使用 where 子句查找两列的差异 - MySQL

Question

我正在尝试使用 WHERE 子句以及它们的总和来获取两列（OTHER AND SENIOR）的差异。公式工作正常，但是，我没有得到我想要的结果。

我尝试使用 INNER JOIN 来做到这一点，不幸的是它不是我期望的结果

这是我的查询：

select a.ID as EMPLOYEE, 
sum(a.amount) as other,
sum(b.amount) as senior, 
(sum(a.amount) - sum(b.amount)) as result 
from gndsale a 
INNER join gndsale 
b ON a.ID= b.ID 
where a.TYPE = "5" and b.seniortype = "10" 
group by a.ID

结果：

我想要做的是获得 Column (Other) where type = 5 和 Column (Senior) where Seniortype = 10

这是我对高级的查询：

 select ID as EMPLOYEE, sum(amount) 
 from gndsale 
 where seniortype = "10" 
 group by ID

结果：

这是我对其他的查询：

select ID as employee, sum(amount) 
from gndsale 
where type = "5" 
group by ID

结果应该是

9907 = 530
9912 = 63.71

谁能帮我解决这个问题？ :(

样本/预期输出：

Answer 1

您为该栏选择了一个非常糟糕的名称。尽管名称为ID，但它并不是用于标识表中记录的ID。

您可能正在寻找类似以下的内容，即每个 ID 的聚合，您分别对类型 5 和高级类型 10 求和：

select 
  a.ID as employee, 
  coalesce(sum(case when type = 5 then amount end), 0) as other, 
  coalesce(sum(case when seniortype = 10 then amount end), 0) as senior, 
  coalesce(sum(case when type = 5 then amount end), 0) -
  coalesce(sum(case when seniortype = 10 then amount end), 0) as result
from gndsale
group by a.id
having sum(type = 5) > 0 and sum(seniortype = 10) > 0;

(HAVING 子句确保只获取同时具有 type = 5 和 Seniortype = 10 的记录的 ID。我们在这里使用 MySQL 的 true = 1 / false = 0。如果您还需要其他 ID，请将其删除。如果你保留它并且值不能为空，那么你可以删除合并。）

Answer 2

你可以试试这个：

SELECT 
    a.`employee`,
    sum(a.`amount`) - sum(b.`amount`) as total
FROM `gndsale` a
JOIN `gndsale` b
ON a.`employee` = b.`employee` and b.`senior_type` = "10"
WHERE a.`type` = "5"
GROUP BY a.`employee`
ORDER BY a.`employee`;

它只是连接你的两个查询。

Answer 3

请试试这个查询

select employee, other, senior, (other-senior) AS result
From (
    select ID as employee,
           sum(if(type=5,amount,0)) AS other,
           sum(if(seniortype =10,amount,0)) AS senior 
           from gndsale group by ID
) a