快速版以下哪项是最好的?为什么? (或者有更好的方法):
SELECT FLOOR(LOG10(Number))+1 AS NumLength FROM Table
SELECT LEN(CONVERT(VARCHAR, Number)) AS NumLength FROM Table
SELECT LEN(CAST(Number AS VARCHAR(10))) AS NumLength FROM Table
更详细一点:
我希望找出最有效的机制来计算整数的字符串表示的长度(更具体地说,是一个自然数 - 总是> 0)。
我正在使用MS SQL Server(2005)。
我已经提出了上面的3个解决方案,所有这些解决方案似乎都运行良好。
我知道第三个版本可能存在非常大的整数问题,但是现在我们可以假设“数字”长度不超过9位小数。
更详细: (您无需阅读此位来回答我的问题)
此查询在事务处理环境中大量使用
到目前为止,我已经假设“数字”总是正好是6位数
但是,现在我必须更新代码以支持4到9位数字。
此SQL是识别卡片卡方案的条件的一部分。
完整查询尝试在开始和结束范围内找到与卡号开头匹配的记录。
所以完整的SQL条件是这样的:
WHERE
-- Start and End match
((Start=End OR End=0) AND (Start=CAST(LEFT('<card number>', FLOOR(LOG10(Start))+1) AS BIGINT))) OR
-- Start != End
-- >= Start
(Start<=CAST(LEFT('<card number>', FLOOR(LOG10(Start))+1) AS BIGINT) AND
-- <= End
End>=CAST(LEFT('<card number>', FLOOR(LOG10(Start))+1) AS BIGINT))
注意:
我可以重新设计表以使用VARCHAR而不是INT。这将允许我使用“LEN(开始)”而不是“FLOOR(LOG10(开始))+ 1)”然而条件将有更多的CAST。
我宁愿继续处理INT,因为数据库模式将保持不变,无论如何处理INT应该比VARCHAR更快。
如果我将字段更改为VARCHAR,我的情况可能是:
WHERE
-- Start and End match
((Start=End OR LEN(End)=0) AND (Start=LEFT('<card number>', LEN(Start)))) OR
-- Start != End
-- >= Start
(CAST(Start AS BIGINT)<=CAST(LEFT('<card number>', LEN(Start)) AS BIGINT) AND
-- <= End
CAST(End AS BIGINT)>=CAST(LEFT('<card number>', LEN(Start)) AS BIGINT))
非常感谢任何帮助,
戴夫
答案 0 :(得分:2)
在我的机器上,版本2和3出现大致相同并击败其他两个。
编辑虽然刚刚发现我的原始测试对CASE
有点不公平,因为按升序排列语句意味着只有10个可能的数字符合第一个条件并提前退出。我在下面添加了一个额外的测试。您也可以尝试嵌套CASE
语句来进行二分查找。
SET NOCOUNT ON
SET STATISTICS TIME ON
PRINT 'Test 1';
WITH E00(N) AS (SELECT 1 UNION ALL SELECT 1),
E02(N) AS (SELECT 1 FROM E00 a, E00 b),
E04(N) AS (SELECT 1 FROM E02 a, E02 b),
E08(N) AS (SELECT 1 FROM E04 a, E04 b),
E16(N) AS (SELECT 1 FROM E08 a, E08 b),
E32(N) AS (SELECT 1 FROM E16 a, E16 b),
cteTally(N) AS (SELECT ROW_NUMBER() OVER (ORDER BY N) FROM E32)
SELECT MAX(FLOOR(LOG10(N))+1)
FROM cteTally
WHERE N <= 10000000;
PRINT 'Test 2';
WITH E00(N) AS (SELECT 1 UNION ALL SELECT 1),
E02(N) AS (SELECT 1 FROM E00 a, E00 b),
E04(N) AS (SELECT 1 FROM E02 a, E02 b),
E08(N) AS (SELECT 1 FROM E04 a, E04 b),
E16(N) AS (SELECT 1 FROM E08 a, E08 b),
E32(N) AS (SELECT 1 FROM E16 a, E16 b),
cteTally(N) AS (SELECT ROW_NUMBER() OVER (ORDER BY N) FROM E32)
SELECT MAX(LEN(CONVERT(VARCHAR, N)))
FROM cteTally
WHERE N <= 10000000;
PRINT 'Test 3';
WITH E00(N) AS (SELECT 1 UNION ALL SELECT 1),
E02(N) AS (SELECT 1 FROM E00 a, E00 b),
E04(N) AS (SELECT 1 FROM E02 a, E02 b),
E08(N) AS (SELECT 1 FROM E04 a, E04 b),
E16(N) AS (SELECT 1 FROM E08 a, E08 b),
E32(N) AS (SELECT 1 FROM E16 a, E16 b),
cteTally(N) AS (SELECT ROW_NUMBER() OVER (ORDER BY N) FROM E32)
SELECT MAX(LEN(CAST(N AS VARCHAR(10))))
FROM cteTally
WHERE N <= 10000000;
PRINT 'Test 4';
WITH E00(N) AS (SELECT 1 UNION ALL SELECT 1),
E02(N) AS (SELECT 1 FROM E00 a, E00 b),
E04(N) AS (SELECT 1 FROM E02 a, E02 b),
E08(N) AS (SELECT 1 FROM E04 a, E04 b),
E16(N) AS (SELECT 1 FROM E08 a, E08 b),
E32(N) AS (SELECT 1 FROM E16 a, E16 b),
cteTally(N) AS (SELECT ROW_NUMBER() OVER (ORDER BY N) FROM E32)
SELECT MAX(CASE
WHEN N < 10 THEN 1
WHEN N < 100 THEN 2
WHEN N < 1000 THEN 3
WHEN N < 10000 THEN 4
WHEN N < 100000 THEN 5
WHEN N < 1000000 THEN 6
WHEN N < 10000000 THEN 7
WHEN N < 100000000 THEN 8
END)
FROM cteTally
WHERE N <= 10000000;
PRINT 'Test 5';
WITH E00(N) AS (SELECT 1 UNION ALL SELECT 1),
E02(N) AS (SELECT 1 FROM E00 a, E00 b),
E04(N) AS (SELECT 1 FROM E02 a, E02 b),
E08(N) AS (SELECT 1 FROM E04 a, E04 b),
E16(N) AS (SELECT 1 FROM E08 a, E08 b),
E32(N) AS (SELECT 1 FROM E16 a, E16 b),
cteTally(N) AS (SELECT ROW_NUMBER() OVER (ORDER BY N) FROM E32)
SELECT MAX(CASE
WHEN N >= 100000000 THEN NULL
WHEN N >= 10000000 THEN 8
WHEN N >= 1000000 THEN 7
WHEN N >= 100000 THEN 6
WHEN N >= 10000 THEN 5
WHEN N >= 1000 THEN 4
WHEN N >= 100 THEN 3
WHEN N >= 10 THEN 2
ELSE 1
END )
FROM cteTally
WHERE N <= 10000000;
在我的机器上运行示例的结果是
Test 1
CPU time = 9422 ms, elapsed time = 9523 ms.
Test 2
CPU time = 7021 ms, elapsed time = 7130 ms.
Test 3
CPU time = 6864 ms, elapsed time = 7006 ms.
Test 4
CPU time = 9328 ms, elapsed time = 9456 ms.
Test 5
CPU time = 6989 ms, elapsed time = 7358 ms.
答案 1 :(得分:1)
要回答您的问题,第二个版本更清楚您实际需要什么。想想在六个月内看到这段代码的人会想到:他们是否会意识到第一个版本正在尝试获得以十进制表示的数字的长度,或者他们会认为你正在执行一些模糊的数学运算,他们可以'找到需要的文件?
更一般地说,您应该考虑将这些值存储为字符数据,因为它们并不代表您的真实“数字”(您不是根据相对值进行比较,而是您没有执行算术,等等。)。您可以使用CHECK
约束来确保字段中只有数字。
我不清楚为什么将它们存储为字符数据需要在查询中进行转换,假设您是一致的。也没有理由认为处理int
s会比varchar
更快,特别是如果两种情况都涉及转换。