14. 類似正則表達式的字元處理問題

SQL Serve提供了簡單的字元模糊匹配功能，比如：like, patindex，不過對於某些字元處理場景還顯得並不足夠，日常碰到的幾個問題有：

• 1. 同一個字元/字元串，出現了多少次
• 2. 同一個字元，第N次出現的位置
• 3. 多個相同字元連續，合併為一個字元
• 4. 是否為有效IP/身份證號/手機號等

一. 同一個字元/字元串，出現了多少次

同一個字元，將其替換為空串，即可計算

declare @text varchar(1000)
declare @str  varchar(10)
set @text = 'ABCBDBE'
set @str = 'B'

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

同一個字元串，仍然是替換，因為是多個字元，方法1替換後需要做一次除法；方法2替換時增加一個字元，則不需要

--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. 同一個字元/字元串，第N次出現的位置

SQL SERVER定位字元位置的函數為CHARINDEX：

CHARINDEX ( expressionToFind , expressionToSearch [ , start_location ] )

可以從指定位置起開始檢索，但是不能取第N次出現的位置，需要自己寫SQL來補充，有以下幾種思路：

1. 自定義函數, 迴圈中每次為charindex加一個計數，直到為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. 通過CTE，對待處理的整個表欄位操作, 遞歸中每次為charindex加一個計數，直到為N

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

create table #T
(
source_string nvarchar(4000)
)

insert into #T values (N'我們我們')
insert into #T values (N'我我哦我')

declare @sub_string nvarchar(1024)
declare @nth        int
set @sub_string = N'我們'
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
--我們我們    3    2

3. 藉助數字表 (tally table)，到不同起點位置去做charindex，需要先自己構造個數字表

--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

--以上數字表創建一次即可，不需要每次都重覆創建

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. 通過CROSS APPLY結合charindex，適用於N值較小的時候，因為CROSS APPLY的次數要隨著N的變大而增加，語句也要做相應的修改

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里有內置的函數，但T-SQL中並沒有

--FINDSTRING in SQL Server 2005 SSIS
FINDSTRING([yourColumn], "|", 2),

--TOKEN in SQL Server 2012 SSIS
TOKEN(Col1,"|",3)

註：不難發現，這些方法和字元串拆分的邏輯是類似的，只不過一個是定位，一個是截取，如果要獲取第N個字元左右的一個/多個字元，有了N的位置，再結合substring去截取即可；

三. 多個相同字元連續，合併為一個字元

最常見的就是把多個連續的空格合併為一個空格，解決思路有兩個：

1. 比較容易想到的就是用多個replace

但是究竟需要replace多少次並不確定，所以還得迴圈多次才行

--把兩個連續空格替換成一個空格，然後迴圈，直到charindex檢查不到兩個連續空格
declare @str varchar(100)
set @str='abc        abc     kljlk     kljkl'
while(charindex('  ',@str)>0)
begin
    select @str=replace(@str,'  ',' ')
end
select @str

2. 按照空格把字元串拆開

對每一段拆分開的字元串trim或者replace後，再用一個空格連接，有點繁瑣，沒寫代碼示例，如何拆分字元串可參考：“第N次出現的位置”；

四. 是否為有效IP/身份證號/手機號等

類似IP/身份證號/手機號等這些字元串，往往都有自身特定的規律，通過substring去逐位或逐段判斷是可以的，但SQL語句的方式往往性能不佳，建議嘗試正則函數，見下。

五. 正則表達式函數

1. Oracle

從10g開始，可以在查詢中使用正則表達式，它通過一些支持正則表達式的函數來實現：

Oracle 10 g

REGEXP_LIKE

REGEXP_REPLACE

REGEXP_INSTR

REGEXP_SUBSTR

Oracle 11g (新增)

REGEXP_COUNT

Oracle用REGEXP函數處理上面幾個問題：

(1) 同一個字元/字元串，出現了多少次

select length(regexp_replace('123-345-566', '[^-]', '')) from dual;
select REGEXP_COUNT('123-345-566', '-') from dual; --Oracle 11g

(2) 同一個字元/字元串，第N次出現的位置

不需要正則，ORACLE的instr可以直接查找位置：

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

n表示從第n個字元開始搜索，預設值為1，m表示第m次出現，預設值為1。

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

(3) 多個相同字元連續，合併為一個字元

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

(4) 是否為有效IP/身份證號/手機號等

--是否為有效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])$');
--是否為有效身份證/手機號，暫未舉例

2. SQL Server

目前最新版本為SQL Server 2017，還沒有對REGEXP函數的支持，需要通用CLR來擴展，如下為CLR實現REG_REPLACE：

--1. 開啟 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

--首次創建時，應該是從dll文件創建
--CREATE ASSEMBLY [RegexUtility] FROM 'C:\xxxxxxx.dll'
--GO

--創建好後，可以生成腳本出來部署到其他支持CLR的SQL Server上
CREATE ASSEMBLY [RegexUtility]
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WITH PERMISSION_SET = SAFE

GO

--3. 創建 CLR 函數
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替換多個空格為一個空格
select dbo.regex_replace('agc f   f  ','\s+',' ');

註：通過CLR實現更多REGEXP函數，如果有高級語言開發能力，可以自行開發；或者直接使用一些開源貢獻也行，比如：http://devnambi.com/2016/sql-server-regex/

小結：

1. 非正則SQL語句的思路，對不同資料庫往往都適用；

2. 正則表達式中的規則(pattern) 在不同開發語言里，有很多語法是相通的，通常是遵守perl或者linux shell中的sed等工具的規則；

3. 從性能上來看，通用SQL判斷 > REGEXP函數 > 自定義SQL函數。

參考：

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

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