SQL Server 类似正则表达式的字符处理问题

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)
--4

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' , '1'
GO
RECONFIGURE
GO
EXEC sp_configure 'clr enabled' , '1'
GO
RECONFIGURE
GO
EXEC sp_configure 'show advanced options' , '0';
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

View Code

--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]','');