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：
我可以在包括从疾病表中总结“疾病时间”在内的订单上，总结每个工人（在order_status表中）的“总时间”。我也已正确选择了彼此的工人订单（LNAME，FNAME）订单（DESC_ORDER和NUMBER_ORDER）和“ TOTAL TIME”。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 |
+---------+---------+---------------+------------+------------+--------------+
Okey，但另一方面，该订单已于2019年3月23日完成。保罗·金（Paul PAUL KING）也在25-03-2019患病，在他正在执行的此命令期间不应该增加他的患病时间。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 |
+---------+---------+---------------+------------+------------+--------------+
我想知道这与代码问题有关吗？
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
有人想出如何在订单期间结束之前进行汇总吗？有可能吗我正在搜索任何想法，但我不是mysql专家。谢谢您的帮助。
答案 0 :(得分：1)
您的sickAgg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表已经具有SICKNESS_TIME列。您必须定义的条件是sickAgg需要引用END_DATE表中的order_status列，以便它仅返回END_DATE之前的值。在下面尝试以下查询：
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;
我已经完全为您的sickAgg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，而且我相信在较新的MySQL中，还有更好的方法。但是我认为您了解您最初编写的内容，最好先使用您首先了解的内容，而不是获取全新的查询。