0

我有一个简单的学生表。

name  | amount | vdate
Josh  | 15     | 01.01.2020
Steve | 25     | 05.04.2008
Josh  | 40     | 01.01.2022

我想要做的是彼此减去 Josh 值。

我写了这个查询，但它不起作用

   select name , sum(b.amount-a.amount) diff from  
     select name,amount from student a where name = 'Josh' and vdate='01.01.2020'
     union all
     select name,amount from student b where name = 'Josh' and vdate = '01.01.2022')
   group by name

预期结果是：

name | diff
Josh | 25
Steve| 25
4

2 回答 2

1

你可以试试这段代码，

    select 
    fname,
    abs(sum(amount2)) amount
from
(
WITH
student(fname,amount,vdate) AS (
            SELECT 'Josh' ,15, to_date('01102017','ddmmyyyy')  from dual
  UNION ALL SELECT 'Steve',25,  to_date('01102017','ddmmyyyy') from dual
  UNION ALL SELECT 'Josh' ,40   ,to_date('01102019','ddmmyyyy')from dual
)


select 
    h.fname,
    h.amount,
    decode((ROW_NUMBER() OVER(PARTITION BY fname order by vdate desc)),1,amount,amount* -1) amount2

    from student h
)
group by 
    fname
    ;

我假设您获得的人的金额更大并减去其他值，您可以通过修改分区窗口中的 order by 子句来选择更大的日期，即

decode((ROW_NUMBER() OVER(PARTITION BY fname order by vdate desc)),1,amount,amount * -1) amount2
于 2021-10-16T12:03:35.043 回答
0

你可以试试这个（我不知道它有什么意义......）：计算到目前为止找到的行数fname（“名称”是保留字，我不使用它）。如果这样得到的行号是奇数，则使用负数，否则使用正数。最后，对这些正/负行求和。

WITH
indata(fname,amount) AS (
            SELECT 'Josh' ,15 
  UNION ALL SELECT 'Steve',25
  UNION ALL SELECT 'Josh' ,40 
)
,
alternate AS (
  SELECT
    fname
  , CASE ROW_NUMBER() OVER(PARTITION BY fname) % 2 
      WHEN 1 THEN amount * -1 -- when odd then negative
      ELSE        amount      -- else positive
    END  AS amount
  FROM indata
)
SELECT
  fname
, ABS(SUM(amount)) AS amount -- absolute value
FROM alternate
GROUP BY fname;
-- out  fname | amount 
-- out -------+--------
-- out  Josh  |     25
-- out  Steve |     25
于 2021-10-15T14:32:01.437 回答