Question

MySQL版本为 8.0.16

我想获取给定日期的总休息时间，表格如下：

id | employee_id  | check_in             | check_out           
--------------------------------------------------------------
 1 | 2103         | 2019-09-15 07:30:00  | 2019-09-15 09:20:00  
 2 | 2103         | 2019-09-15 10:35:00  | null
 3 | 2103         | 2019-09-16 08:00:00  | 2019-09-16 10:00:00
 4 | 2103         | 2019-09-16 11:00:00  | 2019-09-16 18:00:00
 5 | 2095         | 2019-09-16 08:30:00  | 2019-09-16 18:30:00

我的预期结果

id | employee_id  | check_in             | check_out          |  breaks 
--------------------------------------------------------------------------
 1 | 2103         | 2019-09-15 07:30:00  | null               |  75
 3 | 2103         | 2019-09-16 08:00:00  | 2019-09-16 18:00:00|  60
 5 | 2095         | 2019-09-16 08:30:00  | 2019-09-16 18:30:00|  0

我的查询仅返回雇员ID，日期，first_check_in，last_check_out列

SELECT 
    a1.employee_id,
    cast(a1.check_in as date) AS date,
    MIN(a1.check_in) AS first_check_in,
    CASE WHEN a2.employee_id IS NULL 
            THEN MAX(a1.check_out)     
        ELSE NULL
    END AS last_check_out
    FROM  attendance_employees AS a1  
    LEFT JOIN attendance_employees AS a2   
        ON a1.employee_id = a2.employee_id
        AND  CAST(a1.check_in AS date) =  CAST(a2.check_in AS date)
        AND a2.check_out IS NULL
    WHERE  1 = 1
    AND DATE(a1.check_in) in ('2019-09-15', '2019-09-16')
    GROUP BY  a1.employee_id, 
    CAST(a1.check_in AS date)
    ORDER BY  a1.employee_id,    
    date

有没有一种方法可以计算MySQL的中断时间？谢谢

更新我的完整查询，其中包含预期工作时间和休息时间范围的时间段

SELECT  CONCAT(IFNULL(employees.first_name,''),' ',IFNULL(employees.surname,'')) as full_name,
        teams.description,
        time_periods.start_time as start,
        time_periods.end_time as end,
        time_periods.time_period_type,
        time_groups.name,
        cast(a1.check_in as date) AS date,
        MIN(a1.check_in) AS first_check_in,
        CASE WHEN a2.employee_id IS NULL 
                THEN MAX(a1.check_out)     
            ELSE NULL
        END AS last_check_out
        FROM  attendance_employees AS a1  
        LEFT JOIN attendance_employees AS a2   
            ON a1.employee_id = a2.employee_id
            AND  CAST(a1.check_in AS date) =  CAST(a2.check_in AS date)
            AND a2.check_out IS NULL
        JOIN employees
            ON employees.id = a1.employee_id
        JOIN teams
            ON employees.team_id = teams.id
        JOIN time_groups
            ON teams.time_group_id = time_groups.id
        LEFT JOIN day_time_group_time_period
            ON time_groups.id = day_time_group_time_period.time_group_id
            AND day_time_group_time_period.day_id = 3
        LEFT JOIN time_periods
            ON day_time_group_time_period.time_period_id = time_periods.id
        WHERE  1 = 1
        AND DATE(a1.check_in) in ('2019-09-15', '2019-09-16') 
        GROUP BY  a1.employee_id, 
        CAST(a1.check_in AS date),    
        a2.employee_id,
        time_periods.start_time,
        time_periods.end_time,
        time_periods.time_period_type
        ORDER BY  a1.employee_id,    
        date

给出以下结果：

full_name| start   | end     | type| name     | date      | first_check_in     | last_check_out
-----------------------------------------------------------------------------------------------------
EMP1     | 08:00:00| 16:30:00| 1   | Day Shift| 2019-09-16| 2019-09-16 08:45:00| 2019-09-16 19:30:00
EMP1     | 10:00:00| 10:30:00| 2   | Day Shift| 2019-09-16| 2019-09-16 08:45:00| 2019-09-16 19:30:00
EMP1     | 12:00:00| 12:30:00| 2   | Day Shift| 2019-09-16| 2019-09-16 08:45:00| 2019-09-16 19:30:00
EMP2     | 08:00:00| 16:30:00| 1   | Day Shift| 2019-09-15| 2019-09-15 07:30:00| NULL
EMP2     | 10:00:00| 10:30:00| 2   | Day Shift| 2019-09-15| 2019-09-15 07:30:00| NULL
EMP2     | 12:00:00| 12:30:00| 2   | Day Shift| 2019-09-15| 2019-09-15 07:30:00| NULL

哪里类型1用于预期的工作时间范围，并且类型2用于预期的休息时间范围

Answer 1

在您最初的问题中，可以使用以下方法：

SELECT   dt.employee_id,
         Min(dt.check_in) AS check_in,
         CASE WHEN COUNT(dt.check_out) = COUNT(*) THEN MAX(dt.check_out)
              ELSE NULL
         END AS check_out,
         Sum(dt.break_interval) AS break_interval
FROM     (
                  SELECT   employee_id,
                           check_in,
                           check_out,
                           timestampdiff(minute, Lag(check_out) over w, check_in) AS break_interval
                  FROM     attendance_employees
                  WHERE    check_in BETWEEN '2019-09-15 00:00:00' AND      '2019-09-16 23:59:59'
                  WINDOW w AS (partition BY employee_id, date(check_in) ORDER BY check_in ASC) ) dt
GROUP BY dt.employee_id,
         date(dt.check_in);

| employee_id | check_in            | check_out           | break_interval |
| ----------- | ------------------- | ------------------- | -------------- |
| 2095        | 2019-09-16 08:30:00 | 2019-09-16 18:30:00 |                |
| 2103        | 2019-09-15 07:30:00 |                     | 75             |
| 2103        | 2019-09-16 08:00:00 | 2019-09-16 18:00:00 | 60             |

View on DB Fiddle

详细信息：

在MySQL 8+中，我们可以使用诸如LAG()之类的Analytic / Window函数来计算同一日期的雇员的前次 check_out时间。为此，我们的分区窗口将位于雇员上方，日期为check_in。分区窗口将按check_in时间排序，以便LAG()仅返回前一个条目。
我们可以使用TimeStampDiff()函数在当前行的检入时间和前一行的检出时间之间以分钟为单位确定break_interval。
一旦我们确定了每一行的这些值；我们可以在子查询（Derived Table）中使用它们，并对特定日期的员工进行break_interval的汇总。另外，如果在特定日期的员工最后一个条目中没有check_out，那么check_out时间将需要一些特殊的处理，因为您需要在那里NULL和MAX(..)将不会返回NULL。
您还可以通过在DATE()时避免使用check_in函数并使用Range条件来进行查询sargeable。

有没有一种方法可以计算first_check_in和last_check_out之间的总休息时间？

1 个答案: