Sql Server 2008中的字母数字排序

Question

任何人都可以帮我解决这个问题吗？
我有一个动态的模式列表，我想要排序，它包含字母数字值和字母。

CREATE TABLE dbo.Pattern (Pattern varchar(50) NULL)
INSERT INTO dbo.Pattern (Pattern) VALUES ('A11')
INSERT INTO dbo.Pattern (Pattern) VALUES ('A12')
INSERT INTO dbo.Pattern (Pattern) VALUES ('A8')
INSERT INTO dbo.Pattern (Pattern) VALUES ('A2')
INSERT INTO dbo.Pattern (Pattern) VALUES ('B6')
INSERT INTO dbo.Pattern (Pattern) VALUES ('B21')
INSERT INTO dbo.Pattern (Pattern) VALUES ('B10')
INSERT INTO dbo.Pattern (Pattern) VALUES ('B3')
INSERT INTO dbo.Pattern (Pattern) VALUES ('B100')
INSERT INTO dbo.Pattern (Pattern) VALUES ('B2')
INSERT INTO dbo.Pattern (Pattern) VALUES ('AA')
INSERT INTO dbo.Pattern (Pattern) VALUES ('BA')
INSERT INTO dbo.Pattern (Pattern) VALUES ('A20')
INSERT INTO dbo.Pattern (Pattern) VALUES ('AB')
INSERT INTO dbo.Pattern (Pattern) VALUES ('BB')

SELECT Pattern FROM dbo.Pattern ORDER BY Pattern

DROP Table dbo.Pattern

结果显示如下：

 A11
 A12
 A2
 A20
 A8
 AA
 AB
 B10
 B100
 B2
 B21
 B3
 B6
 BA
 BB

但我要展示的就是这样的结果：

AA
A1
A2
A8
A11
A12
A20
AB
BA
B2
B3
B6
B10
B21
B100
BB

Answer 1

对于您的示例数据，这很接近：

order by left(pattern, patindex('%[0-9]%', pattern)),
         patindex('%[0-9]%', pattern),
         len(pattern) asc,
         pattern

但是，你希望所有的alpha都是最后的，所以需要一个case （我认为）：

order by left(pattern, patindex('%[0-9]%', pattern)),
         (case when pattern like '%[0-9]%'
               then patindex('%[0-9]%', pattern)
               else 999
          end),
         len(pattern) asc,
         pattern

Answer 2

SELECT Pattern
FROM dbo.Pattern
ORDER BY CASE WHEN PATINDEX('%[0-9]%', Pattern) > 0
              THEN LEFT(Pattern, PATINDEX('%[0-9]%', Pattern)-1)
              ELSE Pattern
         END,
         CASE WHEN PATINDEX('%[0-9]%', Pattern) > 0
              THEN CONVERT(INT, SUBSTRING(Pattern, PATINDEX('%[0-9]%', Pattern), LEN(Pattern)))
              ELSE 0
         END

请在发布前搜索网络和Stack Overflow。

参考： http ： //www.essentialsql.com/use-sql-server-to-sort-alphanumeric-values/

Answer 3

SELECT Pattern
FROM dbo.Pattern 
ORDER BY LEFT(Pattern,1), 
        CASE WHEN SUBSTRING(Pattern,2,LEN(Pattern)) LIKE '%[0-9]%' THEN CAST(SUBSTRING(Pattern,2,LEN(Pattern)) as int) 
            WHEN SUBSTRING(Pattern,2,LEN(Pattern)) = 'A' THEN 0
            ELSE 10000000 END,
            SUBSTRING(Pattern,2,LEN(Pattern))

将输出：

Pattern
AA
A2
A8
A11
A12
A20
AB
BA
B2
B3
B6
B10
B21
B100
BB

Answer 4

我会分离alpha和num部分：

ORDER BY
    CASE WHEN PATINDEX('%[0-9]%', Pattern)=0 THEN 1 ELSE 0 END,--Put no-nums last
    CASE WHEN PATINDEX('%[0-9]%', Pattern) != 0 THEN LEFT(Pattern, PATINDEX('%[0-9]%', Pattern)-1) ELSE Pattern END,
    CASE WHEN PATINDEX('%[0-9]%', Pattern) != 0 THEN SUBSTRING(Pattern, PATINDEX('%[0-9]%', Pattern), LEN(Pattern)) END

Answer 5

使用交叉应用来简化行内计算

select pattern
from pattern
cross apply (
    select leftLen = isnull(nullif(patindex('%[0-9]%', pattern),0) - 1, len(pattern))
          ,totalLen = len(pattern)  
    ) c
order by 
 case leftLen when totalLen then 2 else 1 end,
 left(Pattern, leftLen),
 cast(right(Pattern, totalLen-leftLen) as int)