Question

我之前在帖子中提出了一个后续问题。（SQL: Overflow error for type INT）。我现在遇到的问题是将@ SUMX2Y和@ SUMX2X2的结果转换为FLOAT。我尝试过多种不同的方法，但仍有问题。有谁知道我在这里做错了什么？我没有包括我的转换尝试，但它们包括CAST和CONVERT，直到等式的各个级别。

CREATE TABLE #Set1
(
    X BIGINT,
    Y BIGINT
)
INSERT INTO #Set1 
VALUES (220, 630)
      ,(350, 940)
      ,(450, 1140)


DECLARE @SUMX BIGINT
DECLARE @SUMY BIGINT
DECLARE @SUMX2 BIGINT
DECLARE @SUMY2 BIGINT
DECLARE @SUMX3 BIGINT
DECLARE @SUMX4 BIGINT
DECLARE @SUMXX BIGINT
DECLARE @SUMXY_ BIGINT
DECLARE @SUMXY BIGINT
DECLARE @SUMXX2 BIGINT
DECLARE @SUMX2Y BIGINT
DECLARE @SUMX2Y_ BIGINT
DECLARE @SUMX2X2 BIGINT
DECLARE @N INT

    SET @N = (SELECT COUNT(*) FROM #Set1)
    SET @SUMY = (SELECT SUM(Y) FROM #Set1)
    SET @SUMX2 = (SELECT SUM(POWER(X,2)) FROM #Set1)
    SET @SUMX4 = (SELECT SUM(POWER(X,4)) FROM #Set1)

    SET @SUMX2Y_ = (SELECT SUM(POWER(X,2)*Y) FROM #Set1)

    SET @SUMX2Y = @SUMX2Y_ - (@SUMX2*@SUMY)/(@N)
    SET @SUMX2X2 = @SUMX4 - (POWER(@SUMX2,2)/@N)

    PRINT @SUMX2Y
    PRINT @SUMX2X2

Answer 1

我最后只是将BIGINT改为FLOAT。我不确定它是否会导致问题。我们会看到

Answer 2

使用NUMERIC（18,2）或您想要保留的精度。使用float，这将保留小数精度值，直到您显式传递小数值

下面的代码可以使用

CREATE TABLE #Set1
(
    X NUMERIC(18,2),
    Y NUMERIC(18,2)
)
INSERT INTO #Set1 
VALUES (220, 630)
      ,(350, 940)
      ,(450, 1140)


DECLARE @SUMX NUMERIC(18,2)
DECLARE @SUMY NUMERIC(18,2)
DECLARE @SUMX2 NUMERIC(18,2)
DECLARE @SUMY2 NUMERIC(18,2)
DECLARE @SUMX3 NUMERIC(18,2)
DECLARE @SUMX4 NUMERIC(18,2)
DECLARE @SUMXX NUMERIC(18,2)
DECLARE @SUMXY_ NUMERIC(18,2)
DECLARE @SUMXY NUMERIC(18,2)
DECLARE @SUMXX2 NUMERIC(18,2)
DECLARE @SUMX2Y NUMERIC(18,2)
DECLARE @SUMX2Y_ NUMERIC(18,2)
DECLARE @SUMX2X2 NUMERIC(18,2)
DECLARE @N NUMERIC(18,2)

    SET @N = (SELECT COUNT(*) FROM #Set1)
    SET @SUMY = (SELECT SUM(Y) FROM #Set1)
    SET @SUMX2 = (SELECT SUM(POWER(X,2)) FROM #Set1)
    SET @SUMX4 = (SELECT SUM(POWER(X,4)) FROM #Set1)

    SET @SUMX2Y_ = (SELECT SUM(POWER(X,2)*Y) FROM #Set1)

    SET @SUMX2Y = @SUMX2Y_ - (@SUMX2*@SUMY)/(@N)
    SET @SUMX2X2 = @SUMX4 - (POWER(@SUMX2,2)/@N)

    PRINT @SUMX2Y
    PRINT @SUMX2X2

使用以下脚本进行Float＆amp;数字差异

DECLARE @SUMX NUMERIC(18,2) = 100 

DECLARE @SUMX1 FLOAT = 100 

SELECT @SUMX,@SUMX1

分割时，舍入我的INT结果

2 个答案: