SQL Server 类似正则表达式的字符处理问题

时间：2024-12-17 09:37:56

SQL Serve提供了简单的字符模糊匹配功能，比如：like, patindex，不过对于某些字符处理场景还显得并不足够，日常碰到的几个问题有：

  • 1. 同一个字符/字符串，出现了多少次
  • 2. 同一个字符，第N次出现的位置
  • 3. 多个相同字符连续，合并为一个字符
  • 4. 是否为有效IP/身份证号/手机号等

  • 5. 去除所有数字/字母

 

一. 同一个字符/字符串，出现了多少次

同一个字符，将其替换为空串，即可计算

declare @text varchar(1000)

declare @str  varchar(10)

set @text = 'ABCBDBE'

set @str = 'B'

select len(@text) - len(replace(@text,@str,''))

同一个字符串，仍然是替换，因为是多个字符，方法1替换后需要做一次除法；方法2替换时增加一个字符，则不需要

--方法1

declare @text varchar(1000)

declare @str  varchar(10)

set @text = 'ABBBCBBBDBBBE'

set @str = 'BBB'

select (len(@text) - len(replace(@text,@str,'')))/len(@str)

--方法2

declare @text varchar(1000)

declare @str  varchar(10)

set @text = 'ABBBCBBBDBBBE'

set @str = 'BBB'

select len(replace(@text,@str,@str+'_')) - len(@text)

二. 同一个字符/字符串，第N次出现的位置

SQL SERVER定位字符位置的函数为CHARINDEX：

CHARINDEX ( expressionToFind , expressionToSearch [ , start_location ] )

可以从指定位置起开始检索，但是不能取第N次出现的位置，需要自己写SQL来补充，有以下几种思路：

1. 自定义函数, 循环中每次为charindex加一个计数，直到为N

if object_id('NthChar','FN') is not null

    drop function Nthchar

GO

create function NthChar

(

@source_string as nvarchar(4000),

@sub_string    as nvarchar(1024),

@nth           as int

)

returns int

as

begin

    declare @postion int

    declare @count   int 

    set @postion = CHARINDEX(@sub_string, @source_string)

    set @count = 0 

    while @postion > 0

    begin

        set @count = @count + 1

        if @count = @nth

        begin

            break

        end

        set @postion = CHARINDEX(@sub_string, @source_string, @postion + 1)

    End

    return @postion

end

GO

--select dbo.NthChar('abcabc','abc',2)

--

2. 通过CTE，对待处理的整个表字段操作, 递归中每次为charindex加一个计数，直到为N

if  object_id('tempdb..#T') is not null

    drop table #T

create table #T

(

source_string nvarchar(4000)

)

insert into #T values (N'我们我们')

insert into #T values (N'我我哦我')

declare @sub_string nvarchar(1024)

declare @nth        int

set @sub_string = N'我们'

set @nth = 2

;with T(source_string, starts, pos, nth)

as (

    select source_string, 1, charindex(@sub_string, source_string), 1 from #t

    union all

    select source_string, pos + 1, charindex(@sub_string, source_string, pos + 1), nth+1 from T

    where pos > 0

)

select

    source_string, pos, nth

from T

where pos <> 0

  and nth = @nth

order by source_string, starts

--source_string    pos    nth

--我们我们    3    2

3. 借助数字表 (tally table)，到不同起点位置去做charindex，需要先自己构造个数字表

--numbers/tally table

IF EXISTS (select * from dbo.sysobjects where id = object_id(N'[dbo].[Numbers]') and OBJECTPROPERTY(id, N'IsUserTable') = 1)

    DROP TABLE dbo.Numbers

--===== Create and populate the Tally table on the fly

 SELECT TOP 1000000

        IDENTITY(int,1,1) AS number

   INTO dbo.Numbers

   FROM master.dbo.syscolumns sc1,

        master.dbo.syscolumns sc2

--===== Add a Primary Key to maximize performance

  ALTER TABLE dbo.Numbers

        ADD CONSTRAINT PK_numbers_number PRIMARY KEY CLUSTERED (number)

--===== Allow the general public to use it

  GRANT SELECT ON dbo.Numbers TO PUBLIC

--以上数字表创建一次即可，不需要每次都重复创建

DECLARE @source_string    nvarchar(4000),

        @sub_string       nvarchar(1024),

        @nth              int

SET @source_string = 'abcabcvvvvabc'

SET @sub_string = 'abc'

SET @nth = 2 

;WITH T

AS

(

SELECT ROW_NUMBER() OVER(ORDER BY number) AS nth,

       number AS [Position In String]

  FROM dbo.Numbers n

 WHERE n.number <= LEN(@source_string)

   AND CHARINDEX(@sub_string, @source_string, n.number)-number = 0

   ----OR

   --AND SUBSTRING(@source_string,number,LEN(@sub_string)) = @sub_string

)

SELECT * FROM T WHERE nth = @nth

4. 通过CROSS APPLY结合charindex，适用于N值较小的时候，因为CROSS APPLY的次数要随着N的变大而增加，语句也要做相应的修改

declare @T table

(

source_string nvarchar(4000)

)

insert into @T values

('abcabc'),

('abcabcvvvvabc')

declare @sub_string nvarchar(1024)

set @sub_string = 'abc'

select source_string,

       p1.pos as no1,

       p2.pos as no2,

       p3.pos as no3

from @T

cross apply (select (charindex(@sub_string, source_string))) as P1(Pos)

cross apply (select (charindex(@sub_string, source_string, P1.Pos+1))) as P2(Pos)

cross apply (select (charindex(@sub_string, source_string, P2.Pos+1))) as P3(Pos)

5. 在SSIS里有内置的函数，但T-SQL中并没有

--FINDSTRING in SQL Server 2005 SSIS

FINDSTRING([yourColumn], "|", 2),

--TOKEN in SQL Server 2012 SSIS

TOKEN(Col1,"|",3)

注：不难发现，这些方法和字符串拆分的逻辑是类似的，只不过一个是定位，一个是截取，如果要获取第N个字符左右的一个/多个字符，有了N的位置，再结合substring去截取即可；

三. 多个相同字符连续，合并为一个字符

最常见的就是把多个连续的空格合并为一个空格，解决思路有两个：

1. 比较容易想到的就是用多个replace

但是究竟需要replace多少次并不确定，所以还得循环多次才行

--把两个连续空格替换成一个空格，然后循环，直到charindex检查不到两个连续空格

declare @str varchar(100)

set @str='abc        abc     kljlk     kljkl'

while(charindex('  ',@str)>0)

begin

    select @str=replace(@str,'  ',' ')

end

select @str

2. 按照空格把字符串拆开

对每一段拆分开的字符串trim或者replace后，再用一个空格连接，有点繁琐，没写代码示例，如何拆分字符串可参考：“第N次出现的位置”；

四. 是否为有效IP/身份证号/手机号等

类似IP/身份证号/手机号等这些字符串，往往都有自身特定的规律，通过substring去逐位或逐段判断是可以的，但SQL语句的方式往往性能不佳，建议尝试正则函数，见下。

五. 去除所有数字/字母

上面定位第N个字符的方式也都可以尝试使用，只是在定位的同时将数字/字母替换掉即可；

另外也可以通过模糊匹配来做：

--去除所有非数字

declare @pos smallint

declare @string varchar(100)

set @string = '1109A><":{$%^&*4DSE2@!~$%^&567KJHGT'

--如果是1102e53这样的字符串，会被isnumeric认为是科学计数法数字，加上e0就会返回0

--如果是普通数字，加上e0也不会影响返回1

while isnumeric(@string+'e0') = 0 --也可同下用patindex判断

begin

    set @pos = (select patindex('%[^0-9]%',@string))

    set @string = (select replace(@string,substring(@string,@pos,1),''))

end

select @string

--去除所有数字

declare @pos smallint

declare @string varchar(100)

set @string = '1109A><":{$%^&*4DSE2@!~$%^&567KJHGT'

while patindex('%[0-9]%',@string) > 0

begin

    set @pos = (select patindex('%[0-9]%',@string))

    set @string = (select replace(@string,substring(@string,@pos,1),''))

end

select @string

--去除所有字母

declare @pos smallint

declare @string varchar(100)

set @string = '1109A><":{$%^&*4DSE2@!~$%^&567KJHGT'

while patindex('%[A-Za-z]%',@string) > 0

begin

    set @pos = (select patindex('%[A-Za-z]%',@string))

    set @string = (select replace(@string,substring(@string,@pos,1),''))

end

select @string

六. 正则表达式函数

1. Oracle

从10g开始，可以在查询中使用正则表达式，它通过一些支持正则表达式的函数来实现：

Oracle 10 g

REGEXP_LIKE

REGEXP_REPLACE

REGEXP_INSTR

REGEXP_SUBSTR

Oracle 11g (新增)

REGEXP_COUNT

Oracle用REGEXP函数处理上面几个问题：

(1) 同一个字符/字符串，出现了多少次

select length(regexp_replace('123-345-566', '[^-]', '')) from dual;

select REGEXP_COUNT('123-345-566', '-') from dual; --Oracle 11g

(2) 同一个字符/字符串，第N次出现的位置

不需要正则，ORACLE的instr可以直接查找位置：

instr('source_string','sub_string' [,n][,m])

n表示从第n个字符开始搜索，缺省值为1，m表示第m次出现，缺省值为1。

select instr('abcdefghijkabc','abc', 1, 2) position from dual;

(3) 多个相同字符连续，合并为一个字符

select regexp_replace(trim('agc f   f  '),'\s+',' ') from dual;

(4) 是否为有效IP/身份证号/手机号等

--是否为有效IP

WITH IP

AS(

SELECT '10.20.30.40' ip_address FROM dual UNION ALL

SELECT 'a.b.c.d' ip_address FROM dual UNION ALL

SELECT '256.123.0.254' ip_address FROM dual UNION ALL

SELECT '255.255.255.255' ip_address FROM dual

)

SELECT *

FROM IP

WHERE REGEXP_LIKE(ip_address, '^(([0-9]{1}|[0-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])\.){3}([0-9]{1}|[0-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])$');

--是否为有效身份证/手机号，暂未举例

(5) 去除所有数字/字母

--去除非数字

SELECT regex_replace('1109A><":{$%^&*4DSE2@!~$%^&567KJHGT','[^0-9]','') FROM dual;

--去除数字

SELECT regex_replace('1109A><":{$%^&*4DSE2@!~$%^&567KJHGT','[0-9]','') FROM dual;

--去除字母

SELECT regex_replace('1109A><":{$%^&*4DSE2@!~$%^&567KJHGT','[A-Za-z]','') FROM dual;

2. SQL Server

目前最新版本为SQL Server 2017，还没有对REGEXP函数的支持，需要通用CLR来扩展，如下为CLR实现REG_REPLACE：

--1. 开启 CLR

EXEC sp_configure 'show advanced options' , ''

GO

RECONFIGURE

GO

EXEC sp_configure 'clr enabled' , ''

GO

RECONFIGURE

GO

EXEC sp_configure 'show advanced options' , '';

GO
--首次创建时，应该是从dll文件创建

--CREATE ASSEMBLY [RegexUtility] FROM 'C:\xxxxxxx.dll'

--GO

--创建好后，可以生成脚本出来部署到其他支持CLR的SQL Server上

CREATE ASSEMBLY [RegexUtility]

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WITH PERMISSION_SET = SAFE

GO
--3. 创建 CLR 函数

CREATE FUNCTION [dbo].[regex_replace](@input [nvarchar](4000), @pattern [nvarchar](4000), @replacement [nvarchar](4000))

RETURNS [nvarchar](4000) WITH EXECUTE AS CALLER, RETURNS NULL ON NULL INPUT

AS

EXTERNAL NAME [RegexUtility].[RegexUtility].[RegexReplaceDefault]

GO
--4. 使用regex_replace替换多个空格为一个空格

select dbo.regex_replace('agc f   f  ','\s+',' ');
--5. 使用regex_replace去除数字/字母

--去除非数字

select dbo.regex_replace('1109A><":{$%^&*4DSE2@!~$%^&567KJHGT','[^0-9]','');

--去除数字

select dbo.regex_replace('1109A><":{$%^&*4DSE2@!~$%^&567KJHGT','[0-9]','');

--去除字母

select dbo.regex_replace('1109A><":{$%^&*4DSE2@!~$%^&567KJHGT','[A-Za-z]','');

注：通过CLR实现更多REGEXP函数，如果有高级语言开发能力，可以自行开发；或者直接使用一些开源贡献也行，比如：http://devnambi.com/2016/sql-server-regex/

简单的正则表达式符号使用说明：

\f -> 匹配一个换页

\n -> 匹配一个换行符

\r -> 匹配一个回车符

\t -> 匹配一个制表符

\v -> 匹配一个垂直制表符

\s 匹配任何空白字符，包括空格、制表符、换页符等等, 等价于[ \f\n\r\t\v]

\s+ 匹配任意多个空白字符

\S 匹配非空白字符

^ 匹配最前边的字符

$ 与^类似,匹配最末的字符

+ 匹配+号前面的字符1次或n次.等价于{1,}

* 匹配*前面的字符0次或n次

? 匹配?前面的字符0次或1次

. (小数点)匹配除换行符外的所有单个的字符

小结：

1. 非正则SQL语句的思路，对不同数据库往往都适用；

2. 正则表达式中的规则(pattern) 在不同开发语言里，有很多语法是相通的，通常是遵守perl或者linux shell中的sed等工具的规则；

3. 从性能上来看，非循环写法的通用SQL判断 > REGEXP函数 > 自定义SQL函数；

4. SQL SERVER 除了CLR的方式扩展正则表达式函数，也可以开启OLE Automation通过VBS来扩展正式表达式函数，曾经测试过性能，比CLR要差很多。

参考：

regular expression enhancements in 11g

http://www.oracle-developer.net/display.php?id=508

Using Regular Expressions With Oracle Database

https://docs.oracle.com/cd/B12037_01/appdev.101/b10795/adfns_re.htm

Replace non numeric characters in string

https://www.sqlservercentral.com/Forums/Topic470379-338-1.aspx

IsNumeric, IsInt, IsNumber

https://www.tek-tips.com/faqs.cfm?fid=6423

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