it-swarm.com.de

So teilen Sie einen durch Kommas getrennten Wert in Spalten auf

Ich habe so einen Tisch

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

Ich möchte die durch Kommas getrennte Zeichenfolge in zwei Spalten trennen

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

Ich brauche nur zwei feste zusätzliche Spalten 

102
Gurru

Ihr Zweck kann mit folgender Abfrage gelöst werden: 

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

Es gibt keine Readymade-Split-Funktion in SQL Server, daher müssen Sie eine benutzerdefinierte Funktion erstellen.

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)[email protected])
            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
116

xML-Basisantwort ist einfach und sauber

verweisen das

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)
37
aads
;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

und auch den Link unten als Referenz

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

37
bvr

Ich finde das cool 

SELECT value,
    PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
    PARSENAME(REPLACE(String,',','.'),1) 'Sur Name'
FROM table WITH (NOLOCK)
24
Azar

Mit CROSS APPLY

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
24
Lavisa

Es gibt mehrere Möglichkeiten, dieses Problem zu lösen, und es wurden bereits verschiedene Wege vorgeschlagen. Am einfachsten wäre es, LEFTSUBSTRING und andere Stringfunktionen zu verwenden, um das gewünschte Ergebnis zu erzielen.

_/ Beispieldaten

DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))

INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');

Verwenden von Stringfunktionen wie LEFT

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

Dieser Ansatz schlägt fehl, wenn sich in einem String mehr als 2 Elemente befinden. In einem solchen Szenario können wir einen Splitter verwenden und dann PIVOT verwenden oder die Zeichenfolge in eine XML umwandeln und .nodes verwenden, um Zeichenfolgenelemente abzurufen. Die Lösung von XML wurde von aads und bvr in ihrer Lösung beschrieben. 

Die Antworten auf diese Frage, die einen Splitter verwenden, verwenden alle WHILE, was zum Splitten ineffizient ist. Überprüfen Sie das Leistungsvergleich . Einer der besten Splitter ist DelimitedSplit8K, erstellt von Jeff Moden. Sie können mehr darüber lesen hier

Splitter mit 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

Ausgabe

Value   Fname   Lname
1   Cleo    Smith
2   John    Mathew

DelimitedSplit8K von 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
15
ughai

Probieren Sie dies aus (ändern Sie Instanzen von '' in '' oder 'welches Trennzeichen Sie verwenden möchten)

CREATE FUNCTION dbo.Wordparser
(
  @multiwordstring VARCHAR(255),
  @wordnumber      NUMERIC
)
returns VARCHAR(255)
AS
  BEGIN
      DECLARE @remainingstring VARCHAR(255)
      SET @[email protected]

      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

Verwendungsbeispiel:

SELECT dbo.Wordparser(COLUMN, 1),
       dbo.Wordparser(COLUMN, 2),
       dbo.Wordparser(COLUMN, 3)
FROM   TABLE
15
user7347410

Ich denke, PARSENAME ist die nette Funktion, die für dieses Beispiel zu verwenden ist, wie in diesem Artikel beschrieben: http://www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/

Die PARSENAME-Funktion ist logisch so ausgelegt, dass sie vierteilige Objektnamen analysiert. Das Schöne an PARSENAME ist, dass es nicht nur darauf beschränkt ist, nur vierteilige SQL Server-Objektnamen zu analysieren, sondern alle durch Punkte getrennten Funktions- oder String-Daten parsen.

Der erste Parameter ist das zu analysierende Objekt und der zweite ist der ganzzahlige Wert des zurückzugebenden Objektstücks. Der Artikel befasst sich mit dem Analysieren und Rotieren von begrenzten Daten - den Telefonnummern des Unternehmens -, aber es kann auch zum Parsen von Namen/Nachnamen-Daten verwendet werden. 

Beispiel:

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

In diesem Artikel wird auch die Verwendung eines Common Table Expression (CTE) mit dem Namen 'replaceChars' beschrieben, um PARSENAME mit den durch Trennzeichen ersetzten Werten auszuführen. Ein CTE ist nützlich, um eine temporäre Ansicht oder Ergebnismenge zurückzugeben. 

Danach wurde die UNPIVOT-Funktion verwendet, um einige Spalten in Zeilen umzuwandeln. Die Funktionen SUBSTRING und CHARINDEX wurden zum Bereinigen der Inkonsistenzen in den Daten verwendet, und am Ende wurde die LAG-Funktion (neu für SQL Server 2012) verwendet, da damit auf frühere Datensätze verwiesen werden kann.

10
JasonP

Mit SQL Server 2016 können Sie string_split verwenden, um dies zu erreichen:

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,',')
10

Wir können eine Funktion als diese erstellen

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

Anschließend können wir die CSV-Werte mithilfe einer SELECT-Anweisung in unsere jeweiligen Spalten aufteilen

10
Himansz

Ich denke, dass folgende Funktion für Sie funktionieren wird:

Sie müssen zuerst eine Funktion in SQL erstellen. So was

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)[email protected])       
    END
RETURN
END

Sie können diese Funktion wie folgt aufrufen:

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

Implementierung:

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

Ergebnis wird Ihnen gefallen:

 enter image description here

8
Muhammad Awais

Verwenden Sie die Funktion 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

Ergebnis 

Name    Family
-----   ------
Aria    Karimi
Bab     Karimi
Joe     Karimi
6
Mirak

Mit der Einstellungsfunktion :) 

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

Verwendete zwei Funktionen,
1. substring(string, position, length) ==> gibt den String von Position zu Länge zurück
2. instr(string,pattern) ==> gibt die Position des Musters zurück.

Wenn wir in substring kein length-Argument angeben, wird es bis zum Ende der Zeichenfolge zurückgegeben

5
WoodChopper
select id,SUBSTRING(name,0,charindex(',',name))as firstname
,SUBSTRING(name,charindex(',',name),len(name)+1)as lastname from spilt
5
anonymous

Versuche dies:

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
4
Rangani
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 <[email protected]_END
BEGIN
    SET @OUTPUT=(SUBSTRING (@INPUT,@L_START,1))
    IF @[email protected]_CHAR
    BEGIN
        PRINT @OUTPUT
    END
    SET @[email protected]_START+1
END
4

Diese Funktion ist am schnellsten:

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

Verwendungsbeispiel:

SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
       dbo.F_ExtractSubString(COLUMN, 2, ', '),
       dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM   TABLE
4
Mariano Sedano

Sie können eine eingebaute STRING_SPLIT-Funktion verwenden, die nur unter Kompatibilitätsgrad 130 verfügbar ist. Wenn Ihr Datenbankkompatibilitätsgrad niedriger als 130 ist, kann SQL Server STRING_SPLIT-Funktion nicht finden und ausführen. Sie können einen Kompatibilitätsgrad der Datenbank mit dem folgenden Befehl ändern:

ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130

Syntax

STRING_SPLIT ( string , separator )

siehe Dokumentation hier

3
bwanamaina

Ich hatte ein ähnliches Problem, aber ein komplexes, und da dies der erste Thread ist, den ich zu diesem Thema gefunden habe, habe ich mich entschlossen, meine Ergebnisse zu veröffentlichen. Ich weiß, es ist eine komplexe Lösung für ein einfaches Problem, aber ich hoffe, dass ich anderen helfen kann, die zu diesem Thread gehen und nach einer komplexeren Lösung suchen. Ich musste einen String mit 5 Zahlen (Spaltenname: levelsFeed) aufteilen und jede Zahl in einer separaten Spalte anzeigen Zum Beispiel: 8,1,2,2,2 sollte wie folgt angezeigt werden: 

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

Lösung 1: Verwendung von XML-Funktionen: Diese Lösung ist bei weitem die langsamste Lösung

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)

Lösung 2: Mit Split-Funktion und Pivot. (Die Split-Funktion teilt einen String in Zeilen mit dem Spaltennamen Data auf)

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

Lösung 3: Verwenden von String-Manipulationsfunktionen - am schnellsten mit geringerem Abstand zu Lösung 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

da der levelsFeed 5 string-Werte enthält, musste ich die substring-Funktion für den ersten string verwenden.

ich hoffe, dass meine Lösung anderen hilft, die zu diesem Thread gekommen sind, der nach einer komplexeren Aufteilung nach Spalten-Methoden sucht

3
Yossi
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 
2
vignesh

Möglicherweise finden Sie die Lösung in SQL User Defined Function zum Analysieren einer Trennzeichenfolge hilfreich (aus The Code Project ).

Dies ist der Code-Teil von dieser Seite:

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
2

Das hat für mich funktioniert

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
2
Krishna

mein Tisch:

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

Das Folgende sollte funktionieren, wenn nicht zu viele Spalten vorhanden sind

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', '') 

Ergebnis:

Value  ColOne ColTwo
--------------------
1      Cleo   Smith
2
kolunar

es ist so einfach, Sie können es unter der folgenden Abfrage nehmen:

DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3'
SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))
2
Mehdi najafian
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
1
Glyn
CREATE FUNCTION [dbo].[fn_split_string_to_column] (
    @string NVARCHAR(MAX),
    @delimiter CHAR(1)
    )
RETURNS @out_put TABLE (
    [column_id] INT IDENTITY(1, 1) NOT NULL,
    [value] NVARCHAR(MAX)
    )
AS
BEGIN
    DECLARE @value NVARCHAR(MAX),
        @pos INT = 0,
        @len INT = 0

    SET @string = CASE 
            WHEN RIGHT(@string, 1) != @delimiter
                THEN @string + @delimiter
            ELSE @string
            END

    WHILE CHARINDEX(@delimiter, @string, @pos + 1) > 0
    BEGIN
        SET @len = CHARINDEX(@delimiter, @string, @pos + 1) - @pos
        SET @value = SUBSTRING(@string, @pos, @len)

        INSERT INTO @out_put ([value])
        SELECT LTRIM(RTRIM(@value)) AS [column]

        SET @pos = CHARINDEX(@delimiter, @string, @pos + @len) + 1
    END

    RETURN
END
1
Blixter

Ich habe festgestellt, dass die Verwendung von PARSENAME wie oben bewirkt, dass ein beliebiger Name mit einem Punkt auf Null gesetzt wurde.

Wenn ein Name oder ein Titel gefolgt von einem Punkt vorlag, wird NULL zurückgegeben. 

Ich fand das funktioniert für mich: 

SELECT 
REPLACE(SUBSTRING(FullName, 1,CHARINDEX(',', FullName)), ',','') as Name,
REPLACE(SUBSTRING(FullName, CHARINDEX(',', FullName), LEN(FullName)), ',', '') as Surname
FROM Table1
1
RoadRunner
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
1
Frank
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
1
user7678586

Ich habe oben eine Antwort neu geschrieben und sie verbessert:

CREATE FUNCTION [dbo].[CSVParser]
(
  @s        VARCHAR(255),
  @idx      NUMERIC
)
RETURNS VARCHAR(12)
BEGIN
    DECLARE @comma int
    SET @comma = CHARINDEX(',', @s)
    WHILE 1=1
    BEGIN
        IF @comma=0
            IF @idx=1
                RETURN @s
            ELSE
                RETURN ''

        IF @idx=1
        BEGIN
            DECLARE @Word VARCHAR(12)
            SET @Word=LEFT(@s, @comma - 1)
            RETURN @Word
        END

        SET @s = RIGHT(@s,LEN(@s)[email protected])
        SET @comma = CHARINDEX(',', @s)
        SET @idx = @idx - 1
    END
    RETURN 'not used'
END

Beispielverwendung:

SELECT dbo.CSVParser(COLUMN, 1),
       dbo.CSVParser(COLUMN, 2),
       dbo.CSVParser(COLUMN, 3)
FROM   TABLE
0
Pete Alvin

Sie können die Split-Funktion verwenden.

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
0
vicky
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
0