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126直营-126bet6手机版-126bet8直营网手机版

发布时间:2019-03-19

原标题:14. 类似正则表达式的字符处理问题

一走就数公里远,直到来到一座山脉底下才停下来,看四周的地势,灵气流动很明显是一个风水宝地,虽然不是洞天福地,但绝对是顶级的风水宝地。

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火和冰的极致对碰大量的白雾弥漫全场,可是两宠却还是坚持着进行激烈的对轰,使出火焰车的火精灵挡着暴风雪不断前进,眼看要闯过去的时候伊布夹然换招,刘皓没有命令,这夹如其来的换招让辉夜有点措手不及。
红衣女孩莹抓着电话的手又紧了一次,脑子里飞过许多这段日子里和雪飞鸿在一起的画面,最后用力的咬着唇,细声答道:“他对我很好,和他在一起我很快乐。”

“这可由不得你来决定的,而是由我自己,我想什么时候继续战斗就什么时候继续战斗。”。

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

 

. 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)

 

. 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;

 

. 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”;

 

. 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。

 

. 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

 

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]
FROM 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WITH PERMISSION_SET = SAFE

GO
2. 创建 Assembly
--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+"," ");

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/

 

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,通用SQL判断 > REGEXP函数 > 自定义SQL函数。

 

参考:

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

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

 

编辑:北顺宗石

发布时间:2019-03-19 03:40:56

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