从SQL计算Floor（）时出错

我需要计算工资准备所需的钞票数量。 我对最后一列有问题，即1美元

我将获得1美元=2的值

DECLARE @GrossPay Decimal(12,2)
SET @GrossPay = 132.78

SELECT
    @GrossPay,
    FLOOR(@GrossPay/100) AS USD100,
    FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50) AS USD50,
    FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20) AS USD20,
    FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50 - (FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20))*20)/10) AS USD10,
FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50 - (FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20))*20 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50 - (FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20))*20)/10))*10)/5) AS USD5,
FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50 - (FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20))*20 -(FLOOR((@GrossPay -
FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50 - (FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20))*20)/10))*10-(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50 - (FLOOR((@GrossPay -
FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20))*20 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50 - (FLOOR((@GrossPay -
FLOOR(@GrossPay/100)*100 -(FLOOR((@GrossPay - FLOOR(@GrossPay/100)*100)/50))*50)/20))*20)/10))*10)/5))*5)/1) AS USD1

抱歉这么长时间的询问

您可以复制并粘贴上面的Sql查询并执行它，这样您就理解了我的意思。希望有人能帮我

提前谢谢

考虑使用：

或者，添加

0.5

将使

floor

四舍五入，而不是floor：

SELECT FLOOR(2.78+0.5)

---

如果你意识到你不需要减去之前计算的金额，你的很多表达式都可以简化（目前我唯一找不到合适的表达式是

10美元

）。例如，

USD1

列始终是一个介于0和4之间的值，基于将整个金额除以5的剩余部分

这是模运算符的定义，在SQL中，这是，因此我们可以有：

(CONVERT(int,@GrossPay)%5) as USD1

正如我所说，其他大部分也可以通过使用模来计算（除了10美元，仍在试图对此进行简化）。我还引入了另一个变量，称为

@GrossPayInt

，因为它自动处理舍入：

DECLARE @GrossPay Decimal(12,2) SET @GrossPay = 132.78
DECLARE @GrossPayInt int SET @GrossPayInt = @GrossPay

SELECT @GrossPay,
    @GrossPayInt/100 as USD100,
    (@GrossPayInt%100)/50 as USD50,
    (@GrossPayInt%50)/20 as USD20,
    CASE WHEN (@GrossPayInt%100)/10 IN (1,3,6,8) THEN 1 ELSE 0 END as USD10 as USD10,
    (@GrossPayInt%10)/5 as USD5,
    (@GrossPayInt%5) as USD1

---

如果有人能建议10美元的短表达式，我想这已经完成了。

Ummm….如果你得到

floor（2.78）==2

，那么

floor

函数的工作原理与预期完全一样。请保留返回3的选择部分，只有bozotron才能读取你的代码，无意冒犯：）在你开始思考一定有更好的方法来解决这个问题之前，你需要输入多少次“地板”这个词？谢谢大家给我上的好课，谢谢！

DECLARE @GrossPay Decimal(12,2) SET @GrossPay = 132.78
DECLARE @GrossPayInt int SET @GrossPayInt = @GrossPay

SELECT @GrossPay,
    @GrossPayInt/100 as USD100,
    (@GrossPayInt%100)/50 as USD50,
    (@GrossPayInt%50)/20 as USD20,
    CASE WHEN (@GrossPayInt%100)/10 IN (1,3,6,8) THEN 1 ELSE 0 END as USD10 as USD10,
    (@GrossPayInt%10)/5 as USD5,
    (@GrossPayInt%5) as USD1