如何将逗号分隔的值拆分为列
我有这样的桌子
Value String ------------------- 1 Cleo, Smith 我想将逗号分隔的string分成两列
Value Name Surname ------------------- 1 Cleo Smith
我只需要两个固定的额外的列
试试这个（改变''到'，'或你想要使用的任何分隔符的实例）
CREATE FUNCTION dbo.Wordparser ( @multiwordstring VARCHAR(255), @wordnumber NUMERIC ) returns VARCHAR(255) AS BEGIN DECLARE @remainingstring VARCHAR(255) SET @remainingstring=@multiwordstring DECLARE @numberofwords NUMERIC SET @numberofwords=(LEN(@remainingstring) - LEN(REPLACE(@remainingstring, ' ', '')) + 1) DECLARE @word VARCHAR(50) DECLARE @parsedwords TABLE ( line NUMERIC IDENTITY(1, 1), word VARCHAR(255) ) WHILE @numberofwords > 1 BEGIN SET @word=LEFT(@remainingstring, CHARINDEX(' ', @remainingstring) - 1) INSERT INTO @parsedwords(word) SELECT @word SET @remainingstring= REPLACE(@remainingstring, Concat(@word, ' '), '') SET @numberofwords=(LEN(@remainingstring) - LEN(REPLACE(@remainingstring, ' ', '')) + 1) IF @numberofwords = 1 BREAK ELSE CONTINUE END IF @numberofwords = 1 SELECT @word = @remainingstring INSERT INTO @parsedwords(word) SELECT @word RETURN (SELECT word FROM @parsedwords WHERE line = @wordnumber) END
用法示例：
SELECT dbo.Wordparser(COLUMN, 1), dbo.Wordparser(COLUMN, 2), dbo.Wordparser(COLUMN, 3) FROM TABLE
您的目的可以通过以下查询来解决 –
Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name, Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as Surname from Table1
在sql server中没有现成的Splitfunction，所以我们需要创build用户自定义函数。
CREATE FUNCTION Split ( @InputString VARCHAR(8000), @Delimiter VARCHAR(50) ) RETURNS @Items TABLE ( Item VARCHAR(8000) ) AS BEGIN IF @Delimiter = ' ' BEGIN SET @Delimiter = ',' SET @InputString = REPLACE(@InputString, ' ', @Delimiter) END IF (@Delimiter IS NULL OR @Delimiter = '') SET @Delimiter = ',' --INSERT INTO @Items VALUES (@Delimiter) -- Diagnostic --INSERT INTO @Items VALUES (@InputString) -- Diagnostic DECLARE @Item VARCHAR(8000) DECLARE @ItemList VARCHAR(8000) DECLARE @DelimIndex INT SET @ItemList = @InputString SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0) WHILE (@DelimIndex != 0) BEGIN SET @Item = SUBSTRING(@ItemList, 0, @DelimIndex) INSERT INTO @Items VALUES (@Item) -- Set @ItemList = @ItemList minus one less item SET @ItemList = SUBSTRING(@ItemList, @DelimIndex+1, LEN(@ItemList)-@DelimIndex) SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0) END -- End WHILE IF @Item IS NOT NULL -- At least one delimiter was encountered in @InputString BEGIN SET @Item = @ItemList INSERT INTO @Items VALUES (@Item) END -- No delimiters were encountered in @InputString, so just return @InputString ELSE INSERT INTO @Items VALUES (@InputString) RETURN END -- End Function GO ---- Set Permissions --GRANT SELECT ON Split TO UserRole1 --GRANT SELECT ON Split TO UserRole2 --GO
;WITH Split_Names (Value,Name, xmlname) AS ( SELECT Value, Name, CONVERT(XML,'<Names><name>' + REPLACE(Name,',', '</name><name>') + '</name></Names>') AS xmlname FROM tblnames ) SELECT Value, xmlname.value('/Names[1]/name[1]','varchar(100)') AS Name, xmlname.value('/Names[1]/name[2]','varchar(100)') AS Surname FROM Split_Names
并检查以下链接以供参考
http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html
XML基本答案很简单，干净
参考这个
DECLARE @S varchar(max), @Split char(1), @X xml SELECT @S = 'ab,cd,ef,gh,ij', @Split = ',' SELECT @X = CONVERT(xml,' <root> <myvalue>' + REPLACE(@S,@Split,'</myvalue> <myvalue>') + '</myvalue> </root> ') SELECT Tcvalue('.','varchar(20)'),--retrieve all values at once Tcvalue('(/root/myvalue)[1]','VARCHAR(20)') , --retrieve index 1 only, which is the 'ab' Tcvalue('(/root/myvalue)[2]','VARCHAR(20)'), Tcvalue('(/root/myvalue)[3]','VARCHAR(20)') FROM @X.nodes('/root/myvalue') T(c)
与十字架申请
select ParsedData.* from MyTable mt cross apply ( select str = mt.String + ',,' ) f1 cross apply ( select p1 = charindex( ',', str ) ) ap1 cross apply ( select p2 = charindex( ',', str, p1 + 1 ) ) ap2 cross apply ( select Nmame = substring( str, 1, p1-1 ) , Surname = substring( str, p1+1, p2-p1-1 ) ) ParsedData
我觉得这很酷
SELECT value, PARSENAME(REPLACE(String,',','.'),2) 'Name' , PARSENAME(REPLACE(String,',','.'),1) 'Sur Name' FROM table WITH (NOLOCK)
有多种方法可以解决这个问题，已经提出了很多不同的方法。 最简单的方法是使用LEFT / SUBSTRING和其他string函数来获得所需的结果。
示例数据
DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX)) INSERT INTO @tbl1 VALUES(1,'Cleo, Smith'); INSERT INTO @tbl1 VALUES(2,'John, Mathew');
使用像LEFT这样的string函数
SELECT Value, LEFT(String,CHARINDEX(',',String)-1) as Fname, LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname FROM @tbl1
如果string中有更多2个项目，则此方法失败。 在这种情况下，我们可以使用分离器，然后使用PIVOT或将string转换为XML并使用.nodes获取string项目。 aad和bvr在他们的解决scheme中详细阐述了基于XML的解决scheme。
这个使用splitter的问题的答案，全部使用WHILE ，这是无效的分裂。 检查这个性能比较 。 其中最好的分离器是由Jeff Moden创build的DelimitedSplit8K 。 你可以在这里阅读更多
与PIVOT
DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX)) INSERT INTO @tbl1 VALUES(1,'Cleo, Smith'); INSERT INTO @tbl1 VALUES(2,'John, Mathew'); SELECT t3.Value,[1] as Fname,[2] as Lname FROM @tbl1 as t1 CROSS APPLY [dbo].[DelimitedSplit8K](String,',') as t2 PIVOT(MAX(Item) FOR ItemNumber IN ([1],[2])) as t3
产量
Value Fname Lname 1 Cleo Smith 2 John Mathew
DelimitedSplit8K通过Jeff Moden
CREATE FUNCTION [dbo].[DelimitedSplit8K] /********************************************************************************************************************** Purpose: Split a given string at a given delimiter and return a list of the split elements (items). Notes: 1. Leading a trailing delimiters are treated as if an empty string element were present. 2. Consecutive delimiters are treated as if an empty string element were present between them. 3. Except when spaces are used as a delimiter, all spaces present in each element are preserved. Returns: iTVF containing the following: ItemNumber = Element position of Item as a BIGINT (not converted to INT to eliminate a CAST) Item = Element value as a VARCHAR(8000) Statistics on this function may be found at the following URL: http://www.sqlservercentral.com/Forums/Topic1101315-203-4.aspx CROSS APPLY Usage Examples and Tests: --===================================================================================================================== -- TEST 1: -- This tests for various possible conditions in a string using a comma as the delimiter. The expected results are -- laid out in the comments --===================================================================================================================== --===== Conditionally drop the test tables to make reruns easier for testing. -- (this is NOT a part of the solution) IF OBJECT_ID('tempdb..#JBMTest') IS NOT NULL DROP TABLE #JBMTest ; --===== Create and populate a test table on the fly (this is NOT a part of the solution). -- In the following comments, "b" is a blank and "E" is an element in the left to right order. -- Double Quotes are used to encapsulate the output of "Item" so that you can see that all blanks -- are preserved no matter where they may appear. SELECT * INTO #JBMTest FROM ( --# & type of Return Row(s) SELECT 0, NULL UNION ALL --1 NULL SELECT 1, SPACE(0) UNION ALL --1 b (Empty String) SELECT 2, SPACE(1) UNION ALL --1 b (1 space) SELECT 3, SPACE(5) UNION ALL --1 b (5 spaces) SELECT 4, ',' UNION ALL --2 bb (both are empty strings) SELECT 5, '55555' UNION ALL --1 E SELECT 6, ',55555' UNION ALL --2 b E SELECT 7, ',55555,' UNION ALL --3 b E b SELECT 8, '55555,' UNION ALL --2 b B SELECT 9, '55555,1' UNION ALL --2 EE SELECT 10, '1,55555' UNION ALL --2 EE SELECT 11, '55555,4444,333,22,1' UNION ALL --5 EEEEE SELECT 12, '55555,4444,,333,22,1' UNION ALL --6 EE b EEE SELECT 13, ',55555,4444,,333,22,1,' UNION ALL --8 b EE b EEE b SELECT 14, ',55555,4444,,,333,22,1,' UNION ALL --9 b EE bb EEE b SELECT 15, ' 4444,55555 ' UNION ALL --2 E (w/Leading Space) E (w/Trailing Space) SELECT 16, 'This,is,a,test.' --EEEE ) d (SomeID, SomeValue) ; --===== Split the CSV column for the whole table using CROSS APPLY (this is the solution) SELECT test.SomeID, test.SomeValue, split.ItemNumber, Item = QUOTENAME(split.Item,'"') FROM #JBMTest test CROSS APPLY dbo.DelimitedSplit8K(test.SomeValue,',') split ; --===================================================================================================================== -- TEST 2: -- This tests for various "alpha" splits and COLLATION using all ASCII characters from 0 to 255 as a delimiter against -- a given string. Note that not all of the delimiters will be visible and some will show up as tiny squares because -- they are "control" characters. More specifically, this test will show you what happens to various non-accented -- letters for your given collation depending on the delimiter you chose. --===================================================================================================================== WITH cteBuildAllCharacters (String,Delimiter) AS ( SELECT TOP 256 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789', CHAR(ROW_NUMBER() OVER (ORDER BY (SELECT NULL))-1) FROM master.sys.all_columns ) SELECT ASCII_Value = ASCII(c.Delimiter), c.Delimiter, split.ItemNumber, Item = QUOTENAME(split.Item,'"') FROM cteBuildAllCharacters c CROSS APPLY dbo.DelimitedSplit8K(c.String,c.Delimiter) split ORDER BY ASCII_Value, split.ItemNumber ; ----------------------------------------------------------------------------------------------------------------------- Other Notes: 1. Optimized for VARCHAR(8000) or less. No testing or error reporting for truncation at 8000 characters is done. 2. Optimized for single character delimiter. Multi-character delimiters should be resolvedexternally from this function. 3. Optimized for use with CROSS APPLY. 4. Does not "trim" elements just in case leading or trailing blanks are intended. 5. If you don't know how a Tally table can be used to replace loops, please see the following... http://www.sqlservercentral.com/articles/T-SQL/62867/ 6. Changing this function to use NVARCHAR(MAX) will cause it to run twice as slow. It's just the nature of VARCHAR(MAX) whether it fits in-row or not. 7. Multi-machine testing for the method of using UNPIVOT instead of 10 SELECT/UNION ALLs shows that the UNPIVOT method is quite machine dependent and can slow things down quite a bit. ----------------------------------------------------------------------------------------------------------------------- Credits: This code is the product of many people's efforts including but not limited to the following: cteTally concept originally by Iztek Ben Gan and "decimalized" by Lynn Pettis (and others) for a bit of extra speed and finally redacted by Jeff Moden for a different slant on readability and compactness. Hat's off to Paul White for his simple explanations of CROSS APPLY and for his detailed testing efforts. Last but not least, thanks to Ron "BitBucket" McCullough and Wayne Sheffield for their extreme performance testing across multiple machines and versions of SQL Server. The latest improvement brought an additional 15-20% improvement over Rev 05. Special thanks to "Nadrek" and "peter-757102" (aka Peter de Heer) for bringing such improvements to light. Nadrek's original improvement brought about a 10% performance gain and Peter followed that up with the content of Rev 07. I also thank whoever wrote the first article I ever saw on "numbers tables" which is located at the following URL and to Adam Machanic for leading me to it many years ago. http://sqlserver2000.databases.aspfaq.com/why-should-i-consider-using-an-auxiliary-numbers-table.html ----------------------------------------------------------------------------------------------------------------------- Revision History: Rev 00 - 20 Jan 2010 - Concept for inline cteTally: Lynn Pettis and others. Redaction/Implementation: Jeff Moden - Base 10 redaction and reduction for CTE. (Total rewrite) Rev 01 - 13 Mar 2010 - Jeff Moden - Removed one additional concatenation and one subtraction from the SUBSTRING in the SELECT List for that tiny bit of extra speed. Rev 02 - 14 Apr 2010 - Jeff Moden - No code changes. Added CROSS APPLY usage example to the header, some additional credits, and extra documentation. Rev 03 - 18 Apr 2010 - Jeff Moden - No code changes. Added notes 7, 8, and 9 about certain "optimizations" that don't actually work for this type of function. Rev 04 - 29 Jun 2010 - Jeff Moden - Added WITH SCHEMABINDING thanks to a note by Paul White. This prevents an unnecessary "Table Spool" when the function is used in an UPDATE statement even though the function makes no external references. Rev 05 - 02 Apr 2011 - Jeff Moden - Rewritten for extreme performance improvement especially for larger strings approaching the 8K boundary and for strings that have wider elements. The redaction of this code involved removing ALL concatenation of delimiters, optimization of the maximum "N" value by using TOP instead of including it in the WHERE clause, and the reduction of all previous calculations (thanks to the switch to a "zero based" cteTally) to just one instance of one add and one instance of a subtract. The length calculation for the final element (not followed by a delimiter) in the string to be split has been greatly simplified by using the ISNULL/NULLIF combination to determine when the CHARINDEX returned a 0 which indicates there are no more delimiters to be had or to start with. Depending on the width of the elements, this code is between 4 and 8 times faster on a single CPU box than the original code especially near the 8K boundary. - Modified comments to include more sanity checks on the usage example, etc. - Removed "other" notes 8 and 9 as they were no longer applicable. Rev 06 - 12 Apr 2011 - Jeff Moden - Based on a suggestion by Ron "Bitbucket" McCullough, additional test rows were added to the sample code and the code was changed to encapsulate the output in pipes so that spaces and empty strings could be perceived in the output. The first "Notes" section was added. Finally, an extra test was added to the comments above. Rev 07 - 06 May 2011 - Peter de Heer, a further 15-20% performance enhancement has been discovered and incorporated into this code which also eliminated the need for a "zero" position in the cteTally table. **********************************************************************************************************************/ --===== Define I/O parameters (@pString VARCHAR(8000), @pDelimiter CHAR(1)) RETURNS TABLE WITH SCHEMABINDING AS RETURN --===== "Inline" CTE Driven "Tally Table" produces values from 0 up to 10,000... -- enough to cover NVARCHAR(4000) WITH E1(N) AS ( SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 ), --10E+1 or 10 rows E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max cteTally(N) AS (--==== This provides the "base" CTE and limits the number of rows right up front -- for both a performance gain and prevention of accidental "overruns" SELECT TOP (ISNULL(DATALENGTH(@pString),0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4 ), cteStart(N1) AS (--==== This returns N+1 (starting position of each "element" just once for each delimiter) SELECT 1 UNION ALL SELECT t.N+1 FROM cteTally t WHERE SUBSTRING(@pString,tN,1) = @pDelimiter ), cteLen(N1,L1) AS(--==== Return start and length (for use in substring) SELECT s.N1, ISNULL(NULLIF(CHARINDEX(@pDelimiter,@pString,s.N1),0)-s.N1,8000) FROM cteStart s ) --===== Do the actual split. The ISNULL/NULLIF combo handles the length for the final element when no delimiter is found. SELECT ItemNumber = ROW_NUMBER() OVER(ORDER BY l.N1), Item = SUBSTRING(@pString, l.N1, l.L1) FROM cteLen l ; GO
我认为PARSENAME是用于这个例子的整洁的函数，正如这篇文章所描述的： http : //www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/
PARSENAME函数在逻辑上devise为parsing四部分对象名称。 关于PARSENAME的好处在于，它不仅限于parsingSQL Server四部分对象名称 – 它将parsing由点分隔的任何函数或string数据。
第一个参数是要parsing的对象，第二个参数是要返回的对象片段的整数值。 本文正在讨论parsing和旋转分隔数据 – 公司电话号码，但它也可以用来parsing名称/姓氏数据。
例：
USE COMPANY; SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue';
本文还介绍了使用名为“replaceChars”的公用expression式（CTE），对分隔符replace的值运行PARSENAME。 CTE对于返回临时视图或结果集非常有用。
之后，UNPIVOT函数已经被用来将一些列转换成行; SUBSTRING和CHARINDEX函数已被用于清除数据中的不一致性，最后使用了LAG函数（SQL Server 2012的新function），因为它允许引用以前的logging。
在SQL Server 2016中，我们可以使用string_split来实现这个function：
create table commasep ( id int identity(1,1) ,string nvarchar(100) ) insert into commasep (string) values ('John, Adam'), ('test1,test2,test3') select id, [value] as String from commasep cross apply string_split(string,',')
我们可以创build一个这样的function
CREATE Function [dbo].[fn_CSVToTable] ( @CSVList Varchar(max) ) RETURNS @Table TABLE (ColumnData VARCHAR(100)) AS BEGIN IF RIGHT(@CSVList, 1) <> ',' SELECT @CSVList = @CSVList + ',' DECLARE @Pos BIGINT, @OldPos BIGINT SELECT @Pos = 1, @OldPos = 1 WHILE @Pos < LEN(@CSVList) BEGIN SELECT @Pos = CHARINDEX(',', @CSVList, @OldPos) INSERT INTO @Table SELECT LTRIM(RTRIM(SUBSTRING(@CSVList, @OldPos, @Pos - @OldPos))) Col001 SELECT @OldPos = @Pos + 1 END RETURN END
然后我们可以使用SELECT语句将CSV值分离到我们各自的列中
select id,SUBSTRING(name,0,charindex(',',name))as firstname ,SUBSTRING(name,charindex(',',name),len(name)+1)as lastname from spilt
使用instring函数:)
select Value, substring(String,1,instr(String," ") -1) Fname, substring(String,instr(String,",") +1) Sname from tablename;
使用了两个函数，
1. substring(string, position, length) ==>返回string从positon到长度
2. instr(string,pattern) ==>返回pattern的位置。
如果我们不在substring中提供长度参数，它将返回到string结尾
使用Parsename（）函数
with cte as( select 'Aria,Karimi' as FullName Union select 'Joe,Karimi' as FullName Union select 'Bab,Karimi' as FullName ) SELECT PARSENAME(REPLACE(FullName,',','.'),2) as Name, PARSENAME(REPLACE(FullName,',','.'),1) as Family FROM cte
结果
Name Family ----- ------ Aria Karimi Bab Karimi Joe Karimi
我遇到了类似的问题，但复杂的，因为这是我发现的第一个线程，我决定发布我的发现。 我知道这是一个简单的问题的复杂的解决scheme，但我希望我能帮助其他人谁去这个线程寻找一个更复杂的解决scheme。 我不得不拆分一个包含5个数字的string（列名称：levelsFeed），并在一个单独的列中显示每个数字。 例如：8,1,2,2,2应显示为：
1 2 3 4 5 ------------- 8 1 2 2 2
解决scheme1：使用XMLfunction：这个解决scheme是迄今为止最慢的解决scheme
SELECT Distinct FeedbackID, , Savalue('(/H/r)[1]', 'INT') AS level1 , Savalue('(/H/r)[2]', 'INT') AS level2 , Savalue('(/H/r)[3]', 'INT') AS level3 , Savalue('(/H/r)[4]', 'INT') AS level4 , Savalue('(/H/r)[5]', 'INT') AS level5 FROM ( SELECT *,CAST (N'<H><r>' + REPLACE(levelsFeed, ',', '</r><r>') + '</r> </H>' AS XML) AS [vals] FROM Feedbacks ) as d CROSS APPLY d.[vals].nodes('/H/r') S(a)
解决scheme2：使用分割function和数据透视。 （分割函数将string拆分为列名为Data的行）
SELECT FeedbackID, [1],[2],[3],[4],[5] FROM ( SELECT *, ROW_NUMBER() OVER (PARTITION BY feedbackID ORDER BY (SELECT null)) as rn FROM ( SELECT FeedbackID, levelsFeed FROM Feedbacks ) as a CROSS APPLY dbo.Split(levelsFeed, ',') ) as SourceTable PIVOT ( MAX(data) FOR rn IN ([1],[2],[3],[4],[5]) )as pivotTable
解决scheme3：使用string操作函数 – 在解决scheme2上以最快的速度运行
SELECT FeedbackID, SUBSTRING(levelsFeed,0,CHARINDEX(',',levelsFeed)) AS level1, PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),4) AS level2, PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),3) AS level3, PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),2) AS level4, PARSENAME(REPLACE(SUBSTRING(levelsFeed,CHARINDEX(',',levelsFeed)+1,LEN(levelsFeed)),',','.'),1) AS level5 FROM Feedbacks
因为levelsFeed包含5个string值，我需要使用第一个string的子string函数。
我希望我的解决scheme将帮助其他人得到这个线程寻找一个更复杂的拆分列方法
尝试这个：
declare @csv varchar(100) ='aaa,bb,csda,daass'; set @csv = @csv+','; with cte as ( select SUBSTRING(@csv,1,charindex(',',@csv,1)-1) as val, SUBSTRING(@csv,charindex(',',@csv,1)+1,len(@csv)) as rem UNION ALL select SUBSTRING(a.rem,1,charindex(',',a.rem,1)-1)as val, SUBSTRING(a.rem,charindex(',',a.rem,1)+1,len(A.rem)) from cte a where LEN(a.rem)>=1 ) select val from cte
DECLARE @INPUT VARCHAR (MAX)='N,A,R,E,N,D,R,A' DECLARE @ELIMINATE_CHAR CHAR (1)=',' DECLARE @L_START INT=1 DECLARE @L_END INT=(SELECT LEN (@INPUT)) DECLARE @OUTPUT CHAR (1) WHILE @L_START <=@L_END BEGIN SET @OUTPUT=(SUBSTRING (@INPUT,@L_START,1)) IF @OUTPUT!=@ELIMINATE_CHAR BEGIN PRINT @OUTPUT END SET @L_START=@L_START+1 END
mytable中：
Value ColOne -------------------- 1 Cleo, Smith
如果没有太多的列，以下应该工作
ALTER TABLE mytable ADD ColTwo nvarchar(256); UPDATE mytable SET ColTwo = LEFT(ColOne, Charindex(',', ColOne) - 1); --'Cleo' = LEFT('Cleo, Smith', Charindex(',', 'Cleo, Smith') - 1) UPDATE mytable SET ColTwo = REPLACE(ColOne, ColTwo + ',', ''); --' Smith' = REPLACE('Cleo, Smith', 'Cleo' + ',') UPDATE mytable SET ColOne = REPLACE(ColOne, ',' + ColTwo, ''), ColTwo = LTRIM(ColTwo); --'Cleo' = REPLACE('Cleo, Smith', ',' + ' Smith', '')
结果：
Value ColOne ColTwo -------------------- 1 Cleo Smith
这很容易，你可以通过下面的查询：
DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3' SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))
这对我有效
CREATE FUNCTION [dbo].[SplitString]( @delimited NVARCHAR(MAX), @delimiter NVARCHAR(100) ) RETURNS @t TABLE ( val NVARCHAR(MAX)) AS BEGIN DECLARE @xml XML SET @xml = N'<t>' + REPLACE(@delimited,@delimiter,'</t><t>') + '</t>' INSERT INTO @t(val) SELECT r.value('.','varchar(MAX)') as item FROM @xml.nodes('/t') as records(r) RETURN END
我认为以下function将为您工作：
你必须先在SQL中创build一个函数。 喜欢这个
CREATE FUNCTION [dbo].[fn_split]( @str VARCHAR(MAX), @delimiter CHAR(1) ) RETURNS @returnTable TABLE (idx INT PRIMARY KEY IDENTITY, item VARCHAR(8000)) AS BEGIN DECLARE @pos INT SELECT @str = @str + @delimiter WHILE LEN(@str) > 0 BEGIN SELECT @pos = CHARINDEX(@delimiter,@str) IF @pos = 1 INSERT @returnTable (item) VALUES (NULL) ELSE INSERT @returnTable (item) VALUES (SUBSTRING(@str, 1, @pos-1)) SELECT @str = SUBSTRING(@str, @pos+1, LEN(@str)-@pos) END RETURN END
你可以这样调用这个函数：
select * from fn_split('1,24,5',',')
执行：
Declare @test TABLE ( ID VARCHAR(200), Data VARCHAR(200) ) insert into @test (ID, Data) Values ('1','Cleo,Smith') insert into @test (ID, Data) Values ('2','Paul,Grim') select ID, (select item from fn_split(Data,',') where idx in (1)) as Name , (select item from fn_split(Data,',') where idx in (2)) as Surname from @test
结果会是这样的：
您可能会发现解决scheme在SQL用户定义函数parsing分隔符string有帮助（从代码项目 ）。
这是此页面的代码部分：
CREATE FUNCTION [fn_ParseText2Table] (@p_SourceText VARCHAR(MAX) ,@p_Delimeter VARCHAR(100)=',' --default to comma delimited. ) RETURNS @retTable TABLE([Position] INT IDENTITY(1,1) ,[Int_Value] INT ,[Num_Value] NUMERIC(18,3) ,[Txt_Value] VARCHAR(MAX) ,[Date_value] DATETIME ) AS /* ******************************************************************************** Purpose: Parse values from a delimited string & return the result as an indexed table Copyright 1996, 1997, 2000, 2003 Clayton Groom (<A href="mailto:Clayton_Groom@hotmail.com">Clayton_Groom@hotmail.com</A>) Posted to the public domain Aug, 2004 2003-06-17 Rewritten as SQL 2000 function. Reworked to allow for delimiters > 1 character in length and to convert Text values to numbers 2016-04-05 Added logic for date values based on "new" ISDATE() function, Updated to use XML approach, which is more efficient. ******************************************************************************** */ BEGIN DECLARE @w_xml xml; SET @w_xml = N'<root><i>' + replace(@p_SourceText, @p_Delimeter,'</i><i>') + '</i></root>'; INSERT INTO @retTable ([Int_Value] , [Num_Value] , [Txt_Value] , [Date_value] ) SELECT CASE WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1 THEN CAST(CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC) AS INT) END AS [Int_Value] , CASE WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1 THEN CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC(18, 3)) END AS [Num_Value] , [i].value('.', 'VARCHAR(MAX)') AS [txt_Value] , CASE WHEN ISDATE([i].value('.', 'VARCHAR(MAX)')) = 1 THEN CAST([i].value('.', 'VARCHAR(MAX)') AS DATETIME) END AS [Num_Value] FROM @w_xml.nodes('//root/i') AS [Items]([i]); RETURN; END; GO
我发现，如上所述使用PARSENAME导致任何名称的周期为零。
所以如果名称后面有一个首字母或标题，则它们返回NULL。
我发现这对我有用：
SELECT REPLACE(SUBSTRING(FullName, 1,CHARINDEX(',', FullName)), ',','') as Name, REPLACE(SUBSTRING(FullName, CHARINDEX(',', FullName), LEN(FullName)), ',', '') as Surname FROM Table1
CREATE FUNCTION [dbo].[fnSplit](@sInputList VARCHAR(8000), @sDelimiter VARCHAR(8000) = ',') RETURNS @List TABLE (item VARCHAR(8000)) BEGIN DECLARE @sItem VARCHAR(8000) WHILE CHARINDEX(@sDelimiter, @sInputList, 0) <> 0 BEGIN SELECT @sItem = RTRIM(LTRIM(SUBSTRING(@sInputList, 1, CHARINDEX(@sDelimiter, @sInputList,0) - 1))), @sInputList = RTRIM(LTRIM(SUBSTRING(@sInputList, CHARINDEX(@sDelimiter, @sInputList, 0) + LEN(@sDelimiter),LEN(@sInputList)))) -- Indexes to keep the position of searching IF LEN(@sItem) > 0 INSERT INTO @List SELECT @sItem END IF LEN(@sInputList) > 0 BEGIN INSERT INTO @List SELECT @sInputList -- Put the last item in END RETURN END
你可以试试这个查询：
select substr('abc lmn pqr',1,instr('abc lmn pqr',' ',1)) as first, substr('abc lmn pqr',instr('abc lmn pqr',' ',1,1),instr('abc lmn pqr',' ',-1,2)) as middle, substr('abc lmn pqr',instr('abc lmn pqr',' ',1,2)) as last from dual;
select distinct modelFileId,F4.* from contract cross apply (select XmlList=convert(xml, '<x>'+replace(modelFileId,';','</x><x>')+'</x>').query('.')) F2 cross apply (select mfid1=XmlNode.value('/x[1]','varchar(512)') ,mfid2=XmlNode.value('/x[2]','varchar(512)') ,mfid3=XmlNode.value('/x[3]','varchar(512)') ,mfid4=XmlNode.value('/x[4]','varchar(512)') from XmlList.nodes('x') F3(XmlNode)) F4 where modelFileId like '%;%' order by modelFileId
您可以使用仅在兼容级别130下可用的内置STRING_SPLIT函数。如果您的数据库兼容级别低于130，则SQL Server将无法find并执行STRING_SPLIT函数。 您可以使用以下命令更改数据库的兼容级别：
ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130
句法
STRING_SPLIT ( string , separator )
见这里的文档
ALTER FUNCTION [dbo].[StringListTo] (@StringList Nvarchar(max),@Separators char(1),@start int, @index int ) RETURNS nvarchar(max) AS BEGIN declare @out Nvarchar(max) declare @i int declare @start_old int set @start=@start+1 set @i=1 while(@i<=@index) begin set @start_old=@start set @start=CHARINDEX('.',@StringList,@start+1) if (@start>0) begin set @out=Substring(@StringList,@start_old+1,@start-@start_old-1) end else begin set @out=Substring(@StringList,@start_old+1,len(@StringList)-1) end set @i=@i+1 end RETURN @out END;
You can use split function.
SELECT (select top 1 item from dbo.Split(FullName,',') where id=1 ) as Name, (select top 1 item from dbo.Split(FullName,',') where id=2 ) as Surname, FROM MyTbl
Select distinct PROJ_UID,PROJ_NAME,RES_UID from E2E_ProjectWiseTimesheetActuals where CHARINDEX(','+cast(PROJ_UID as varchar(8000))+',', @params) > 0 and CHARINDEX(','+cast(RES_UID as varchar(8000))+',', @res) > 0
