Question

只是寻找我的问题的任何信息，不一定是确切的答案。

我有一张名为WorkDay的表。

在工作日表中，有6个字段：

Date_of_Workday, Workday_type, Boss, Regular, Tech1, Tech2.

可以为员工分配多个角色，还有一些字段为NULL。

我想写一个查询，它将返回分配给每个工作日的员工总数，但很明显，如果有员工分配给Boss角色以及Tech2角色，那么这只会算作一名员工。

编辑：这是我想要的输出。

+------------+------------+----+
|Date_of_Work| Work_type  |    |
+------------+------------+----+
| 2019-02-09 | AM         |  4 |
| 2019-02-09 | PM         |  4 |
| 2019-02-10 | AM         |  3 |
| 2019-02-10 | PM         |  2 |
| 2019-02-11 | AM         |  4 |
| 2019-02-11 | PM         |  4 |
| 2019-02-12 | AM         |  4 |
| 2019-02-12 | PM         |  4 |
| 2019-02-13 | AM         |  4 |
| 2019-02-13 | PM         |  4 |
 ------------+------------+----

这是SELECT * FROM Workday：

+------------+------------+---------+----------+-----------+-----------+
| DateofWork | workday_typ | Boss   | Regular  |   Tech1   |   Tech2   |
+------------+------------+---------+----------+-----------+-----------+
| 2019-02-09 | AM         | LB1     | AW1      | AE1       | JE1       |
| 2019-02-09 | PM         | AE1     | IM1      | AL1       | BJ1       |
| 2019-02-10 | AM         | AE1     | MM1      | MW1       | NULL      |
| 2019-02-10 | PM         | AE1     | AE1      | EB1       | NULL      |
| 2019-02-11 | AM         | LB1     | AB1      | DJ1       | JP1       |
| 2019-02-11 | PM         | LB1     | JE1      | AB2       | BJ1       |
| 2019-02-12 | AM         | LB1     | NS1      | AE1       | MB1       |
| 2019-02-12 | PM         | LB1     | HP1      | EH1       | AL1       |
| 2019-02-13 | AM         | LB1     | EB2      | ME1       | MM1       |
| 2019-02-13 | PM         | LB1     | ME2      | AB2       | DJ1       |
+------------+------------+---------+----------+-----------+-----------+

我不确定是否要进行此查询。任何信息/指针将不胜感激。

Answer 1

看起来你糟糕的非规范化表格中有多个包含人名的列。您希望计算每天所有这些列中唯一值的数量。

尝试这样的方法来获取每天的独特价值。这将返回一个以Date_of_Workday, Workday_Type, Person为列的结果集。 UNION的使用消除了重复的Person值。

          SELECT Date_of_Workday, Workday_Type, Boss AS Person FROM tbl
          UNION 
          SELECT Date_of_Workday, Workday_Type,  Regular AS Person  FROM tbl
          UNION 
          SELECT Date_of_Workday, Workday_Type, Tech1 AS Person FROM tbl
          UNION 
          SELECT Date_of_Workday, Workday_Type, Tech2 AS Person FROM tbl

然后在聚合查询中使用该子查询，如下所示：（http://sqlfiddle.com/#!9/f1320a/1/0）

SELECT Date_of_Workday, Workday_Type, COUNT(Person) AS PersonCount
  FROM (
          SELECT Date_of_Workday, Workday_Type, Boss AS Person FROM tbl
          UNION 
          SELECT Date_of_Workday, Workday_Type,  Regular AS Person  FROM tbl
          UNION 
          SELECT Date_of_Workday, Workday_Type, Tech1 AS Person FROM tbl
          UNION 
          SELECT Date_of_Workday, Workday_Type, Tech2 AS Person FROM tbl
       ) Persons
 GROUP BY Date_of_Workday, Workday_Type

您需要COUNT(Person)而不是COUNT(*)，否则会计算NULL个值，因此会超出人数。

检索行中的不同值并计算总计

1 个答案: