SQL转换为十进制然后四舍五入为最接近的半数

Question

我有1张桌子：

表格1：

CREATE TABLE #TempTable (Time decimal(10,6), ID int)
INSERT INTO #TempTable (Time, ID) VALUES (0.5,1), (1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,0.10),(10,11)

这给了我：

    select * from #TempTable

+---------+----+
| Time    | ID |
+---------+----+
|0.500000 | 1  |
|1.000000 | 2  |
|2.000000 | 3  |
|3.000000 | 4  |
|4.000000 | 5  |
|5.000000 | 6  |
|6.000000 | 7  |
|7.000000 | 8  |
|8.000000 | 9  |
|9.000000 |10  |
|10.000000|11  |
+---------+----+

我想获取@Number，它基于时间@Time（Col1中的值）。 为了做到这一点，我正在做以下事情：

    SET @Number = (CASE @Time
                        WHEN 00.50 THEN 0.016000
                        WHEN 01.00 THEN 0.013300
                        WHEN 01.50 THEN 0.012650
                        WHEN 02.00 THEN 0.012000
                        WHEN 02.50 THEN 0.011200
                        WHEN 03.00 THEN 0.010400
                        WHEN 03.50 THEN 0.010150
                        WHEN 04.00 THEN 0.009900
                        WHEN 04.50 THEN 0.009700
                        WHEN 05.00 THEN 0.009500
                        WHEN 05.50 THEN 0.009450
                        WHEN 06.00 THEN 0.009400
                        WHEN 06.50 THEN 0.009300
                        WHEN 07.00 THEN 0.009200
                        WHEN 07.50 THEN 0.009200
                        WHEN 08.00 THEN 0.009200
                        WHEN 08.50 THEN 0.009100
                        WHEN 09.00 THEN 0.009000
                        WHEN 09.50 THEN 0.009000
                        WHEN 10.00 THEN 0.009000
                        ELSE 9.999999
                    END); 

问题是@Time是用户输入的数据（强制以以下格式hh.hh [00.50 = 30 minutes]输入，当他们输入值时默认为varchar）。 为了确保@Number从不= 9.999999，除非它> 10.00，否则我尝试执行以下逻辑：

 IF @Time > 00.00 AND @Time <= 10.00 THEN round to .5 and convert to decimal(10,2)
 ELSE convert to decimal(10,6)


    SELECT  convert(decimal(10,6),@TimeA), 
CASE
  WHEN convert(decimal(10,6), @TimeA) >= 0.0 AND convert(decimal(10,6), @TimeA) <= 10.0 THEN convert(decimal(10,2),round(@TimeA * 2,0)/2)
  ELSE convert(decimal(10,6),round(@TimeA * 2,0)/2)
END
    FROM azteca.Table 1 WHERE ID = @ID

SET @Time = @TimeA

上面给了我：nvarchar 04.50无法转换为数据类型int。 这意味着case语句有问题。 我期望以下结果-

User Input: 00.50 @Number = 0.016000
User Input: 01.50 @Number = 0.012650
User Input: 05.78 @Number = 0.009400
User Input: 07.90 @Number = 0.009200
User Input: 09.12 @Number = 0.009000

Answer 1

您的方法似乎非常复杂。 为什么不使用不等式编写CASE ：

DECLARE @Time2 DECIMAL(10, 6);
SET @Time2 = CAST(@Time AS DECIMAL(10, 6));

SET @Number = (CASE WHEN @Time2 <= 00.50 THEN 0.016000
                    WHEN @Time2 <= 01.00 THEN 0.013300
                    WHEN @Time2 <= 01.50 THEN 0.012650
                    . . .
               END);