Question

我显示每日基础数据。 Say One Human（HumanID）每天早晚吃2次。所以我输入这样的数据。

表：报告

-----------------------------------------------
ID  | HumanID |  date      | schedule | amount|  
-----------------------------------------------
1  |  101     | 2016-01-01 | morning  |  10   |
2  |  101     | 2016-01-01 | evening  |  8    |
3  |  102     | 2016-01-01 | morning  |  11   |
4  |  102     | 2016-01-01 | evening  |  9    |
5  |  103     | 2016-01-01 | morning  |  8    |
6  |  103     | 2016-01-01 | evening  |  7    |

我做Mysql查询：

select HumanID, date, 
max(case when schedule = 'morning' then amount end) as morning, 
max(case when schedule = 'evening' then amount end) as evening
from report 
group by HumanID, date;

查询后，结果如下所示

---------------------------------------
HumanID |  date      | morning | evening |  
---------------------------------------
101     | 2016-01-01 |   10    |   8     |
102     | 2016-01-01 |   11    |   9     |
103     | 2016-01-01 |   8     |   7     |

我想在php逻辑中，结果将是这样的

---------------------------------------
HumanID |  date      | morning | evening |  
---------------------------------------
101     | 2016-01-01 |   10    |   8     |
102     | 2016-01-01 |   11    |   9     |
103     | 2016-01-01 |   8     |   7     |
------------------------------------------
Total:               |   29    |   24    |

Answer 1

您可以尝试使用UNION添加此结果：

SELECT CAST(t.humanID as char) as humanID,t.date,
       MAX(.... )
..... -- Rest of your query
UNION ALL
SELECT 'total:' as humanID , null ,
         SUM(case when schedule = 'morning' then amount end) , 
         SUM(case when schedule = 'evening' then amount end) 
FROM report
ORDER BY humanID

Answer 2

E.g：

select HumanID, date, 
Sum(case when schedule = 'morning' then amount end) as morning, 
Sum(case when schedule = 'evening' then amount end) as evening
from report 
group by HumanID, date
With ROLLUP

你会如何整理这个Mysql查询逻辑

2 个答案: