mysql - 在 SQL 中对两行求和（前一行，一个是同一表中的当前行）

Question

我有这张桌子：

Date_on      deposited  withdrawal   in_bank
2012-09-1      3000       2000        50000
2012-09-2/t    4000/t        0        54000
2013-09-3/t    3000       2000        55000

现在我想执行一个查询来添加存款金额并从前几天的条目中减去提款in_bank。我怎样才能做到这一点？任何人都可以帮助我吗？这是我的查询：

select date_on, in_bank,((in_bank+deposited)-withdrawal)
  from  tablename where date_on > '2012-09-01' order by date_on

Answer 1

SELECT today.withdrawal, today.deposited, today.date_on,
  (IFNULL(today.deposited, 0) - IFNULL(today.withdrawal, 0)) + IFNULL((SELECT in_bank FROM tablename AS yesterday WHERE yesterday.date_on = DATE_SUB(today.date_on, INTERVAL 1 DAY)), 0)
FROM tablename AS today
WHERE date_on BETWEEN '2012-09-01' AND '2012-09-02'
ORDER BY date_on ASC

如果找不到前几天的日期，查询将假定余额为零。

添加了一个小提琴 http://sqlfiddle.com/#!2/d8261/14

Answer 2

SELECT 
  date_on, 
  in_bank, 
  (( ISNULL(in_bank, 0) + ISNULL(in_deposited,0)) - ISNULL(withdrawal,0))
FROM tablename 
WHERE date_on > '2012-09-01' 
ORDER BY date_on

Answer 3

尝试这个：

SELECT
  t1.date_on,
  t1.in_bank,
  (
    SELECT in_bank
    FROM
    (
      SELECT *, (@rownum2 := @rownum2 +1) rank
      FROM Table1,(SELECT @rownum2 :=0 ) t
      ORDER BY date_on 
     ) t2 WHERE t1.rank - t2.rank = 1
   ) - t1.withdrawal - deposited "total"
FROM
(
  SELECT *, (@rownum := @rownum +1) rank
  FROM Table1,(SELECT @rownum :=0 ) t
  ORDER BY date_on 
) t1;

SQL 小提琴演示

Answer 4

我对mysql不是很熟悉，但这可能会有所帮助

select account.*,D.expectable from account left join 
(
select deposited - withdrawal + in_bank as expectable,DATE_ADD(Date_on,INTERVAL 1 DAY) as nDate from account
)
D on account.Date_on = D.nDate