如何将“病假”总结到订单结束 - mysql答案

【问题标题】：How to sumarize "sickness time" to the end of order - mysql如何将“病假”总结到订单结束 - mysql
【发布时间】：2019-03-25 15:52:00
【问题描述】：

在这种情况下，我有“疾病”表：

+--------+---------+---------+-------------+---------+-------------------------------+---------------+
| Id_SICK|ID_WORKER| FNAME   | LNAME   | BEGIN_DATE          | END_DATE              | SICKNESS_TIME |
+--------+---------+---------+---------+------------+--------------------+-----------+---------------+
| 6      |   17    | PAUL    | KING    |2019-03-19 07:00:00  |2019-03-20 15:00:00    |    16:00:00   | 
| 7      |   17    | PAUL    | KING    |2019-03-25 07:00:00  |2019-03-25 15:00:00    |    8:00:00    |
+--------+---------+---------+----------------------+--------------------------------+---------------+

“工人”表：

+----------+---------+---------+
|ID_WORKER |  FNAME  | LNAME   |
+----------+---------+----------
| 17       |  PAUL   |  KING   |
| 18       |  SAM    |  BULK   |
+----------+---------+---------+

“订单”表：

+----------+--------------+---------------+
|ID_ORDER  |  DESC_ORDER  | NUMBER_ORDER  |
+----------+--------------+---------------+
| 20       |  TEST        |  TEST         |
+----------+--------------+---------------+

“订单状态”表：

+----------+---------+---------+---------------------+-------------------+------------+
| Id_status|ID_WORKER| ID_ORDER| BEGIN_DATE          | END_DATE          | ORDER_DONE |
+----------+---------+---------+----------+------------+---------+--------------------+
| 47       |   17    |    20   |2019-03-18 06:50:35  |2019-03-18 15:21:32|  NO        |
| 48       |   17    |    20   |2019-03-20 06:44:12  |2019-03-20 15:11:23|  NO        |
| 50       |   17    |    20   |2019-03-22 06:50:20  |2019-03-22 12:22:33|  YES       |
| 51       |   18    |    20   |2019-03-18 06:45:11  |2019-03-18 15:14:45|  NO        |
| 52       |   18    |    20   |2019-03-20 06:50:22  |2019-03-20 15:10:32|  NO        |
| 53       |   18    |    20   |2019-03-22 06:54:11  |2019-03-22 11:23:45|  YES       |
+----------+---------+---------+------------+---------+-------------------+-----------+

我做了什么：

我可以汇总订单上每个其他工人的“总时间”（在 order_status 表中），包括汇总 Sickness 表中的“病假时间”。我也正确地选择了工人（LNAME，FNAME）订单（DESC_ORDER 和 NUMBER_ORDER）和“TOTAL TIME”。我在下面写了mysql命令：

SELECT workers.fname, 
   workers.lname, 
   order_statusAgg.number_order,
   workers.id_worker,
   order_statusAgg.desc_order, 
   SEC_TO_TIME(SUM(order_statusAgg.stime)) AS 'TOTAL TIME', 
   IFNULL(SEC_TO_TIME(SUM(sickAgg.vtime)),'00:00:00') AS 'LEAVE TIME'
FROM workers 
LEFT JOIN (
SELECT leave.id_worker, SUM((datediff(sickness.end_date, sickness.begin_date) + 1) * (time_to_sec(time(sickness.end_date)) - time_to_sec(time(sickness.begin_date)))) AS vtime 
FROM sickness
GROUP BY sickness.id_worker) sickAgg
           ON sickAgg.id_worker = workers.id_worker
   LEFT JOIN (
SELECT order_status.id_worker, orders.number_order, orders.desc_order, 
SUM((Time_to_sec(order_status.end_date) - 
                   Time_to_sec(order_status.begin_date))) AS stime
FROM order_status
       INNER JOIN orders 
           ON orders.id_order = order_status.id_order
GROUP BY order_status.id_worker) order_statusAgg
           ON workers.id_worker = order_statusAgg.id_worker 
WHERE  order_statusAgg.number_order LIKE 'TEST'
GROUP BY workers.id_worker;

然后我得到：

+---------+---------+---------------+------------+------------+--------------+
|  FNAME  | LNAME   |  NUMBER_ORDER | DESC_ORDER | TOTAL TIME | Sickness_time| 
+---------+---------+---------------+------------+------------+--------------+
|  PAUL   |  KING   | TEST          | TEST       | 22:30:21   |   24:00:00   |   
|  SAM    |  BULK   | TEST          | TEST       | 21:19:18   |   00:00:00   |   
+---------+---------+---------------+------------+------------+--------------+

好的，但另一方面，该订单已于 2019 年 3 月 23 日完成。 PAUL KING 在 25-03-2019 也生病了，他的生病时间不应该在他正在执行的这个命令中增加。所以在这种情况下应该是：

+---------+---------+---------------+------------+------------+--------------+
|  FNAME  | LNAME   |  NUMBER_ORDER | DESC_ORDER | TOTAL TIME | Sickness_time| 
+---------+---------+---------------+------------+------------+--------------+
|  PAUL   |  KING   | TEST          | TEST       | 22:30:21   |   16:00:00   |   
|  SAM    |  BULK   | TEST          | TEST       | 21:19:18   |   00:00:00   |   
+---------+---------+---------------+------------+------------+--------------+

我想知道这是否与代码问题有关？

LEFT JOIN (
SELECT leave.id_worker, SUM((datediff(sickness.end_date, sickness.begin_date) + 1) * (time_to_sec(time(sickness.end_date)) - time_to_sec(time(sickness.begin_date)))) AS vtime 
FROM sickness
GROUP BY sickness.id_worker) sickAgg
           ON sickAgg.id_worker = workers.id_worker

有人知道如何在订单持续时间结束之前对其进行汇总吗？这可能吗？我正在寻找任何想法，但我不是 mysql 大师。感谢您的帮助。

【问题讨论】：

首先，修复你的GROUP BY。一旦GROUP BY 被修复，我们将讨论修复你的逻辑。
@Eric 是对的，这就是你应该如何在 SQL 中使用GROUP BY
@Eric 好的，但为什么不应该使用GROUP BY？ ORDER BY 与 GROUP BY 几乎相反
@Prochu1991 我从不说不要使用GROUP BY。我说要正确使用。您阅读过@RaymondNijland 发布的链接吗？

标签： mysql datetime sum

【解决方案1】：

sickAgg 的子查询不再需要此 [SUM((datediff(sickness.end_date, sickness.begin_date) + 1) * (time_to_sec(time(sickness.end_date)) - time_to_sec(time(sickness.begin_date)))) AS vtime] 计算，因为您的 sickness 表已经有 SICKNESS_TIME 列。您必须定义的条件是您的sickAgg 需要引用order_status 表中的END_DATE 列，以便它只返回END_DATE 之前的值。试试下面这个查询：

SELECT workers.fname, 
workers.lname, 
order_statusAgg.number_order,
workers.id_worker,
order_statusAgg.desc_order, 
SEC_TO_TIME(SUM(order_statusAgg.stime)) AS 'TOTAL TIME', 
IFNULL(SEC_TO_TIME(SUM(sickAgg.vtime)),'00:00:00') AS 'SICK TIME'
FROM workers 
LEFT JOIN (SELECT sickness.id_worker, TIME_TO_SEC(sickness_time) AS vtime FROM sickness 
LEFT JOIN
(SELECT id_worker,MIN(begin_date) AS 'MIN_BEGIN_DATE',MAX(end_date) AS 'MAX_END_DATE' 
FROM order_status GROUP BY id_worker) ordstat ON 
sickness.id_worker=ordstat.id_worker 
WHERE sickness.END_DATE <= MAX_END_DATE) sickAgg
       ON sickAgg.id_worker = workers.id_worker
LEFT JOIN (
SELECT order_status.id_worker, orders.number_order, orders.desc_order, 
SUM((TIME_TO_SEC(order_status.end_date) - TIME_TO_SEC(order_status.begin_date))) AS stime 
FROM order_status INNER JOIN orders 
       ON orders.id_order = order_status.id_order
GROUP BY order_status.id_worker) order_statusAgg
       ON workers.id_worker = order_statusAgg.id_worker 
WHERE order_statusAgg.number_order LIKE 'TEST'
GROUP BY workers.id_worker;

我已经为您的sickAgg 完全重建了子查询。看这部分：

SELECT sickness.id_worker, TIME_TO_SEC(sickness_time) AS vtime FROM sickness 
LEFT JOIN
(SELECT id_worker,MIN(begin_date) AS 'MIN_BEGIN_DATE',MAX(end_date) AS 'MAX_END_DATE' 
FROM order_status GROUP BY id_worker) ordstat ON sickness.id_worker=ordstat.id_worker 
WHERE sickness.END_DATE <= MAX_END_DATE

现在，有很多方法可以做到这一点，我相信在较新的 MySQL 中，有更好的方法。但是我认为您最初了解您所写的内容，并且最好使用您首先了解的内容，而不是获得一个全新的查询。

【讨论】：

我只更改了您编辑的部分。相反，TIME_TO_SEC(sickness_time) 我更新为：`SUM((datediff(sickness.END_DATE，sickness.BEGIN_DATE) + 1) * (time_to_sec(time(sickness.END_DATE)) - time_to_sec(time(sickness.BEGIN_DATE))))。因为 SICKNESS_TIME 是从我从更新中显示的那个查询创建的。但它确实工作得很好。谢谢你的帮助！ :)