mysql - 如何将 "sickness time"汇总到命令末尾 - mysql

Question

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

“ worker ”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 worker 的“总时间”(在 order_status 表中)，包括汇总“病假时间”表中的“病假时间”。我也正确地从其他 worker 那里选择了 worker (LNAME、FNAME)订单(DESC_ORDER 和 NUMBER_ORDER)和“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。 PAUL KING 也在 2019 年 3 月 25 日生病了，在他正在执行的订单中不应添加他的生病时间。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   |   
+---------+---------+---------------+------------+------------+--------------+

我想知道这是否与该代码问题有关？

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

有人知道如何处理总结直到订单持续时间结束吗？有可能吗？我正在寻找任何想法，但我不是 mysql 大师。感谢您的帮助。

Answer 1

您的子查询 sickAgg不再需要这个[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 ] 计算自您的sickness表已经有 SICKNESS_TIME柱子。您必须定义的条件是您的 sickAgg需要引用END_DATE专栏在你order_status表，以便它只返回 END_DATE 之前的值。尝试下面的查询:

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 重建了子查询完全。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，我确信在较新的 MySQL 中，有更好的方法。但我认为您了解最初编写的内容，最好先使用您理解的内容，而不是获得全新的查询。