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쉼표로 구분된 값을 열로 분할하는 방법
이런 테이블이 있어요.
| 가치 | 스트링 |
|---|---|
| 1 | 스미스 주 클레오 |
쉼표로 구분된 문자열을 두 열로 구분합니다.
| 가치 | 이름. | 성 |
|---|---|---|
| 1 | 클레오 | 스미스 |
두 개의 고정 추가 열만 필요합니다.
당신의 목적은 다음 쿼리를 사용하여 해결할 수 있습니다.
Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name,
Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as Surname
from Table1
sql 서버에는 readymade Split 함수가 없기 때문에 사용자 정의 함수를 만들어야 합니다.
CREATE FUNCTION Split (
@InputString VARCHAR(8000),
@Delimiter VARCHAR(50)
)
RETURNS @Items TABLE (
Item VARCHAR(8000)
)
AS
BEGIN
IF @Delimiter = ' '
BEGIN
SET @Delimiter = ','
SET @InputString = REPLACE(@InputString, ' ', @Delimiter)
END
IF (@Delimiter IS NULL OR @Delimiter = '')
SET @Delimiter = ','
--INSERT INTO @Items VALUES (@Delimiter) -- Diagnostic
--INSERT INTO @Items VALUES (@InputString) -- Diagnostic
DECLARE @Item VARCHAR(8000)
DECLARE @ItemList VARCHAR(8000)
DECLARE @DelimIndex INT
SET @ItemList = @InputString
SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
WHILE (@DelimIndex != 0)
BEGIN
SET @Item = SUBSTRING(@ItemList, 0, @DelimIndex)
INSERT INTO @Items VALUES (@Item)
-- Set @ItemList = @ItemList minus one less item
SET @ItemList = SUBSTRING(@ItemList, @DelimIndex+1, LEN(@ItemList)-@DelimIndex)
SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
END -- End WHILE
IF @Item IS NOT NULL -- At least one delimiter was encountered in @InputString
BEGIN
SET @Item = @ItemList
INSERT INTO @Items VALUES (@Item)
END
-- No delimiters were encountered in @InputString, so just return @InputString
ELSE INSERT INTO @Items VALUES (@InputString)
RETURN
END -- End Function
GO
---- Set Permissions
--GRANT SELECT ON Split TO UserRole1
--GRANT SELECT ON Split TO UserRole2
--GO
;WITH Split_Names (Value,Name, xmlname)
AS
(
SELECT Value,
Name,
CONVERT(XML,'<Names><name>'
+ REPLACE(Name,',', '</name><name>') + '</name></Names>') AS xmlname
FROM tblnames
)
SELECT Value,
xmlname.value('/Names[1]/name[1]','varchar(100)') AS Name,
xmlname.value('/Names[1]/name[2]','varchar(100)') AS Surname
FROM Split_Names
아래 링크도 참조해 주세요.
http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html
이거 멋진 것 같아
SELECT value,
PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
PARSENAME(REPLACE(String,',','.'),1) 'Surname'
FROM table WITH (NOLOCK)
xml 기반 답변은 단순하고 깨끗합니다.
이것을 참조하다
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)')
FROM @X.nodes('/root/myvalue') T(c)
CROSS APPLY(크로스 적용)
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
이 문제를 해결하는 방법은 여러 가지가 있으며 이미 여러 가지 방법이 제시되어 있습니다. 방법은 '우리'를 사용하는 입니다.LEFTSUBSTRING및 기타 문자열 기능을 사용하여 원하는 결과를 얻을 수 있습니다.
샘플 데이터
DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))
INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');
함수 사용(예: 'String LEFT
SELECT
Value,
LEFT(String,CHARINDEX(',',String)-1) as Fname,
LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname
FROM @tbl1
이 접근법은 String에 2개 이상의 항목이 있는 경우 실패합니다., 「스플리터」를 사용할 수 .PIVOT을 는또스스 or an an an an an an an로 합니다.XML를 사용합니다..nodes이치노 XML아다스 bvr.
하는 이 스플리터를 사용합니다.WHILE분할에는 비효율적입니다.이 퍼포먼스 비교를 확인해 주세요.최고의 스플리터 중 하나는DelimitedSplit8KJeff Moden에 의해 작성되었습니다.자세한 내용은 이쪽에서 확인하실 수 있습니다.
(「」가 있는 경우PIVOT
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
산출량
Value Fname Lname
1 Cleo Smith
2 John Mathew
DelimitedSplit8K Jeff Moden 지음
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
SQL Server 2016에서는 string_split을 사용하여 이를 수행할 수 있습니다.
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,',')
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
SELECT id,
Substring(NAME, 0, Charindex(',', NAME)) AS firstname,
Substring(NAME, Charindex(',', NAME), Len(NAME) + 1) AS lastname
FROM spilt
이 작업을 수행해 보십시오('의 인스턴스를 ''로 변경하거나 사용할 구분 기호를 선택하십시오).
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
사용 예:
SELECT dbo.Wordparser(COLUMN, 1),
dbo.Wordparser(COLUMN, 2),
dbo.Wordparser(COLUMN, 3)
FROM TABLE
PARSENAME은 이 예에서 사용하는 훌륭한 함수라고 생각합니다.이 예에서는 http://www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/ 에서 설명하고 있습니다.
PARSENAME 함수는 4부분의 오브젝트 이름을 해석하도록 논리적으로 설계되어 있습니다.PARSENAME의 장점은 SQL Server의 4부분 오브젝트 이름만 해석하는 것이 아니라 점으로 구분된 함수 또는 문자열 데이터를 해석한다는 것입니다.
첫 번째 파라미터는 해석할 오브젝트이고 두 번째 파라미터는 반환할 오브젝트 조각의 정수값입니다.이 기사에서는 구분된 데이터(회사 전화번호)의 해석과 순환에 대해 설명하고 있지만 이름/스너명 데이터의 해석에도 사용할 수 있습니다.
예:
USE COMPANY;
SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue';
또, 「replaceChars」라고 불리는 Common Table Expression(CTE)을 사용해 딜리미터로 치환된 값에 대해서 PARSENAME 를 실행하는 방법에 대해서도 설명합니다.CTE는 임시 뷰 또는 결과 세트를 반환할 때 유용합니다.
그 후 UNPIVOT 함수를 사용하여 일부 열을 행으로 변환하고 SUBSTRING 및 CHARINDEX 함수를 사용하여 데이터의 불일치를 해소하며, 최종적으로는 이전 레코드를 참조할 수 있도록 LAG 기능(SQL Server 2012에 새롭게 추가)을 사용했습니다.
다음과 같은 함수를 만들 수 있습니다.
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
그런 다음 SELECT 문을 사용하여 CSV 값을 각 컬럼으로 분리할 수 있습니다.
테이블 값 함수를 사용할 수 있습니다.STRING_SPLIT130달러입니다.는 130보다 낮은 SQL Server를 할 수 .STRING_SPLIT할 수 있습니다.다음 명령을 사용하여 데이터베이스의 호환성 수준을 변경할 수 있습니다.
ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130
구문
SELECT * FROM STRING_SPLIT ( string, separator )
다음과 같은 기능이 도움이 될 것 같습니다.
먼저 SQL에서 함수를 만들어야 합니다.이것처럼.
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
이 함수를 다음과 같이 호출할 수 있습니다.
select * from fn_split('1,24,5',',')
구현:
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
결과는 다음과 같습니다.
Parsename() 함수를 사용합니다.
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
결과
Name Family
----- ------
Aria Karimi
Bab Karimi
Joe Karimi
이것을 시험해 보세요.
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
이 기능은 가장 빠릅니다.
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
사용 예:
SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
dbo.F_ExtractSubString(COLUMN, 2, ', '),
dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM TABLE
저는 비슷한 문제를 겪었지만 복잡한 문제를 발견했습니다.그리고 이것이 그 문제에 관해 제가 발견한 첫 번째 스레드이기 때문에 저는 저의 결과를 올리기로 결정했습니다.심플한 문제에 대한 복잡한 해결책이라는 것을 알지만, 이 스레드로 가는 다른 사람들이 더 복잡한 해결책을 찾도록 도울 수 있기를 바랍니다.숫자 5개가 포함된 문자열(열 이름: levelsFeed)을 분할하고 각 숫자를 별도의 열에 표시해야 했습니다.예를 들어 8,1,2,2는 다음과 같이 표시됩니다.
1 2 3 4 5
-------------
8 1 2 2 2
솔루션 1: XML 기능 사용: 이 솔루션은 지금까지 가장 느린 솔루션입니다.
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)
해결책 2: 분할 기능과 피벗을 사용합니다.(분할 함수는 열 이름이 Data인 행으로 문자열을 분할합니다.)
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
솔루션 3: 문자열 조작 기능 사용 - 솔루션 2보다 적은 차이로 가장 빠릅니다.
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
수준부터피드에는 첫 번째 문자열에 대해 하위 문자열 함수를 사용하기 위해 필요한 5개의 문자열 값이 포함되어 있습니다.
내 솔루션이 이 스레드에 도달한 다른 사용자가 더 복잡한 컬럼 분할 방법을 찾는 데 도움이 되기를 바랍니다.
instring 기능 사용 :)
select Value,
substring(String,1,instr(String," ") -1) Fname,
substring(String,instr(String,",") +1) Sname
from tablename;
두 가지 기능을 사용했습니다.
1. substring(string, position, length)==> 문자열이 양에서 길이로 반환됩니다.
2. instr(string,pattern)==> 패턴의 위치를 반환합니다.
부분 문자열에 length 인수를 지정하지 않으면 문자열 끝까지 반환됩니다.
이건 내게 효과가 있었다.
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
mytable:
Value ColOne
--------------------
1 Cleo, Smith
열이 너무 많지 않으면 다음이 작동합니다.
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', '')
결과:
Value ColOne ColTwo
--------------------
1 Cleo Smith
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
구분된 문자열을 구문 분석하는 SQL 사용자 정의 함수의 솔루션이 도움이 될 수 있습니다(코드 프로젝트).
이 페이지의 코드 부분은 다음과 같습니다.
CREATE FUNCTION [fn_ParseText2Table]
(@p_SourceText VARCHAR(MAX)
,@p_Delimeter VARCHAR(100)=',' --default to comma delimited.
)
RETURNS @retTable
TABLE([Position] INT IDENTITY(1,1)
,[Int_Value] INT
,[Num_Value] NUMERIC(18,3)
,[Txt_Value] VARCHAR(MAX)
,[Date_value] DATETIME
)
AS
/*
********************************************************************************
Purpose: Parse values from a delimited string
& return the result as an indexed table
Copyright 1996, 1997, 2000, 2003 Clayton Groom (<A href="mailto:Clayton_Groom@hotmail.com">Clayton_Groom@hotmail.com</A>)
Posted to the public domain Aug, 2004
2003-06-17 Rewritten as SQL 2000 function.
Reworked to allow for delimiters > 1 character in length
and to convert Text values to numbers
2016-04-05 Added logic for date values based on "new" ISDATE() function, Updated to use XML approach, which is more efficient.
********************************************************************************
*/
BEGIN
DECLARE @w_xml xml;
SET @w_xml = N'<root><i>' + replace(@p_SourceText, @p_Delimeter,'</i><i>') + '</i></root>';
INSERT INTO @retTable
([Int_Value]
, [Num_Value]
, [Txt_Value]
, [Date_value]
)
SELECT CASE
WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
THEN CAST(CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC) AS INT)
END AS [Int_Value]
, CASE
WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
THEN CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC(18, 3))
END AS [Num_Value]
, [i].value('.', 'VARCHAR(MAX)') AS [txt_Value]
, CASE
WHEN ISDATE([i].value('.', 'VARCHAR(MAX)')) = 1
THEN CAST([i].value('.', 'VARCHAR(MAX)') AS DATETIME)
END AS [Num_Value]
FROM @w_xml.nodes('//root/i') AS [Items]([i]);
RETURN;
END;
GO
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
위의 답변을 다시 작성하여 개선했습니다.
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)-@comma)
SET @comma = CHARINDEX(',', @s)
SET @idx = @idx - 1
END
RETURN 'not used'
END
사용 예:
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
위와 같이 PARSENAME을 사용하면 마침표를 가진 이름이 모두 무효가 됩니다.
따라서 이름에 이니셜 또는 제목이 있고 그 뒤에 점이 있으면 NULL을 반환합니다.
나는 이것이 나에게 효과가 있다는 것을 알았다.
SELECT
REPLACE(SUBSTRING(FullName, 1,CHARINDEX(',', FullName)), ',','') as Name,
REPLACE(SUBSTRING(FullName, CHARINDEX(',', FullName), LEN(FullName)), ',', '') as Surname
FROM Table1
매우 간단하기 때문에 아래 쿼리를 통해 확인할 수 있습니다.
DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3'
SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))
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
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
질문은 간단하지만 문제는 뜨겁다:)
따라서 string_split()에 대한 래퍼를 만듭니다.이러한 래퍼는 보다 일반적인 방법으로 피벗됩니다.대부분의 CSV 회선에 충분한 값(nn, value1, value2, ..., value50)을 반환하는 테이블 함수입니다.값이 더 많으면 다음 행으로 바꿉니다.nnn은 행 번호를 나타냅니다.세 번째 파라미터 @columnCnt = [ yourNumber ]를 설정하여 특정 위치에서 랩합니다.
alter FUNCTION fn_Split50
(
@str varchar(max),
@delim char(1),
@columnCnt int = 50
)
RETURNS TABLE
AS
RETURN
(
SELECT *
FROM (SELECT
nn = (nn - 1) / @columnCnt + 1,
nnn = 'value' + cast(((nn - 1) % @columnCnt) + 1 as varchar(10)),
value
FROM (SELECT
nn = ROW_NUMBER() over (order by (select null)),
value
FROM string_split(@str, @delim) aa
) aa
where nn > 0
) bb
PIVOT
(
max(value)
FOR nnn IN (
value1, value2, value3, value4, value5, value6, value7, value8, value9, value10,
value11, value12, value13, value14, value15, value16, value17, value18, value19, value20,
value21, value22, value23, value24, value25, value26, value27, value28, value29, value30,
value31, value32, value33, value34, value35, value36, value37, value38, value39, value40,
value41, value42, value43, value44, value45, value46, value47, value48, value49, value50
)
) AS PivotTable
)
사용 예:
select * from dbo.fn_split50('zz1,aa2,ss3,dd4,ff5', ',', DEFAULT)
select * from dbo.fn_split50('zz1,aa2,ss3,dd4,ff5,gg6,hh7,jj8,ww9,qq10', ',', 3)
select * from dbo.fn_split50('zz1,11,aa2,22,ss3,33,dd4,44,ff5,55,gg6,66,hh7,77,jj8,88,ww9,99,qq10,1010', ',',2)
Hope, 도움이 될 것입니다. :)
언급URL : https://stackoverflow.com/questions/10581772/how-to-split-a-comma-separated-value-to-columns
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