Cómo dividir un valor separado por comas en columnas

Tengo una mesa como esta

Value String ------------------- 1 Cleo, Smith 

Quiero separar la cadena delimitada por comas en dos columnas

 Value Name Surname ------------------- 1 Cleo Smith 

Necesito solo dos columnas extra fijas

Su propósito se puede resolver usando la siguiente consulta:

 Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name, Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as Surname from Table1 

No hay una función de división lista en el servidor sql, por lo que necesitamos crear una función definida por el usuario.

 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,'' + REPLACE(Name,',', '') + '') 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 

y también verifique el siguiente enlace para referencia

http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html

La respuesta de la base xml es simple y limpia

referir esto

 DECLARE @S varchar(max), @Split char(1), @X xml SELECT @S = 'ab,cd,ef,gh,ij', @Split = ',' SELECT @X = CONVERT(xml,'  ' + REPLACE(@S,@Split,' ') + '  ') 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) 

Con CROSS se aplica

 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 

Creo que esto es genial

 SELECT value, PARSENAME(REPLACE(String,',','.'),2) 'Name' , PARSENAME(REPLACE(String,',','.'),1) 'Sur Name' FROM table WITH (NOLOCK) 

Pruebe esto (cambie las instancias de ” a ‘,’ o cualquier delimitador que quiera usar)

 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 

Ejemplo de uso:

 SELECT dbo.Wordparser(COLUMN, 1), dbo.Wordparser(COLUMN, 2), dbo.Wordparser(COLUMN, 3) FROM TABLE 

Hay varias formas de resolver esto y ya se han propuesto muchas formas diferentes. Lo más simple sería usar LEFT / SUBSTRING y otras funciones de cuerda para lograr el resultado deseado.

Data de muestra

 DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX)) INSERT INTO @tbl1 VALUES(1,'Cleo, Smith'); INSERT INTO @tbl1 VALUES(2,'John, Mathew'); 

Usar funciones de cadena como LEFT

 SELECT Value, LEFT(String,CHARINDEX(',',String)-1) as Fname, LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname FROM @tbl1 

Este enfoque falla si hay más 2 elementos en una Cadena. En tal escenario, podemos usar un divisor y luego usar PIVOT o convertir la cadena en un XML y usar .nodes para obtener elementos de cadena. XML solución basada en XML ha sido detallada por aads y bvr en su solución.

Las respuestas para esta pregunta que usan divisor, todas usan WHILE que es ineficiente para dividir. Verifique esta comparación de rendimiento . Uno de los mejores divisores es DelimitedSplit8K , creado por Jeff Moden. Puedes leer más sobre esto aquí

Splitter con 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 

Salida

 Value Fname Lname 1 Cleo Smith 2 John Mathew 

DelimitedSplit8K por 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 

Creo que PARSENAME es la función clara para usar en este ejemplo, como se describe en este artículo: http://www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/

La función PARSENAME está diseñada lógicamente para analizar los nombres de objetos de cuatro partes. Lo bueno de PARSENAME es que no se limita a analizar solo los nombres de objetos de cuatro partes de SQL Server: analizará cualquier función o cadena de datos delimitada por puntos.

El primer parámetro es el objeto a analizar, y el segundo es el valor entero de la pieza del objeto a devolver. El artículo analiza el análisis y la rotativa de los datos delimitados: los números de teléfono de la empresa, pero también se puede utilizar para analizar los datos de nombre / apellido.

Ejemplo:

 USE COMPANY; SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue'; 

El artículo también describe el uso de una Expresión común de tabla (CTE) llamada ‘replaceChars’, para ejecutar PARSENAME contra los valores reemplazados por delimitador. Un CTE es útil para devolver una vista temporal o conjunto de resultados.

Después de eso, la función UNPIVOT se ha utilizado para convertir algunas columnas en filas; Las funciones SUBSTRING y CHARINDEX se han usado para limpiar las inconsistencias en los datos, y la función LAG (nueva para SQL Server 2012) se ha utilizado al final, ya que permite la referencia de registros anteriores.

Con SQL Server 2016 podemos usar string_split para lograr esto:

 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,',') 

Podemos crear una función como esta

 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 

A continuación, podemos separar los valores de CSV en nuestras respectivas columnas utilizando una instrucción SELECT

 select id,SUBSTRING(name,0,charindex(',',name))as firstname ,SUBSTRING(name,charindex(',',name),len(name)+1)as lastname from spilt 

Uso de la función de inserción 🙂

 select Value, substring(String,1,instr(String," ") -1) Fname, substring(String,instr(String,",") +1) Sname from tablename; 

Usó dos funciones,
1. substring(string, position, length) ==> devuelve la cadena de la posición a la longitud
2. instr(string,pattern) ==> devuelve la posición del patrón.

Si no proporcionamos el argumento de longitud en la subcadena, regresa hasta el final de la cadena

Use la función Parsename ()

 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 

Resultado

 Name Family ----- ------ Aria Karimi Bab Karimi Joe Karimi 
 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 

Encontré un problema similar pero complejo y dado que este es el primer hilo que encontré sobre ese tema, decidí publicar mi hallazgo. Sé que es una solución compleja a un problema simple, pero espero poder ayudar a otras personas que van a este tema en busca de una solución más compleja. Tuve que dividir una cadena que contenía 5 números (nombre de columna: levelsFeed) y mostrar cada número en una columna separada. por ejemplo: 8,1,2,2,2 se debe mostrar como:

 1 2 3 4 5 ------------- 8 1 2 2 2 

Solución 1: uso de funciones XML: esta solución para la solución más lenta de lejos

 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'' + REPLACE(levelsFeed, ',', '') + ' ' AS XML) AS [vals] FROM Feedbacks ) as d CROSS APPLY d.[vals].nodes('/H/r') S(a) 

Solución 2: utilizando la función de división y el pivote. (la función dividida divide una cadena en filas con el nombre de columna Datos)

 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 

Solución 3: uso de funciones de manipulación de cadenas – más rápido por un pequeño margen sobre la solución 2

 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 

dado que los levelsFeed contienen 5 valores de cadena, necesitaba usar la función de subcadena para la primera cadena.

Espero que mi solución ayude a otros que llegaron a este hilo en busca de métodos más complejos para dividir a columnas

Prueba esto:

 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 

Creo que la siguiente función funcionará para ti:

Primero debe crear una función en SQL. Me gusta esto

 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 

Puede llamar a esta función, así:

 select * from fn_split('1,24,5',',') 

Implementación:

 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 

El resultado le va a gustar esto:

enter image description here

mi mesa:

 Value ColOne -------------------- 1 Cleo, Smith 

Lo siguiente debería funcionar si no hay demasiadas columnas

 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', '') 

Resultado:

 Value ColOne ColTwo -------------------- 1 Cleo Smith 

es tan fácil, puedes tomarlo por debajo de la consulta:

 DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3' SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str))) 

Esto funcionó para mí

 CREATE FUNCTION [dbo].[SplitString]( @delimited NVARCHAR(MAX), @delimiter NVARCHAR(100) ) RETURNS @t TABLE ( val NVARCHAR(MAX)) AS BEGIN DECLARE @xml XML SET @xml = N'' + REPLACE(@delimited,@delimiter,'') + '' INSERT INTO @t(val) SELECT r.value('.','varchar(MAX)') as item FROM @xml.nodes('/t') as records(r) RETURN END 

Puede utilizar una función STRING_SPLIT incorporada que está disponible solo bajo el nivel de compatibilidad 130. Si el nivel de compatibilidad de su base de datos es inferior a 130, SQL Server no podrá encontrar y ejecutar la función STRING_SPLIT . Puede cambiar un nivel de compatibilidad de la base de datos con el siguiente comando:

 ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130 

Sintaxis

 STRING_SPLIT ( string , separator ) 

ver documentación aquí

Puede encontrar la solución en Función definida por el usuario de SQL para analizar una cadena delimitada útil (de The Code Project ).

Esta es la parte del código de esta página:

 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 (Clayton_Groom@hotmail.com) 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'' + replace(@p_SourceText, @p_Delimeter,'') + ''; 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 

Esta función es más rápida:

 CREATE FUNCTION dbo.F_ExtractSubString ( @String VARCHAR(MAX), @NroSubString INT, @Separator VARCHAR(5) ) RETURNS VARCHAR(MAX) AS BEGIN DECLARE @St INT = 0, @End INT = 0, @Ret VARCHAR(MAX) SET @String = @String + @Separator WHILE CHARINDEX(@Separator, @String, @End + 1) > 0 AND @NroSubString > 0 BEGIN SET @St = @End + 1 SET @End = CHARINDEX(@Separator, @String, @End + 1) SET @NroSubString = @NroSubString - 1 END IF @NroSubString > 0 SET @Ret = '' ELSE SET @Ret = SUBSTRING(@String, @St, @End - @St) RETURN @Ret END GO 

Ejemplo de uso:

 SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '), dbo.F_ExtractSubString(COLUMN, 2, ', '), dbo.F_ExtractSubString(COLUMN, 3, ', ') FROM TABLE 

Descubrí que el uso de PARSENAME como el anterior causaba cualquier nombre con un período para anularlo.

Entonces, si hubo una inicial o un título en el nombre seguido de un punto, devuelven NULL.

Encontré que esto funcionó para mí:

 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 
 select distinct modelFileId,F4.* from contract cross apply (select XmlList=convert(xml, ''+replace(modelFileId,';','')+'').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 
 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 
 ALTER function get_occurance_index(@delimiter varchar(1),@occurence int,@String varchar(100)) returns int AS Begin --Declare @delimiter varchar(1)=',',@occurence int=2,@String varchar(100)='a,b,c' Declare @result int ;with T as ( select 1 Rno,0 as row, charindex(@delimiter, @String) pos,@String st union all select Rno+1,pos + 1, charindex(@delimiter, @String, pos + 1), @String from T where pos > 0 ) select @result=pos from T where pos > 0 and rno = @occurence return isnull(@result,0) ENd declare @data as table (data varchar(100)) insert into @data values('1,2,3') insert into @data values('aaa,bbbbb,cccc') select top 3 Substring (data,0,dbo.get_occurance_index( ',',1,data)) ,--First Record always starts with 0 Substring (data,dbo.get_occurance_index( ',',1,data)+1,dbo.get_occurance_index( ',',2,data)-dbo.get_occurance_index( ',',1,data)-1) , Substring (data,dbo.get_occurance_index( ',',2,data)+1,len(data)) , -- Last record cant be more than len of actual data data From @data 

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 
 enter code here USE TRIAL GO CREATE TABLE DETAILS ( ID INT, NAME VARCHAR(50), ADDRESS VARCHAR(50) ) INSERT INTO DETAILS VALUES (100, 'POPE-JOHN-PAUL','VATICAN CIT|ROME|ITALY') ,(240, 'SIR-PAUL-McARTNEY','NEWYORK CITY|NEWYORK|USA') ,(460,'BARRACK-HUSSEIN-OBAMA','WHITE HOUSE|WASHINGTON|USA') ,(700, 'PRESIDENT-VLADAMIR-PUTIN','RED SQUARE|MOSCOW|RUSSIA') ,(950, 'NARENDRA-DAMODARDAS-MODI','10 JANPATH|NEW DELHI|INDIA') select [ID] ,[NAME] ,[ADDRESS] ,REPLACE(LEFT(NAME, CHARINDEX('-', NAME)),'-',' ') as First_Name ,CASE WHEN CHARINDEX('-',REVERSE(NAME))+ CHARINDEX('-',NAME) < LEN(NAME) THEN SUBSTRING(NAME, CHARINDEX('-', (NAME)) + 1, LEN(NAME) - CHARINDEX('-', REVERSE(NAME)) - CHARINDEX('-', NAME)) ELSE 'NULL' END AS Middle_Name ,REPLACE(REVERSE( SUBSTRING( REVERSE(NAME), 1, CHARINDEX('-',REVERSE(NAME)))), '-','') AS Last_Name ,REPLACE(LEFT(ADDRESS, CHARINDEX('|', ADDRESS)),'|',' ') AS Locality ,CASE WHEN CHARINDEX('|',REVERSE(ADDRESS))+ CHARINDEX('|',ADDRESS) < LEN(ADDRESS) THEN SUBSTRING(ADDRESS, CHARINDEX('|', (ADDRESS))+1, LEN(ADDRESS)-CHARINDEX('|', REVERSE(ADDRESS))-CHARINDEX('|',ADDRESS)) ELSE 'Null' END AS STATE ,REPLACE(REVERSE(SUBSTRING(REVERSE(ADDRESS),1 ,CHARINDEX('|',REVERSE(ADDRESS)))),'|','') AS Country FROM DETAILS SELECT CHARINDEX('-', REVERSE(NAME)) AS LAST,CHARINDEX('-',NAME)AS FIRST, LEN(NAME) AS LENGTH FROM DETAILS SELECT SUBSTRING(NAME, CHARINDEX('-', (NAME))+1, LEN(NAME) -CHARINDEX('-', REVERSE(NAME)) - CHARINDEX('-', NAME)) FROM DETAILS --LET ME KNOW IF YOU HAVE ANY DOUBTS UNDERSTANDING THE CODE