Question

我试图将以下每个总和作为字段，我只获得外部的一个，其他两个返回未定义的索引。

如何将它们作为字段放在外面？

SELECT 
(SELECT SUM(amount) AS income
    FROM transaction 
    WHERE 
    type = '1' && state = 'a') 

- (SELECT SUM(amount) AS expense 
    FROM transaction 
    WHERE 
    type = '2' && state = 'a') 
    AS balance

余额有效，但对于收入和支出，我得到了未定义的索引。

Answer 1

您可以使用case和sum来获取每个结果：

select sum(case when `type` = '1' then `amount` else 0 end) as `income`
     , sum(case when `type` = '2' then `amount` else 0 end) as `expense `
     , sum((case when `type` = '2' then -1 else 1 end) * `amount`) as `difference`
from `transaction`
where `type` in ('1', '2')
where `state` = 'a'

Answer 2

使用条件聚合：

SELECT SUM(CASE WHEN type = '1' AND state = 'a' THEN amount ELSE 0
           END) as income,
       SUM(CASE WHEN type = '2' AND state = 'a' THEN amount ELSE 0
           END) as balance
FROM transaction;

如果你仍然想要差异，那么你需要重复逻辑：

SELECT SUM(CASE WHEN type = '1' AND state = 'a' THEN amount ELSE 0
           END) as income,
       SUM(CASE WHEN type = '2' AND state = 'a' THEN amount ELSE 0
           END) as expense,
       (SUM(CASE WHEN type = '1' AND state = 'a' THEN amount ELSE 0 END) -
        SUM(CASE WHEN type = '2' AND state = 'a' THEN amount ELSE 0 END) 
       ) as balance
FROM transaction;

编辑：

实际上，最好将常见条件分解为WHERE子句：

SELECT SUM(CASE WHEN type = '1' THEN amount ELSE 0 END) as income,
       SUM(CASE WHEN type = '2' THEN amount ELSE 0 END) as expense,
       (SUM(CASE WHEN type = '1' THEN amount ELSE 0 END) -
        SUM(CASE WHEN type = '2' THEN amount ELSE 0 END) 
       ) as balance
FROM transaction
WHERE state = 'a' AND type IN ('1', '2');

这可以利用transaction(state, type, amount)上的索引。

Answer 3

您无法访问income和expense，因为它们仅在各自的子查询中定义，而不是在顶级查询中定义，后者只有一个字段 - 它们之间的差异。我放入一个子查询，然后选择它们和它们之间的区别。为了简化此查询并删除对子查询的需求，您可以通过向where中的case表达式提取不同的条件来模拟两个不同sum子句的行为参数：

SELECT income, expense, income - expense AS balance
FROM   (SELECT SUM(CASE type WHEN 1 THEN amount ELSE NULL) AS income,
               SUM(CASE type WHEN 2 THEN amount ELSE NULL) AS expense,
        FROM   `transaction`
        WHERE  state = 'a') t

MySQL将一个查询中的SUM（）函数与字段相乘

3 个答案: