Date | InTime | OutTime |
+--------+--------------+-----------------------+-----------------------+
| userA | "2019-04-22" | "2019-04-22 08:48:23" | "2019-04-22 17:47:51" |
+--------+--------------+-----------------------+-----------------------+
| userB | "2019-04-22" | "2019-04-22 08:19:52" | "2019-04-22 18:06:27" |
+--------+--------------+-----------------------+-----------------------+
| userC | "2019-04-22" | "2019-04-22 08:27:28" | "2019-04-22 17:19:25" |
+--------+--------------+-----------------------+-----------------------+

提前致谢。

最佳答案

min()max()group by 一起使用

select Userid , Date, min(`time`) as intime,max(`time`) as outtime
from tablename
group by Userid , Date

关于mysql - 如何将mysql行转换为没有标签的列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55810379/

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