Question

这一定是一件非常容易的事情，但我无法完成任务，也无法在任何地方找到。这就是我的select代码：

SELECT *,
    (SELECT SUM(x) FROM a WHERE (...)) AS sum_x,
    (SELECT SUM(y) FROM b WHERE (...)) AS sum_y
FROM table

我想使用(sum_x - sum_y)再创建一个字段，但是当我尝试做这样的事情时

SELECT *,
    (SELECT SUM(x) FROM a WHERE (...)) AS sum_x,
    (SELECT SUM(y) FROM b WHERE (...)) AS sum_y,
    (sum_x - sum_y) AS sum_c
FROM table

它对我生气并说Unknown column 'sum_x' in 'field list'。

我在这里缺少什么？

Answer 1

SQL通常不允许您在定义它们的同一级别引用列别名。您需要使用子查询执行此操作：

select t.*, (sum_x - sum_y) as sum_c
from (SELECT *,
             (SELECT SUM(x) FROM a WHERE (...)) AS sum_x,
             (SELECT SUM(y) FROM b WHERE (...)) AS sum_y
      FROM table
     ) t;

然而，MySQL倾向于实现子查询，因此存在性能损失。实现的开销可能少于两次子查询计算：

      SELECT *,
             (SELECT SUM(x) FROM a WHERE (...)) AS sum_x,
             (SELECT SUM(y) FROM b WHERE (...)) AS sum_y,
             ((SELECT SUM(x) FROM a WHERE (...)) -
              (SELECT SUM(y) FROM b WHERE (...))
             ) as sum_c
      FROM table;

只进行一次计算也更容易维护和理解。

在MySQL中，你也可以这样做：

      SELECT *,
             @sum_x := (SELECT SUM(x) FROM a WHERE (...)) AS sum_x,
             @sum_y := (SELECT SUM(y) FROM b WHERE (...)) AS sum_y,
             (@sum_x - @sum_y) as sum_c
      FROM table cross join
           (select @sum_x := 0, @sum_y := 0) const

MySQL不保证在分配后评估最后一个表达式。它似乎在实践中始终如此。

Answer 2

试试这个

SELECT *,
    (SELECT SUM(x) FROM a WHERE (...)) AS sum_x,
    (SELECT SUM(y) FROM b WHERE (...)) AS sum_y,
    (SELECT SUM(x) FROM a WHERE (...) - (SELECT SUM(y) FROM b WHERE (...))) AS sum_c
FROM table

Answer 3

只需使用子查询而不是使用别名

SELECT *,
    (SELECT SUM(x) FROM a WHERE (...)) AS sum_x,
    (SELECT SUM(y) FROM b WHERE (...)) AS sum_y,
    (
          (SELECT SUM(x) FROM a WHERE (....)) - (SELECT SUM(y) FROM b WHERE (...))
    ) AS sum_c
FROM table

如何在mysql的同一查询中使用手动创建的字段？

3 个答案: