Date | Column_A | Column_B
Day 1 | 5 | 7
Day 2 | -3 | 7 + (-3) = 4
Day 3 | 8 | 4 + 8 = 12
Day 4 | -21 | 12 + (-21) -> 0 (see formula on row n)
Day n-1 | ... | Column_B(n-1)
Day n |Column_A(n) | IF(Column_B(n-1) + Column_A(n) >= 0, Column_B(n-1) +
Column_A(n), 0)
如何在Mysql中填充Column_B?
最佳答案 您可以将ID作为ID添加到表中吗?
如果是的话……您可以使用以下代码:
DECLARE @count INT;
DECLARE @i INT;
DECLARE @temp INT;
SET @count =( select count(*) from myTBL)
SET @i=2
WHILE (@i <=@count)
BEGIN
SET @temp=(select Column_A from myTBL where ID=@i)+ (select Column_B from myTBL where ID=@i-1);
IF (@temp>0) update myTBL set Column_B=@temp where ID=@i
ELSE update myTBL set Column_B=0 where ID=@i
SET @i = @i+1
END