如何将“生病时间”汇总到医嘱结束-mysql

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   |
+----------+---------+---------+

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

订单表：

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

+----------+--------------+---------------+
|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       |
+----------+---------+---------+------------+---------+-------------------+-----------+

我所做的：

我可以在订单上的订单状态表中汇总其他员工的总时间，包括汇总疾病表中的疾病时间。我已经正确地选择了每个工人的LNAME、FNAME orders DESC_ORDER和NUMBER_ORDER以及订单总时间。我在下面编写了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，该订单于2019年3月23日完成。保罗·金（PAUL KING）在2019年3月25日也生病了，他的生病时间不应该在他所做的这一顺序中增加。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

是否有人知道如何在订单期限结束前进行汇总？有可能吗？我在寻找任何想法，但我不是mysql大师。感谢您的帮助。

您的sickAgg子查询不再需要此[SumdateDiffSight.end_date，Sick.begin_date+1*time_to_SectimeSight.end_date-time_to_SectimeSight.begin_date AS vtime]计算，因为您的疾病表已经有Sight_time列。您必须定义的条件是您的sickAgg需要参考订单状态表中的END_DATE列，以便它只返回END_日期之前的值。请尝试以下查询：

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;

---

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。但我认为您理解最初编写的内容，最好先使用您理解的内容，而不是得到一个全新的查询

首先，通过以下方式修复您的组。一旦“分组依据”被修复，我们将讨论如何修复您的逻辑。@Eric是对的，这是您在中应该如何使用的SQL@Eric好的，但是为什么不应该使用分组方式呢？按几乎与组相反的顺序排列BY@Prochu1991我从来没有说过不要分组使用。我说要好好使用它。你看过@RaymondNijland发布的链接了吗？我只修改了你编辑的部分内容。相反，我将时间更新为：`SumDateDiffSight.END_DATE，Sight.BEGIN_DATE+1*TIME_TO_SectimeSight.END_DATE-TIME_TO_SectimeSight.BEGIN_DATE。因为时间是根据我在更新中显示的查询创建的。但它确实非常有效。谢谢你的帮助！：