SQL Server相当于MySQL中的substring_index函数

时间:2014-05-25 11:18:30

标签: mysql sql-server-2012 porting

我正在尝试将查询从MySQL移植到SQL SERVER 2012.

我如何编写MySQL的substring_index()的等价物?

MySQL SUBSTRING_INDEX()在指定的分隔符出现次数之前返回给定字符串的子字符串。

SUBSTRING_INDEX(str,delim,count)

SELECT SUBSTRING_INDEX('www.somewebsite.com','.',2);

输出:'www.somewebsite'

7 个答案:

答案 0 :(得分:5)

尝试基于T-SQL和XQuery((root/row)[position() <= sql:variable("@count")])的解决方案:

T-SQL标量函数:

CREATE FUNCTION dbo.SUBSTRING_INDEX
(
    @str NVARCHAR(4000),
    @delim NVARCHAR(1),
    @count INT
)
RETURNS NVARCHAR(4000)
WITH SCHEMABINDING
BEGIN
    DECLARE @XmlSourceString XML;
    SET @XmlSourceString = (SELECT N'<root><row>' + REPLACE( (SELECT @str AS '*' FOR XML PATH('')) , @delim, N'</row><row>' ) + N'</row></root>');

    RETURN STUFF
    (
        ((
            SELECT  @delim + x.XmlCol.value(N'(text())[1]', N'NVARCHAR(4000)') AS '*'
            FROM    @XmlSourceString.nodes(N'(root/row)[position() <= sql:variable("@count")]') x(XmlCol)
            FOR XML PATH(N''), TYPE
        ).value(N'.', N'NVARCHAR(4000)')), 
        1, 1, N''
    );
END
GO

SELECT dbo.SUBSTRING_INDEX(N'www.somewebsite.com', N'.', 2) AS Result;

输出:

/*
Result
---------------
www.somewebsite
*/

TSQL内联表值函数:

CREATE FUNCTION dbo.SUBSTRING_INDEX
(
    @str NVARCHAR(4000),
    @delim NVARCHAR(1),
    @count INT
)
RETURNS TABLE
AS 
RETURN
    WITH Base
    AS 
    (
        SELECT XmlSourceString = CONVERT(XML, (SELECT N'<root><row>' + REPLACE( (SELECT @str AS '*' FOR XML PATH('')) , @delim, N'</row><row>' ) + N'</row></root>'))
    )   
    SELECT STUFF
    (
        ((
            SELECT  @delim + x.XmlCol.value(N'(text())[1]', N'NVARCHAR(4000)') AS '*'
            FROM    Base b 
            CROSS APPLY b.XmlSourceString.nodes(N'(root/row)[position() <= sql:variable("@count")]') x(XmlCol)
            FOR XML PATH(N''), TYPE
        ).value(N'.', N'NVARCHAR(4000)')), 
        1, 1, N''
    ) AS Result;
GO

SELECT  *
FROM    (
    SELECT N'www.somewebsite.com' UNION ALL 
    SELECT N'www.yahoo.com' UNION ALL 
    SELECT N'www.outlook.com'
) a(Value)
CROSS APPLY dbo.SUBSTRING_INDEX(a.Value, N'.', 2) b;

输出:

/*
Value               Result
------------------- ---------------
www.somewebsite.com www.somewebsite
www.yahoo.com       www.yahoo
www.outlook.com     www.outlook
*/

答案 1 :(得分:3)

我最近需要这个,所以我编写了以下存储函数。最后是一系列测试,以确保它完全像MySql函数那样运行(在那里运行相同的测试后,预期的结果是从MySql复制的):

-- Function to reproduce the useful functionality of SUBSTRING_INDEX from MySql
CREATE FUNCTION dbo.SUBSTRING_INDEX(@InString  NVARCHAR(Max),
                                    @Delimiter NVARCHAR(Max),
                                    @Count     INT)
RETURNS NVARCHAR(200)
AS
BEGIN
    DECLARE @Pos INT;
    DECLARE @DelimiterOffsets TABLE
    (
         i      INT IDENTITY(1, 1) NOT NULL,
         offset INT NOT NULL
    );

    -- If @Count is zero, we return '' as per spec
    IF @Count = 0
    BEGIN
        RETURN '';
    END;

    DECLARE @OrigLength      INT = LEN(@InString);
    DECLARE @DelimiterLength INT = LEN(@Delimiter);

    -- Prime the pump.
    SET @Pos = Charindex(@Delimiter, @InString, 1);

    -- If the delimiter does not exist in @InString, return the whole string
    IF @Pos = 0
    BEGIN
        RETURN @InString;
    END;

    -- Put all delimiter offsets into @DelimiterOffsets, they get numbered automatically.
    DECLARE @CurrentOffset INT = 0;
    WHILE @Pos > 0
    BEGIN
        SET @CurrentOffset = @Pos;

        INSERT INTO @DelimiterOffsets
                    (offset)
             VALUES (@CurrentOffset);

        SET @Pos = Charindex(@Delimiter, @InString, @CurrentOffset + @DelimiterLength);
    END;

    -- This number is guaranteed to be > 0.
    DECLARE @DelimitersFound INT = (SELECT Count(*) FROM @DelimiterOffsets);

    -- If they requested more delimiters than were found, return the whole string, as per spec.
    IF Abs(@Count) > @DelimitersFound
    BEGIN
        RETURN @InString;
    END;

    DECLARE @StartSubstring INT = 0;
    DECLARE @EndSubstring   INT = @OrigLength;

    -- OK, now return the part they requested
    IF @Count > 0
    BEGIN
        SET @EndSubstring = (SELECT offset 
                               FROM @DelimiterOffsets 
                              WHERE i = @Count);
    END
    ELSE
    BEGIN
        SET @StartSubstring = (SELECT offset + @DelimiterLength 
                                 FROM @DelimiterOffsets 
                                WHERE i = (@DelimitersFound + @Count + 1));
    END;

    RETURN Substring(@InString, @StartSubstring, @EndSubstring);
END; 

Go 

GRANT EXECUTE ON [dbo].SUBSTRING_INDEX TO PUBLIC;

-- Tests
DECLARE @TestResults TABLE (i int, answer nVarChar(MAX), expected nVarChar(MAX));

insert into @TestResults
select * from  
(
    (SELECT  1 as i, [dbo].SUBSTRING_INDEX(N'www.somewebsite.com', N'.', 2)    as r, 'www.somewebsite'     as e) UNION
    (SELECT  2 as i, [dbo].SUBSTRING_INDEX(N'www.yahoo.com', N'.', 2)          as r, 'www.yahoo'           as e) UNION
    (SELECT  3 as i, [dbo].SUBSTRING_INDEX(N'www.outlook.com', N'.', 2)        as r, 'www.outlook'         as e) UNION
    (SELECT  4 as i, [dbo].SUBSTRING_INDEX(N'www.somewebsite.com', N'.', -2)   as r, 'somewebsite.com'     as e) UNION
    (SELECT  5 as i, [dbo].SUBSTRING_INDEX(N'www.yahoo.com', N'.', -2)         as r, 'yahoo.com'           as e) UNION
    (SELECT  6 as i, [dbo].SUBSTRING_INDEX(N'www.outlook.com', N'.', -2)       as r, 'outlook.com'         as e) UNION
    (select  7 as i, [dbo].SUBSTRING_INDEX('hi.you.com','.',2)                 as r, 'hi.you'              as e) UNION
    (select  8 as i, [dbo].SUBSTRING_INDEX('hi.you.com','.',-1)                as r, 'com'                 as e) UNION
    (select  9 as i, [dbo].SUBSTRING_INDEX(N'prueba','ue',1)                   as r, 'pr'                  as e) UNION
    (select 10 as i, [dbo].SUBSTRING_INDEX(N'prueba','ue',-1)                  as r, 'ba'                  as e) UNION
    (select 11 as i, [dbo].SUBSTRING_INDEX(N'prueba','ue',0)                   as r, ''                    as e) UNION
    (SELECT 12 as i, [dbo].SUBSTRING_INDEX(N'wwwxxxoutlookxxxcom', N'xxx', 2)  as r, 'wwwxxxoutlook'       as e) UNION
    (SELECT 13 as i, [dbo].SUBSTRING_INDEX(N'wwwxxxoutlookxxxcom', N'xxx', -2) as r, 'outlookxxxcom'       as e) UNION
    (SELECT 14 as i, [dbo].SUBSTRING_INDEX(N'wwwxxxoutlookxxxcom', N'xxx', 5)  as r, 'wwwxxxoutlookxxxcom' as e) UNION
    (SELECT 15 as i, [dbo].SUBSTRING_INDEX(N'wwwxxxoutlookxxxcom', N'xxx', -5) as r, 'wwwxxxoutlookxxxcom' as e)
) as results;

select tr.i,
       tr.answer,
       tr.expected,
       CASE WHEN tr.answer = tr.expected THEN 'Test Succeeded' ELSE 'Test Failed' END testState
  from @TestResults tr
 order by i;

这是一个受Bogdan Sahlean答案启发的版本,使用SQL Server的XML功能进行解析和组合:

CREATE FUNCTION dbo.SUBSTRING_INDEX(@InString  NVARCHAR(Max),
                                    @Delimiter NVARCHAR(Max),
                                    @Count     INT)
RETURNS NVARCHAR(200)
AS
BEGIN
    -- If @Count is zero, we return '' as per spec
    IF @Count = 0
    BEGIN
        RETURN '';
    END;

    -- First we let the XML parser break up the string by @Delimiter.
    -- Each parsed value will be <piece>[text]</piece>.
    DECLARE @XmlSourceString XML = (select N'<piece>' + REPLACE( (SELECT @InString AS '*' FOR XML PATH('')) , @Delimiter, N'</piece><piece>' ) + N'</piece>');

    -- This will contain the final requested string.
    DECLARE @Results nVarChar(MAX);

    ;WITH Pieces(RowNumber, Piece) as
    (
        -- Take each node in @XmlSourceString, and return it with row numbers
        -- which will identify each piece and give us a handle to change the
        -- order, depending on the direction of search.
        SELECT  row_number() over(order by x.XmlCol) as RowNumber,
                @Delimiter + x.XmlCol.value(N'(text())[1]', N'NVARCHAR(MAX)') AS '*'
          FROM  @XmlSourceString.nodes(N'(piece)') x(XmlCol)
    ), orderedPieces(RowNumber, Piece) as
    (
        -- Order the pieces normally or reversed depending on whether they want
        -- the first @Count pieces or the last @Count pieces.
        select TOP (ABS(@Count)) 
               RowNumber, 
               Piece
          from Pieces
         ORDER BY CASE WHEN @Count < 0 THEN RowNumber END DESC ,
                  CASE WHEN @Count > 0 THEN RowNumber END ASC
    ), combinedPieces(result) as
    (
        -- Now combine the pieces back together, ordering them by
        -- the original order.  There will always
        -- be an extra @Delimiter on the front of the string.
        select CAST(Piece AS VARCHAR(100))
          from OrderedPieces
         order by RowNumber
           FOR XML PATH(N'')
    )
    -- Finally, strip off the extra delimiter using STUFF and store the string in @Results.
    select @Results = STUFF(result, 1, LEN(@Delimiter), '') from combinedPieces;

    return @Results;
END;

运行测试会产生以下结果:

i  answer              expected             testState
1  www.somewebsite     www.somewebsite      Test Succeeded
2  www.yahoo           www.yahoo            Test Succeeded
3  www.outlook         www.outlook          Test Succeeded
4  somewebsite.com     somewebsite.com      Test Succeeded
5  yahoo.com           yahoo.com            Test Succeeded
6  outlook.com         outlook.com          Test Succeeded
7  hi.you              hi.you               Test Succeeded
8  com                 com                  Test Succeeded
9  pr                  pr                   Test Succeeded
10 ba                  ba                   Test Succeeded
11                                          Test Succeeded
12 wwwxxxoutlook       wwwxxxoutlook        Test Succeeded
13 outlookxxxcom       outlookxxxcom        Test Succeeded
14 wwwxxxoutlookxxxcom wwwxxxoutlookxxxcom  Test Succeeded
15 wwwxxxoutlookxxxcom wwwxxxoutlookxxxcom  Test Succeeded

答案 2 :(得分:1)

    CREATE FUNCTION FN_SUBSTRING_INDEX
    (
       @TEXTO NVARCHAR(200),
      @SUBSTRING_INDEX NVARCHAR(10),
       @DESPLAZAMIENTO INT
    )
    RETURNS NVARCHAR(200)
    AS

    BEGIN


        DECLARE @indiceSubstring INT
        DECLARE @RESULTADO NVARCHAR(200)
        SELECT @indiceSubstring = CHARINDEX(@SUBSTRING_INDEX,@TEXTO)

        IF @DESPLAZAMIENTO > 0
        BEGIN
            SELECT @RESULTADO=SUBSTRING(@TEXTO,@indiceSubstring+@DESPLAZAMIENTO+1,LEN(@TEXTO))
        END 
        ELSE
        BEGIN
            SELECT @RESULTADO=SUBSTRING(@TEXTO,0,@indiceSubstring-@DESPLAZAMIENTO-1)
        END 

    RETURN @RESULTADO
    END
    GO

到右边 选择dbo.FN_SUBSTRING_INDEX(N'prueba','ue',1);

到左边 选择dbo.FN_SUBSTRING_INDEX(N'prueba','ue', - 1);

答案 3 :(得分:0)

试试这个......

CREATE FUNCTION SubString_Index
(
   @ExistingString NVARCHAR(200),
   @BreakPoint NVARCHAR(10),
   @number INT
)
RETURNS NVARCHAR(200)
AS
BEGIN
DECLARE @Count INT
DECLARE @Substring NVARCHAR(200)
DECLARE @ssubstring NVARCHAR(200)
SET @ssubstring=@ExistingString
DECLARE @scount INT
SET @scount=0
DECLARE @sscount INT
SET @sscount=0
WHILE(@number>@scount)
    BEGIN
            Select @Count=CHARINDEX(@BreakPoint,@ExistingString)
            Select @ExistingString=SUBSTRING(@ExistingString,@Count+1,LEN(@ExistingString))
            Select @scount=@scount+1 
            select @sscount=@sscount+@Count
    END

SELECT @Substring=SUBSTRING(@ssubstring,0,@sscount)

RETURN @Substring
END
GO


select dbo.SubStringIndex('hi.you.com','.',1)

答案 4 :(得分:0)

此Microsoft sql函数的工作原理与mysql中的substring_index函数完全相同

/** This microsoft sql function Works exactly like substring_index function in mysql **/
  
CREATE FUNCTION SubString_Index
(
   @ExistingString NVARCHAR(MAX),
   @BreakPoint NVARCHAR(MAX),
   @number INT
)
RETURNS NVARCHAR(MAX)
AS
BEGIN
DECLARE @Count INT
DECLARE @SubstringLength INT
DECLARE @Substring NVARCHAR(MAX)
DECLARE @ssubstring NVARCHAR(MAX)
SET @ssubstring=@ExistingString
DECLARE @scount INT
SET @scount=0
DECLARE @sscount INT
SET @sscount=0
DECLARE @number2 INT
DECLARE @occurence INT
SET @occurence=LEN(@ExistingString) - LEN(REPLACE(@ExistingString, @BreakPoint, ''))
If @number<0
     BEGIN
        SET @number2= @occurence-(-1*@number)+1
     END
If @number>0
     BEGIN
        SET @number2=@number
     END
WHILE(@number2>@scount)
    BEGIN
            Select @Count=CHARINDEX(@BreakPoint,@ExistingString)
            Select @SubstringLength=@Count+LEN(@BreakPoint) 
            Select @ExistingString=SUBSTRING(@ExistingString,@SubstringLength,LEN(@ExistingString)-@Count)
            Select @scount=@scount+1 
            select @sscount=@sscount+@Count
    END
If @number<0
     BEGIN
        if (@number = -1) and (@sscount+LEN(@BreakPoint)) = (LEN(@ssubstring)+1)
            BEGIN
               SELECT @Substring=''
            END
        else if @occurence = 0
            BEGIN
               SELECT @Substring=''
            END
        else
            BEGIN
               SELECT @Substring=SUBSTRING(@ssubstring, @sscount+LEN(@BreakPoint), LEN(@ssubstring))
            END
     END
If @number>0
    if @occurence = 0
            BEGIN
               SELECT @Substring=''
            END
        else
            BEGIN
               SELECT @Substring=SUBSTRING(@ssubstring,0,@sscount)
            END

RETURN @Substring
END
GO

答案 5 :(得分:0)

使用 Python 获取索引 -2

use my_guitar_shop;
go  


EXEC sp_execute_external_script 
@language =N'Python',
@script=N'

import pandas as pd
import numpy as np
x = np.array(InputDataSet["email_address"], dtype = str)
broken = np.char.split(x, sep = ".")
OutputDataSet = pd.DataFrame([".".join(e[-2:]) if len(e)>2 else ".".join(e) for e in broken])
', 
@input_data_1 = N'SELECT email_address  from customers;'
WITH RESULT SETS(([indexed_email] nvarchar(250)));

答案 6 :(得分:0)

我最好的选择是:

select substring(email,(charindex('@', email,1)+1),100) from yourtable;

假设 TLD.EXT 最多 100 个字符。你可以随意增加。

祝你好运