我有xdateList表包含:
+---------+
xDateList
+---------+
2018-11-01
2018-11-02
2018-11-03
2018-11-04
2018-11-05
还有表ScanLog
--------------------------------------
ID Name ScanDate Code
--------------------------------------
1 John 2018-11-02 07:00:00 IN
1 John 2018-11-02 10:00:00 OUT
1 John 2018-11-04 08:00:00 IN
1 John 2018-11-04 12:00:00 OUT
我已经尝试过了,但是它不能显示xDateList上的所有记录,它只显示表ScanLog上的记录
select xDateList.date,
scanlog.name,
MIN(scanlog.scandate) AS `IN`,
MAX(scanlog.scandate) AS `OUT`
from scanlog
left JOIN xDateList ON xDateList.date = date(scanlog.scandate)
where scanlog.id='1'
GROUP BY DATE(scanlog.scandate)
我想要这样的结果
--------------------------------------------
Date ID Name In Out
--------------------------------------------
2018-11-01 1 John
2018-11-02 1 John 07:00:00 10:00:00
2018-11-03 1 John
2018-11-04 1 John 08:00:00 12:00:00
2018-11-05 1 John
谢谢您的帮助
答案 0 :(得分:0)
您需要更改LEFT JOIN
中表的顺序。始终记住这一点,以便考虑特定表中的所有行;那个特定的表应该是Join中最左边的表。
此外,每当进行LEFT JOIN时,应在ON
子句中指定右侧表的条件;否则,WHERE子句中的条件可以将其有效地改为INNER JOIN。
此外,在这种情况下,GROUP BY
应该在xDateList.date
上以显示与xDateList.date
值相对应的所有行。并且,我们需要确保SELECT
子句中也指定GROUP BY
列表中所有未聚合的列。请检查:Error related to only_full_group_by when executing a query in MySql
SELECT xDateList.date,
scanlog.name,
MIN(scanlog.scandate) AS `IN`,
MAX(scanlog.scandate) AS `OUT`
FROM xDateList
LEFT JOIN scanlog
ON xDateList.date = date(scanlog.scandate) AND
scanlog.id='1'
GROUP BY xDateList.date, scanlog.name
结果
| date | name | IN | OUT |
| ---------- | ---- | ------------------- | ------------------- |
| 2018-11-01 | | | |
| 2018-11-02 | John | 2018-11-02 07:00:00 | 2018-11-02 10:00:00 |
| 2018-11-03 | | | |
| 2018-11-04 | John | 2018-11-04 08:00:00 | 2018-11-04 12:00:00 |
| 2018-11-05 | | | |