Question

我有一张表，需要调用select两次，具有两个不同的where条件，然后将其放回一个。

如果我分别运行select`s，它会给我正确的结果，但是在一个具有内部联接的查询中没有选择，第二个select被加倍。

表格如下：

amount | paydate    | ...
-337   | 2018-01-10 | ...
1467   | 2018-01-10 | ...
-322   | 2018-02-10 | ...
1452   | 2018-02-10 | ...

结果应为：

January  | 1467 | -337
February | 1452 | -322

但是我得到

January  | 1467 | -322
February | 1452 | -322

这是我的查询

SELECT 
    Q1.month, 
    Q1.ttlincome, 
    Q1.paydate,
    Q2.ttldebid
FROM (
        SELECT MONTHNAME(`paydate`) as month, 
        (`amount`) AS ttlincome, 
        paydate, propertiseid
                FROM `test`
            WHERE `test`.`amount` > 0 
              AND `test`.`paydate` BETWEEN '2018-01-01' AND '2018-12-31' 
            GROUP BY YEAR(`test`.`paydate`), MONTH(`test`.`paydate`)
    ) AS Q1
INNER JOIN (
        SELECT (`amount`) AS ttldebid,
        FROM `test`
            WHERE `test`.`amount` < 0 
              AND `test`.`paydate` BETWEEN '2018-01-01' AND '2018-12-31'
            GROUP BY YEAR(`test`.`paydate`), MONTH(`test`.`paydate`)
    ) AS Q2
GROUP BY YEAR(Q1.`paydate`), MONTH(Q1.`paydate`)

也许有更好的方法可以使它工作？

Answer 1

为什么不这样做：

select monthname(paydate),
       max(amount), min(amount)
from t
where paydate >= '2018-01-01' and paydate < '2019-01-01'
group by monthname(paydate)
order by min(paydate);

或者，如果您可以拥有两行以上：

select monthname(paydate),
       sum(case when amount > 0 then amount end),
       sum(case when amount < 0 then amount end)
from t
where paydate >= '2018-01-01' and paydate < '2019-01-01'
group by monthname(paydate)
order by min(paydate);

Answer 2

我认为您的查询应该是这样

SELECT 
    Q1.month, 
    Q1.ttlincome, 
    Q1.paydate,
    Q2.ttldebid
FROM (
        SELECT distinct MONTHNAME(`paydate`) as month, 
        (`amount`) AS ttlincome, 
        paydate, propertiseid
                FROM `test`
            WHERE `test`.`amount` > 0 
              AND `test`.`paydate` BETWEEN '2018-01-01' AND '2018-12-31' 

    ) AS Q1
INNER JOIN (
        SELECT distinct (`amount`) AS ttldebid,MONTHNAME(`paydate`) as month
        FROM `test`
            WHERE `test`.`amount` < 0 
              AND `test`.`paydate` BETWEEN '2018-01-01' AND '2018-12-31'

    ) AS Q2
on Q1.month=Q2.month

选择内部联接给出不正确的结果

2 个答案: