sql-server - separado - separar datos de una misma columna sql

Cómo dividir un valor separado por comas en columnas

Question

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

Answer 1

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

Answer 2

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,'','')

Answer 3

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.

Answer 4

Creo que esto es genial

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

Answer 5

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:

Answer 6

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,
, S.a.value(''(/H/r)[1]'', ''INT'') AS level1
, S.a.value(''(/H/r)[2]'', ''INT'') AS level2
, S.a.value(''(/H/r)[3]'', ''INT'') AS level3
, S.a.value(''(/H/r)[4]'', ''INT'') AS level4
, S.a.value(''(/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)

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

Answer 7

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 here

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 b b (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 E E
SELECT 10, ''1,55555'' UNION ALL --2 E E
SELECT 11, ''55555,4444,333,22,1'' UNION ALL --5 E E E E E
SELECT 12, ''55555,4444,,333,22,1'' UNION ALL --6 E E b E E E
SELECT 13, '',55555,4444,,333,22,1,'' UNION ALL --8 b E E b E E E b
SELECT 14, '',55555,4444,,,333,22,1,'' UNION ALL --9 b E E b b E E E b
SELECT 15, '' 4444,55555 '' UNION ALL --2 E (w/Leading Space) E (w/Trailing Space)
SELECT 16, ''This,is,a,test.'' --E E E E
) 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,t.N,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

Answer 8

La respuesta de la base xml es simple y limpia

referir this

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 T.c.value(''.'',''varchar(20)''),--retrieve all values at once
T.c.value(''(/root/myvalue)[1]'',''VARCHAR(20)'') , --retrieve index 1 only, which is the ''ab''
T.c.value(''(/root/myvalue)[2]'',''VARCHAR(20)''),
T.c.value(''(/root/myvalue)[3]'',''VARCHAR(20)'')
FROM @X.nodes(''/root/myvalue'') T(c)

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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 (<A href="mailto:[email protected]">[email protected]</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

Answer 13

Puede intentar esta consulta:

 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;

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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)))

Answer 19

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

Answer 20

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''<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

Answer 21

I found that using PARSENAME as above caused any name with a period to get nulled.

So if there was an initial or a title in the name followed by a dot they return NULL.

I found this worked for me:

SELECT
REPLACE(SUBSTRING(FullName, 1,CHARINDEX('','', FullName)), '','','''') as Name,
REPLACE(SUBSTRING(FullName, CHARINDEX('','', FullName), LEN(FullName)), '','', '''') as Surname
FROM Table1

Answer 22

Puedes usar la función dividida.

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

Answer 23

This function is most fast:

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

Answer 24

;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

y también verifique el siguiente enlace para referencia

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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