Sil*_*ust 83 t-sql sql-server precision sqldatatypes sql-server-2016
我有一些客户收到奇怪的账单.我能够找出核心问题:
SELECT 199.96 - (0.0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4)))) -- 200 what the?
SELECT 199.96 - (0.0 * FLOOR(1.0 * CAST(199.96 AS DECIMAL(19, 4)))) -- 199.96
SELECT 199.96 - (0.0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * 199.96)) -- 199.96
SELECT 199.96 - (CAST(0.0 AS DECIMAL(19, 4)) * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4)))) -- 199.96
SELECT 199.96 - (CAST(0.0 AS DECIMAL(19, 4)) * FLOOR(1.0 * CAST(199.96 AS DECIMAL(19, 4)))) -- 199.96
SELECT 199.96 - (CAST(0.0 AS DECIMAL(19, 4)) * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * 199.96)) -- 199.96
-- It gets weirder...
SELECT (0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4)))) -- 0
SELECT (0 * FLOOR(1.0 * CAST(199.96 AS DECIMAL(19, 4)))) -- 0
SELECT (0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * 199.96)) -- 0
-- so... ... 199.06 - 0 equals 200... ... right???
SELECT 199.96 - 0 -- 199.96 ...NO....
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谁有线索,这里到底发生了什么?我的意思是,它肯定与十进制数据类型有关,但我无法真正地围绕它...
关于数字文字的数据类型存在很多混淆,所以我决定展示实线:
PS.SharePrice - (CAST((@InstallmentCount - 1) AS DECIMAL(19, 4)) * CAST(FLOOR(@InstallmentPercent * PS.SharePrice) AS DECIMAL(19, 4))))
PS.SharePrice DECIMAL(19, 4)
@InstallmentCount INT
@InstallmentPercent DECIMAL(19, 4)
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我确保DECIMAL(19, 4)
在将其应用于外部上下文之前,具有不同类型的操作数的每个操作的结果被明确地转换.
然而,结果仍然存在200.00
.
我现在已经创建了一个可以在您的计算机上执行的简化示例.
DECLARE @InstallmentIndex INT = 1
DECLARE @InstallmentCount INT = 1
DECLARE @InstallmentPercent DECIMAL(19, 4) = 1.0
DECLARE @PS TABLE (SharePrice DECIMAL(19, 4))
INSERT INTO @PS (SharePrice) VALUES (599.96)
-- 2000
SELECT
IIF(@InstallmentIndex < @InstallmentCount,
FLOOR(@InstallmentPercent * PS.SharePrice),
1999.96)
FROM @PS PS
-- 2000
SELECT
IIF(@InstallmentIndex < @InstallmentCount,
FLOOR(@InstallmentPercent * CAST(599.96 AS DECIMAL(19, 4))),
1999.96)
FROM @PS PS
-- 1996.96
SELECT
IIF(@InstallmentIndex < @InstallmentCount,
FLOOR(@InstallmentPercent * 599.96),
1999.96)
FROM @PS PS
-- Funny enough - with this sample explicitly converting EVERYTHING to DECIMAL(19, 4) - it still doesn't work...
-- 2000
SELECT
IIF(@InstallmentIndex < @InstallmentCount,
FLOOR(@InstallmentPercent * CAST(199.96 AS DECIMAL(19, 4))),
CAST(1999.96 AS DECIMAL(19, 4)))
FROM @PS PS
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现在我有东西......
-- 2000
SELECT
IIF(1 = 2,
FLOOR(CAST(1.0 AS decimal(19, 4)) * CAST(199.96 AS DECIMAL(19, 4))),
CAST(1999.96 AS DECIMAL(19, 4)))
-- 1999.9600
SELECT
IIF(1 = 2,
CAST(FLOOR(CAST(1.0 AS decimal(19, 4)) * CAST(199.96 AS DECIMAL(19, 4))) AS INT),
CAST(1999.96 AS DECIMAL(19, 4)))
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到底是什么 - 地板应该返回一个整数.这里发生了什么?:-D
我想我现在真的把它归结为本质:-D
-- 1.96
SELECT IIF(1 = 2,
CAST(1.0 AS DECIMAL (36, 0)),
CAST(1.96 AS DECIMAL(19, 4))
)
-- 2.0
SELECT IIF(1 = 2,
CAST(1.0 AS DECIMAL (37, 0)),
CAST(1.96 AS DECIMAL(19, 4))
)
-- 2
SELECT IIF(1 = 2,
CAST(1.0 AS DECIMAL (38, 0)),
CAST(1.96 AS DECIMAL(19, 4))
)
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Sta*_*dii 77
我需要首先解开这个,以便我可以看到发生了什么:
SELECT 199.96 -
(
0.0 *
FLOOR(
CAST(1.0 AS DECIMAL(19, 4)) *
CAST(199.96 AS DECIMAL(19, 4))
)
)
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现在让我们看看SQL Server在减法操作的每一侧使用的确切类型:
SELECT SQL_VARIANT_PROPERTY (199.96 ,'BaseType'),
SQL_VARIANT_PROPERTY (199.96 ,'Precision'),
SQL_VARIANT_PROPERTY (199.96 ,'Scale')
SELECT SQL_VARIANT_PROPERTY (0.0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4))) ,'BaseType'),
SQL_VARIANT_PROPERTY (0.0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4))) ,'Precision'),
SQL_VARIANT_PROPERTY (0.0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4))) ,'Scale')
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结果:
numeric 5 2 numeric 38 1
所以199.96
是numeric(5,2)
和较长Floor(Cast(etc))
的numeric(38,1)
.
结果精度和减法运算的规模(即:) 的规则e1 - e2
如下所示:
精度: max(s1,s2)+ max(p1-s1,p2-s2)+ 1
比例: max(s1,s2)
评估如下:
精度: max(1,2)+ max(38-1,5-2)+ 1 => 2 + 37 + 1 => 40
比例: max(1,2)=> 2
您还可以使用规则链接来确定numeric(38,1)
首先出现的位置(提示:您将两个精度值乘以19).
但:
- 结果精度和标度的绝对最大值为38.当结果精度大于38时,它减少到38,并减小相应的标度以试图防止结果的整数部分被截断.在某些情况下,例如乘法或除法,为了保持小数精度,比例因子不会降低,尽管可以提高溢出误差.
哎呀.精度为40.我们必须减少它,因为降低精度应始终切断最低有效数字,这意味着缩小比例.表达式的最终得到的类型将是numeric(38,0)
,这对于199.96
回合200
.
您可以通过移动和巩固可能解决这个问题CAST()
,从大的表达式中操作一个 CAST()
围绕整个表达式的结果.所以这:
SELECT 199.96 -
(
0.0 *
FLOOR(
CAST(1.0 AS DECIMAL(19, 4)) *
CAST(199.96 AS DECIMAL(19, 4))
)
)
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变为:
SELECT CAST( 199.96 - ( 0.0 * FLOOR(1.0 * 199.96) ) AS decimial(19,4))
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我甚至可以删除外部演员阵容.
我们在这里学习我们应该选择类型来匹配我们现在实际拥有的精度和规模,而不是预期的结果.仅仅考虑大精度数是没有意义的,因为SQL Server会在算术运算期间改变这些类型以试图避免溢出.
Sal*_*n A 20
密切关注以下语句涉及的数据类型:
SELECT 199.96 - (0.0 * FLOOR(CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4))))
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NUMERIC(19, 4) * NUMERIC(19, 4)
是NUMERIC(38, 7)
(见下文)
FLOOR(NUMERIC(38, 7))
是NUMERIC(38, 0)
(见下文)0.0
是 NUMERIC(1, 1)
NUMERIC(1, 1) * NUMERIC(38, 0)
是 NUMERIC(38, 1)
199.96
是 NUMERIC(5, 2)
NUMERIC(5, 2) - NUMERIC(38, 1)
是NUMERIC(38, 1)
(见下文)这就解释了为什么你最终得到200.0
(十进制后的一位数,而不是零)而不是199.96
.
笔记:
FLOOR
返回小于或等于指定数值表达式的最大整数,结果与输入的类型相同.它为INT返回INT,为FLOAT返回FLOAT,为NUMERIC(x,y)返回NUMERIC(x,0).
根据算法:
Run Code Online (Sandbox Code Playgroud)Operation | Result precision | Result scale* e1 * e2 | p1 + p2 + 1 | s1 + s2 e1 - e2 | max(s1, s2) + max(p1-s1, p2-s2) + 1 | max(s1, s2)
*结果精度和比例的绝对最大值为38.当结果精度大于38时,它减少到38,并减小相应的比例以试图防止结果的整数部分被截断.
该描述还包含在加法和乘法运算中如何精确缩小比例的细节.基于该描述:
NUMERIC(19, 4) * NUMERIC(19, 4)
是NUMERIC(39, 8)
和夹紧NUMERIC(38, 7)
NUMERIC(1, 1) * NUMERIC(38, 0)
是NUMERIC(40, 1)
和夹紧NUMERIC(38, 1)
NUMERIC(5, 2) - NUMERIC(38, 1)
是NUMERIC(40, 2)
和夹紧NUMERIC(38, 1)
这是我尝试在JavaScript中实现该算法.我已经针对SQL Server交叉检查了结果.它回答了你问题的本质部分.
// https://docs.microsoft.com/en-us/sql/t-sql/data-types/precision-scale-and-length-transact-sql?view=sql-server-2017
function numericTest_mul(p1, s1, p2, s2) {
// e1 * e2
var precision = p1 + p2 + 1;
var scale = s1 + s2;
// see notes in the linked article about multiplication operations
var newscale;
if (precision - scale < 32) {
newscale = Math.min(scale, 38 - (precision - scale));
} else if (scale < 6 && precision - scale > 32) {
newscale = scale;
} else if (scale > 6 && precision - scale > 32) {
newscale = 6;
}
console.log("NUMERIC(%d, %d) * NUMERIC(%d, %d) yields NUMERIC(%d, %d) clamped to NUMERIC(%d, %d)", p1, s1, p2, s2, precision, scale, Math.min(precision, 38), newscale);
}
function numericTest_add(p1, s1, p2, s2) {
// e1 + e2
var precision = Math.max(s1, s2) + Math.max(p1 - s1, p2 - s2) + 1;
var scale = Math.max(s1, s2);
// see notes in the linked article about addition operations
var newscale;
if (Math.max(p1 - s1, p2 - s2) > Math.min(38, precision) - scale) {
newscale = Math.min(precision, 38) - Math.max(p1 - s1, p2 - s2);
} else {
newscale = scale;
}
console.log("NUMERIC(%d, %d) + NUMERIC(%d, %d) yields NUMERIC(%d, %d) clamped to NUMERIC(%d, %d)", p1, s1, p2, s2, precision, scale, Math.min(precision, 38), newscale);
}
function numericTest_union(p1, s1, p2, s2) {
// e1 UNION e2
var precision = Math.max(s1, s2) + Math.max(p1 - s1, p2 - s2);
var scale = Math.max(s1, s2);
// my idea of how newscale should be calculated, not official
var newscale;
if (precision > 38) {
newscale = scale - (precision - 38);
} else {
newscale = scale;
}
console.log("NUMERIC(%d, %d) + NUMERIC(%d, %d) yields NUMERIC(%d, %d) clamped to NUMERIC(%d, %d)", p1, s1, p2, s2, precision, scale, Math.min(precision, 38), newscale);
}
/*
* first example in question
*/
// CAST(1.0 AS DECIMAL(19, 4)) * CAST(199.96 AS DECIMAL(19, 4))
numericTest_mul(19, 4, 19, 4);
// 0.0 * FLOOR(...)
numericTest_mul(1, 1, 38, 0);
// 199.96 * ...
numericTest_add(5, 2, 38, 1);
/*
* IIF examples in question
* the logic used to determine result data type of IIF / CASE statement
* is same as the logic used inside UNION operations
*/
// FLOOR(DECIMAL(38, 7)) UNION CAST(1999.96 AS DECIMAL(19, 4)))
numericTest_union(38, 0, 19, 4);
// CAST(1.0 AS DECIMAL (36, 0)) UNION CAST(1.96 AS DECIMAL(19, 4))
numericTest_union(36, 0, 19, 4);
// CAST(1.0 AS DECIMAL (37, 0)) UNION CAST(1.96 AS DECIMAL(19, 4))
numericTest_union(37, 0, 19, 4);
// CAST(1.0 AS DECIMAL (38, 0)) UNION CAST(1.96 AS DECIMAL(19, 4))
numericTest_union(38, 0, 19, 4);
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