Question

我只是感到困惑。已经尝试搜索整个网站或谷歌，但没有找到“最近的”解决方案。

好吧，假设我有这种表结构。

id date       finger_id finger_time is_enter
1  2017-03-30  2          09:00       1
2  2017-03-30  2          17:13       0
3  2017-03-31  4          09:10       1
4  2017-03-31  3          09:01       1
5. 2017-03-31  3          17:00       0

我想让表格如下所示。

date       finger_id  enter_time  exit_time
2017-03-30   2        09:00       17:13
2017-03-30   4        09:10
2017-03-31   3        09:10       17:00

我已经制作了sql语句，但它变成了这样。

date       finger_id  enter_time   exit_time
2017-03-30  2         09:00
2017-03-30  2                       17:13
2017-03-31  4         09:10
2017-03-31  3         09:01
2017-03-31  3                       17:00

我只是想知道如何通过finger_id列在同一日期将is_enter 1与is_enter 0合并。

这是我的sql查询参考。

SELECT * 
FROM `tbl_fingerprint` 
    LEFT JOIN `tbl_employee` ON `tbl_employee`.`fingerprint_id`=`tbl_fingerprint`.`fingerprint_id` 
    LEFT JOIN `tbl_position` ON `tbl_position`.`position_id`=`tbl_employee`.`position_id` 
WHERE `fingerprint_date` >= '2017-03-01' 
AND `fingerprint_date` <= '2017-04-01' 
GROUP BY `tbl_fingerprint`.`fingerprint_id`, 
         `tbl_fingerprint`.`fingerprint_date`, 
         `tbl_fingerprint`.`is_enter` 
ORDER BY `fingerprint_date` ASC LIMIT 30

感谢您的帮助。

Answer 1

您可以执行group by date和finger_id字段，并在聚合函数中使用条件表达式（case或if()）来获得预期结果。聚合函数中的条件语句只有在is_enter字段中设置了正确的值时才确保它们返回值。我遗漏了员工的详细信息，因为这些不构成您问题的一部分：

SELECT date, fingerprint_id, max(if(is_enter=1,finger_time,null) as enter_time, max(if(is_enter=0,finger_time,null) as exit_time
FROM `tbl_fingerprint` 
WHERE `fingerprint_date` >= '2017-03-01' 
AND `fingerprint_date` <= '2017-04-01' 
GROUP BY `tbl_fingerprint`.`fingerprint_id`, 
         `tbl_fingerprint`.`fingerprint_date`,  
ORDER BY `fingerprint_date` ASC LIMIT 30

Answer 2

SELECT * FROM `tbl_fingerprint`
LEFT JOIN `tbl_employee` ON `tbl_employee`.`fingerprint_id`=`tbl_fingerprint`.`fingerprint_id`
LEFT JOIN `tbl_position` ON `tbl_position`.`position_id`=`tbl_employee`.`position_id`
LEFT JOIN (SELECT * FROM tbl_fingerprint WHERE is_enter = 0) a
    ON a.finger_id = tbl_fingerprint.finger_id AND a.date = tbl_fingerprint.date
WHERE `fingerprint_date` >= '2017-03-01' AND `fingerprint_date` <= '2017-04-01' AND tbl_fingerprint.is_enter = 1
GROUP BY `tbl_fingerprint`.`fingerprint_id`, `tbl_fingerprint`.`fingerprint_date`, `tbl_fingerprint`.`is_enter`
ORDER BY `fingerprint_date` ASC LIMIT 30

Answer 3

试试这个（如果finger_time属于time类型，这将有效）： -

SELECT date, finger_id, min(finger_time) enter_time, if (min(finger_time) = max(finger_time), null, max(finger_time)) exit_time FROM xyz group by finger_id, date

Answer 4

SELECT a1.*, a3.time as time_out FROM attendance as a1
INNER JOIN (SELECT MIN(id) as id FROM attendance where is_enter = '1' group by date, f_id ) as a2
ON a2.id = a1.id
LEFT JOIN attendance as a3 ON a3.date = a1.date AND a1.f_id = a3.f_id and a3.is_enter = '0'

您可能需要将日期转换为不包含时间部分或使用yyyy-mm-dd格式转换为char

Mysql查询将两个条件合并为一行

4 个答案: