如何从其他表正确总结为“疾病时间”-MySQL

时间:2019-03-19 07:05:38

标签: mysql datetime difference

在mysql数据库中,我创建了“疾病”表:

    +--------+---------+---------+-------------+---------+-------------------------------+
    | Id_SICK|ID_WORKER| FNAME   | LNAME   | BEGIN_DATE          | END_DATE              |
    +--------+---------+---------+---------+------------+--------------------+-----------+
    | 4      |   26    | ANDREW  | WORM    |2019-03-19 07:00:00  |2019-03-19 15:00:00    |  
    +--------+---------+---------+----------------------+--------------------+-----------+  
    | 5      |   25    | ADAM    | GAX     |2019-03-21 07:00:00  |2019-03-21 15:00:00    |  
    +--------+---------+---------+----------------------+--------------------------------+  

“工人”表:

+--------+---------+---------+--
|ID_WORKER |  FNAME  | LNAME   |
+----------+---------+----------
| 25       |  ADAM   |  GAX    |
+----------+---------+----------
| 26       |  ANDREW |  WORM   |
+----------+---------+----------

“订单”表:

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

和“订单状态”表:

+--------+---------+---------+---------+-------------+--------+----------+------------+
| Id_status|ID_WORKER| ID_ORDER| BEGIN_DATE          | END_DATE          | ORDER_DONE |
+----------+---------+---------+----------+------------+---------+--------------------+
| 47       |   25    |    20   |2019-03-18 06:50:35  |2019-03-18 15:21:32|  NO        |
+----------+---------+---------+------------+---------+-------------------+-----------+ 
| 48       |   25    |    20   |2019-03-20 06:44:12  |2019-03-20 15:11:23|  NO        |
+----------+---------+---------+------------+---------+-------------------+-----------+ 
| 50       |   25    |    20   |2019-03-22 06:50:20  |2019-03-22 12:22:33|  YES        |
+----------+---------+---------+------------+---------+-------------------+-----------+ 
| 51       |   26    |    20   |2019-03-18 06:45:11  |2019-03-18 15:14:45|  NO        |
+----------+---------+---------+------------+---------+-------------------+-----------+ 
| 52       |   26    |    20   |2019-03-20 06:50:22  |2019-03-20 15:10:32|  NO       |
+----------+---------+---------+------------+---------+-------------------+-----------+ 
| 53       |   25    |    20   |2019-03-22 06:54:11  |2019-03-22 11:23:45|  YES       |
+----------+---------+---------+------------+---------+-------------------+-----------+ 

我想在订单上总结每个工人的“总计时间”(在order_status表中),包括总结疾病表中的“疾病时间”。我已正确选择了来自其他工作人员的工作人员(LNAME,FNAME)订单(DESC_ORDER和NUMBER_ORDER)和“ TOTAL TIME”。但是我无法总结疾病时间。我在下面编写了mysql命令:

SELECT workers.FNAME, workers.LNAME, orders.NUMBER_ORDER, orders.DESC_ORDER, SEC_TO_TIME(SUM(TIME_TO_SEC(order_status.END_DATE) - TIME_TO_SEC(order_status.BEGIN_DATE))) AS 'TOTAL TIME', SEC_TO_TIME(SUM(TIME_TO_SEC(sickness.END_DATE) - TIME_TO_SEC(sickness.BEGIN_DATE))) AS 'SICKNESS TIME' FROM order_status INNER JOIN workers ON workers.ID_WORKER = order_status.ID_WORKER INNER JOIN orders ON orders.ID_ORDER = order_status.ID_ORDER INNER JOIN sickness ON sickness.ID_WORKER = workers.ID_WORKER WHERE orders.NUMBER_ORDER LIKE 'TEST' GROUP BY workers.ID_WORKER

然后我得到了结果:

+--------+---------+---------+-------+------------+------------+-------------+
|  FNAME  | LNAME   |  NUMBER_ORDER  | DESC_ORDER | TOTAL TIME | SICKNESS_TIME|
+----------+---------+---------------+------------+------------+-------------+
|  ADAM   |  GAX    | TEST           | TEST       | 22:25:38   |   24:00:00   |
+----------+---------+---------------+------------+------------+-------------+
|  ANDREW |  WORM   | TEST           | TEST       | 22:52:12   |   24:00:00   |
+----------+---------+---------------+------------+------------+-------------+

因为“疾病时间不正确”是因为对ID_SICK进行分组后的“疾病”是

+--------+---------+-----+
| Id_SICK| SICKNESS TIME |
+--------+---------+-----+
| 4      |   08:00:00    |  
+--------+---------+-----+ 
| 5      |   08:00:00    |   
+--------+---------+-----+  

例如,我也总结了“总时间+病假时间”

TOTAL TIME: 22:25:38 
SICKNESS TIME: 8:00:00

TOTAL + SICKNESS TIME : 22;25:38 + 8:00:00 = 30:25:38

有人可以帮助我如何处理吗?我应该写什么样的mysql查询?有任何想法吗?谢谢你的帮助:)

1 个答案:

答案 0 :(得分:2)

这实际上是预期的。上面的查询正在联接4个表:

  • order_status
  • 疾病
  • 工人
  • 订单

工人”表与“ order_status ”具有 1:N 关系。

工人”也与疾病有 1:N 关系,即使这是 1:1 ,实际上也没有关系。< / p>

该查询正在上述表之间创建笛卡尔乘积,并且每个表中的重复列值可以位于上述查询的结果集中。

删除group bysum以查看结果集。

例如,在您的id_worker 25的示例中,您有3个order_status行与1个疾病行(该行值将重复3次)联接,与1个工人行(同样将重复3次)联接,并与1个订单联接行(相同)。

因此聚合函数sum组合重复值。

这仅在结果集包含用于聚合函数的所有列的唯一行时才有效

要解决此问题,请使用子查询来汇总结果:

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', 
       SEC_TO_TIME(SUM(sicknessAgg.stime)) AS 'SICKNESS TIME' 
FROM   workers 
INNER JOIN (
SELECT sickness.id_worker, SUM((Time_to_sec(sickness.end_date) - 
                       Time_to_sec(sickness.begin_date))) AS stime
FROM sickness
GROUP BY sickness.id_worker
) sicknessAgg
               ON sicknessAgg.id_worker = workers.id_worker
       INNER 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

请注意,在这种情况下,您需要从 workers 表开始,因为您要汇总order_statussickness表中不同数量的行。

参考文献:

How to join three tables to get Sum

MySQL JOIN with multiple tables and SUMS

How get the sum of two tables value using inner join