4

我有一个名为“Jrl”的表，包含三列：

code  AS varchar(15)
total AS numeric(13,2)
rem   AS numeric(13,4)

为了论证的缘故，我们假设表中只有一行的值为“001”、400.00 和 52.1745。

考虑以下查询：

SELECT code, total - rem
 FROM Jrl

它返回一行 '001' 和347.8255。这是对的。

如果我按如下方式更改查询（这实际上是我在代码中需要的查询）：

SELECT code, SUM(total) - SUM(rem)
 FROM Jrl
 GROUP BY code

它返回一行，其中包含“001”和347.83（即，比例尺为 2 而不是 4）。

现在根据http://msdn.microsoft.com/en-us/library/ms190476%28v=sql.90%29.aspx上的文档，数字表达式（减法）的类型应该是 numeric(16,4 )，显然不是。（我在 SQL Server 2005 和 2008 R2 上得到了相同的行为。）

有人可以告诉我那里发生了什么吗？

顺便提一句。我确实找到了解决方法，但我不喜欢它，这就是我发布这个问题的原因。解决方法是添加显式强制转换：

SELECT code, CAST(SUM(total) AS numeric(13,4)) - SUM(rem)
 FROM Jrl
 GROUP BY code
4

3 回答 3

4

1）请运行此脚本并阅读我的评论。

2）我希望这个答案会对你有所帮助。

3) SUM()-SUM() 的精度为 2，因为您选择先求和 (SUM(total)和SUM(rem))，然后再减去 ( SUM(total) - SUM(rem))。

4）我的建议是使用SELECT t.code, SUM(t.total - t.rem) AS diff ...（先减去然后求和）。

5) 你可以阅读我对这个问题的回答SQL Numeric data type truncating value? ：

DECLARE @Test TABLE(
    code  varchar(15),
    total numeric(13,2),
    rem   numeric(13,4)
);

INSERT  @Test (code, total, rem)
VALUES  ('001', 11.78, 5.6789);

--Test [1]
SELECT  dt.*,
        SQL_VARIANT_PROPERTY(dt.diff, 'BaseType') AS diff_BaseType,
        SQL_VARIANT_PROPERTY(dt.diff, 'Precision') AS diff_Precision,
        SQL_VARIANT_PROPERTY(dt.diff, 'Scale') AS diff_Scale
FROM
(
        SELECT  t.code, t.total - t.rem AS diff
        FROM    @Test t
) dt;

/*
Operation: e1 - e2
Result precision: max(s1, s2) + max(p1-s1, p2-s2) + 1 = max(2,4) + max(13-2, 13-4) + 1 = 4 + 11 + 1 = 16
Result scale: max(s1, s2) = max(2, 4) = 4
*/

--Test [2]
SELECT  dt.*,
        SQL_VARIANT_PROPERTY(dt.diff, 'BaseType') AS diff_BaseType,
        SQL_VARIANT_PROPERTY(dt.diff, 'Precision') AS diff_Precision,
        SQL_VARIANT_PROPERTY(dt.diff, 'Scale') AS diff_Scale
FROM
(
        SELECT  t.code, SUM(t.total - t.rem) AS diff
        FROM    @Test t
        GROUP BY t.code
) dt;

/*
Operation: SUM(e1 - e2)
Result precision: 38--For SUM function, I think (it's just a hipotese), SQL Server choose the maximum precision to prevent the overflow error
                    Argument:
                    DECLARE @t TABLE (Col NUMERIC(2,1)); INSERT @t VALUES (1);
                    SELECT  SQL_VARIANT_PROPERTY(SUM(t.Col), 'Precision') FROM @t t;
                    Result: precision = 38 (maximum DECIMAL/NUMERIC precision)
Result scale: the same scale as (e1-e2)= 4 (please see Test [1])
*/

--Test [3]
SELECT  dt.*,
        SQL_VARIANT_PROPERTY(dt.SUM_total, 'BaseType')  AS SUM_total_BaseType,
        SQL_VARIANT_PROPERTY(dt.SUM_total, 'Precision') AS SUM_total_Precision,
        SQL_VARIANT_PROPERTY(dt.SUM_total, 'Scale')     AS SUM_total_Scale,

        SQL_VARIANT_PROPERTY(dt.SUM_rem, 'BaseType')    AS SUM_rem_BaseType,
        SQL_VARIANT_PROPERTY(dt.SUM_rem, 'Precision')   AS SUM_rem_Precision,
        SQL_VARIANT_PROPERTY(dt.SUM_rem, 'Scale')       AS SUM_rem_Scale,

        SQL_VARIANT_PROPERTY(dt.diff, 'BaseType')       AS diff_BaseType,
        SQL_VARIANT_PROPERTY(dt.diff, 'Precision')      AS diff_Precision,
        SQL_VARIANT_PROPERTY(dt.diff, 'Scale')          AS diff_Scale
FROM
(
        SELECT  t.code, 
                SUM(t.total) AS SUM_total, SUM(t.rem) AS SUM_rem, SUM(t.total) - SUM(t.rem) AS diff
        FROM    @Test t
        GROUP BY t.code
) dt;

/*
Operation: SUM(total) (<> e1 + e2 + ...)
Result precision: 38--I think SQL Server choose the maximum precision to prevent the overflow error
Result scale: the same precision as total= 2
*/


/*
Operation: SUM(rem) (<> e1 + e2 + ...)
Result precision: 38--I think SQL Server choose the maximum precision to prevent the overflow error
Result scale: the same precision as rem= 4
*/

/*
Operation: SUM(total) - SUM(rem) = e1 - e2
Result precision: max(s1, s2) + max(p1-s1, p2-s2) + 1 = max(2,4) + max(38-2, 38-4) + 1 = 4 + 36 + 1 = 41 
but max. precision is 38 so result precision = 38

Calculated result scale: max(s1, s2) = 4 
but because the real precision for result (41) is greater than maximum precision (38)
SQL Server choose to decrease the precision of the result to 2 (please see Test [3] - diff_Scale).
In this case (the real precision for result is greater than maximum precision) I think the 
expression for result's precision is max(s1, s2) - (real precision - maximum precision) + 1 = 4 - (41 - 38) + 1 = 4 - 3 + 1 = 2
For example you could try to modify the definition of total column to `total numeric(13,1)` 
and you will see that the precision for SUM(total) - SUM(rem) becomes 4 - 4(4+37+1=42) + 1 = 1
*/

结果：

--Test [1] SELECT t.code, t.total - t.rem AS diff
code diff   diff_BaseType  diff_Precision diff_Scale
---- ------ -------------- -------------- ----------
001  6.1011 numeric        16             4

--Test [2] SELECT t.code, SUM(t.total - t.rem) AS diff
code diff   diff_BaseType diff_Precision diff_Scale
---- ------ ------------- -------------- ----------
001  6.1011 numeric       38             4

--Test [3] SELECT t.code, ..., SUM(t.total) - SUM(t.rem) AS diff
code SUM_total SUM_rem diff SUM_total_BaseType SUM_total_Precision SUM_total_Scale SUM_rem_BaseType SUM_rem_Precision SUM_rem_Scale diff_BaseType diff_Precision diff_Scale
---- --------- ------- ---- ------------------ ------------------- --------------- ---------------- ------------------------------- ------------- -------------- ----------
001  11.78     5.6789  6.10 numeric            38                  2               numeric          38                4             numeric       38             2
于 2012-05-01T10:50:43.127 回答
1

那是因为total AS numeric(13,2)场

由于您要减去具有不同精度的两个字段的总和，因此 sql server 会以最小的精度显示结果。

如果你这样做：

 create table jrl2(
code  varchar(15),
total numeric(13,4),
rem   numeric(13,4)
)
insert into jrl2 values ('001', 400.00 , 52.1745)

select * from jrl2
SELECT code, total - rem  FROM Jrl

SELECT code, SUM(total) - SUM(rem)
 FROM Jrl2
 GROUP BY code

你会得到：347.8255

于 2012-05-01T09:33:39.187 回答
1

安德烈亚斯，

问题是 SUM() 的返回类型使用 38 的最大精度。（请参阅联机丛书中 SUM 下的“返回类型”部分：http: //msdn.microsoft.com/en-us/library/ms187810% 28v=sql.90%29.aspx。）

您的“总计”列的类型是数字（13,2），因此 SUM（总计）的结果类型是（不幸的是）数字（38,2）。当操作数的精度为 38 时，e1 + e2 的比例（再次不幸）不是 max(s1,s2)。

这在 BOL 的脚注中提到：http: //msdn.microsoft.com/en-us/library/ms190476.aspx。*结果精度和小数位数的绝对最大值为 38。当结果精度大于 38 时，相应的小数位数会减小，以防止结果的整数部分被截断。

另请参阅http://support.microsoft.com/kb/281341。

于 2012-05-01T13:50:54.383 回答