Question

大家好，这个例子将输出什么mysql查询...

            'table 1
            | id |date        |
            | 1  |01/01/2015  |
            | 2  |01/02/2015  |
            | 3  |01/01/2015  |
            | 4  |01/02/2015  | 

            'table 2
            |id |table1_id| value1| value2| value3|value4|
            | 1 |   1     |  5    |    2  |   4   |  4   |
            | 2 |   2     |  40   |    3  |   2   |  3   |  
            | 3 |   3     |   5   |    2  |   4   |  5   |
            | 4 |   4     |       |    4  |   2   |  3   |

样品

$result = ($value1 * $value2) + ($value3 + $value4); 

            if($value1 == ""){

            }else{
            $result = ($value1 * $value2) + ($value3 + $value4);
            }

与同一日期相加... 这是最终输出....

           date             finalvalue
           01/01/2015          37
           01/02/2015          125

先谢谢..

Answer 1

我想你想做点什么：

SELECT table1.date, SUM(table2.value1) + SUM(table2.value2) + SUM(table2.value3) + SUM(table2.value4) AS finalvalue
FROM table1
INNER JOIN table2 ON table1.id = table2.table1_id
GROUP BY table1.date

不要在PHP中进行总结，你可以在SQL查询中更有效地完成它。

Answer 2

我没试过这个（因为我没有设置mysql），但你可以尝试类似下面的内容：

SELECT t.date, t.$value1, t.$value2, t.$value3, t.$value4 FROM (SELECT t1.id, t1.date, $value1, $value2, $value3, $value4 from table t1, table t2 where t1.id == t2.table1_id group by t1.id) ) as t GROUP By t.date;

参考：https://dev.mysql.com/doc/refman/5.0/en/from-clause-subqueries.html来自SubQuery

mysql查询

2 个答案: