Question

我有一个这样的表，我需要将其拆分以更好地分析数据

ID | doc | name    | price | pay 
1  | doc1| PERSON1 | 1     | 1
2  | doc2| PERSON1 | 10    | 0
3  | doc3| PERSON2 | 12    | 1
4  | doc4| PERSON2 | 1     | 0
5  | doc5| PERSON2 | 15    | 0
6  | doc6| PERSON3 | 2     | 1
7  | doc7| PERSON3 | 23    | 0
8  | doc8| PERSON3 | 3     | 0
9  | doc9| PERSON4 | 8     | 0

，我需要这样的输出，但我不知道该怎么做！

 name         | price | pay 
 PERSON1      |       | 
 Doc1         | 1     | 1
 Doc2         | 10    | 0
 Total payed  | 1     |
 Total per pay| 10    | 
 Total        | 11    |
 PERSON2      |       | 
 Doc3         | 12    | 1
 Doc4         | 1     | 0
 Doc5         | 15    | 0
 Total payed  | 12    | 
 Total per pay| 16    | 
 Total        | 28    |
 PERSON3      |       | 
 Doc6         | 2     | 1
 Doc7         | 23    | 0
 Doc8         | 3     | 0
 Total payed  | 2     | 
 Total per pay| 26    | 
 Total        | 28    |
 PERSON4      |       | 
 Doc9         | 8     | 0
 Total payed  | 0     | 
 Total per pay| 8     | 
 Total        | 8     |
 Payed        | 15    |
 PER PAY      | 60    |
 Total SUM    | 75    |

这可能吗？

Answer 1

如果第二种格式更好，我无法理解您如何分析数据。您可以这样做，但这很痛苦：

select doc, price, pay
from ((select doc, price, pay, 2 as ord, name
       from t
      ) union all
      (select distinct name as doc, null, null, 1, name
       from t
       group by name
      ) union all
      (select 'total paid', sum(pay * price), null, 3, name
       from t
       group by name
      ) union all
      (select 'total not paid', sum( (1 - pay) * price), null, 3, name
       from t
       group by name
      ) union all
      (select 'total', sum(pay), null, 5, name
       from t
       group by name
      )
     ) t
order by name, ord;

Answer 2

类似这样的东西：

select t.empid, 
       coalesce(INTIME, OUTIME) as INTIME,
       coalesce(OUTIME, INTIME) as OUTIME
from ( <query here> 
     ) t;

这是一个完整的工作示例，可以为您提供输出。您可以对其进行一些更改以匹配您的真实数据。

根据列组织表中的数据

2 个答案: