Sql server 如何将逗号分隔的值拆分为列
我有一张这样的桌子Sql server 如何将逗号分隔的值拆分为列,sql-server,sql-server-2008,csv,Sql Server,Sql Server 2008,Csv,我有一张这样的桌子 Value String ------------------- 1 Cleo, Smith 我想把逗号分隔的字符串分成两列 Value Name Surname ------------------- 1 Cleo Smith 我只需要两个固定的额外列您的目的可以通过以下查询解决- Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name, Subs
Value String
-------------------
1 Cleo, Smith
Value Name Surname
-------------------
1 Cleo Smith
我只需要两个固定的额外列您的目的可以通过以下查询解决-
Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name,
Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as Surname
from Table1
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
您可以从中找到有帮助的解决方案 这是本页的代码部分：
CREATE FUNCTION [fn_ParseText2Table]
(@p_SourceText VARCHAR(MAX)
,@p_Delimeter VARCHAR(100)=',' --default to comma delimited.
)
RETURNS @retTable
TABLE([Position] INT IDENTITY(1,1)
,[Int_Value] INT
,[Num_Value] NUMERIC(18,3)
,[Txt_Value] VARCHAR(MAX)
,[Date_value] DATETIME
)
AS
/*
********************************************************************************
Purpose: Parse values from a delimited string
& return the result as an indexed table
Copyright 1996, 1997, 2000, 2003 Clayton Groom (<A href="mailto:Clayton_Groom@hotmail.com">Clayton_Groom@hotmail.com</A>)
Posted to the public domain Aug, 2004
2003-06-17 Rewritten as SQL 2000 function.
Reworked to allow for delimiters > 1 character in length
and to convert Text values to numbers
2016-04-05 Added logic for date values based on "new" ISDATE() function, Updated to use XML approach, which is more efficient.
********************************************************************************
*/
BEGIN
DECLARE @w_xml xml;
SET @w_xml = N'<root><i>' + replace(@p_SourceText, @p_Delimeter,'</i><i>') + '</i></root>';
INSERT INTO @retTable
([Int_Value]
, [Num_Value]
, [Txt_Value]
, [Date_value]
)
SELECT CASE
WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
THEN CAST(CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC) AS INT)
END AS [Int_Value]
, CASE
WHEN ISNUMERIC([i].value('.', 'VARCHAR(MAX)')) = 1
THEN CAST([i].value('.', 'VARCHAR(MAX)') AS NUMERIC(18, 3))
END AS [Num_Value]
, [i].value('.', 'VARCHAR(MAX)') AS [txt_Value]
, CASE
WHEN ISDATE([i].value('.', 'VARCHAR(MAX)')) = 1
THEN CAST([i].value('.', 'VARCHAR(MAX)') AS DATETIME)
END AS [Num_Value]
FROM @w_xml.nodes('//root/i') AS [Items]([i]);
RETURN;
END;
GO
基于xml的答案简单明了 提及
我们可以创建这样一个函数
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
SELECT value,
PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
PARSENAME(REPLACE(String,',','.'),1) 'Sur Name'
FROM table WITH (NOLOCK)
Value ColOne
--------------------
1 Cleo, Smith
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', '')
Value ColOne ColTwo
--------------------
1 Cleo Smith
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
Name Family
----- ------
Aria Karimi
Bab Karimi
Joe Karimi
我发现如上所述使用PARSENAME会导致任何带句点的名称都被置为空 因此，如果名称中有首字母或标题，后跟一个点，则返回NULL 我发现这对我有用：
SELECT
REPLACE(SUBSTRING(FullName, 1,CHARINDEX(',', FullName)), ',','') as Name,
REPLACE(SUBSTRING(FullName, CHARINDEX(',', FullName), LEN(FullName)), ',', '') as Surname
FROM Table1
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
有多种方法可以解决这个问题，已经提出了许多不同的方法。最简单的方法是使用LEFT/SUBSTRING和其他字符串函数来实现所需的结果 样本数据 使用像LEFT这样的字符串函数 如果字符串中有两个以上的项，则此方法将失败。 在这种情况下，我们可以使用拆分器，然后使用PIVOT或将字符串转换为XML，然后使用.nodes获取字符串项。aads和bvr在其解决方案中详细介绍了基于XML的解决方案 这个问题的答案都使用了拆分器，而拆分效率很低。看看这个。最好的拆分器之一是由Jeff Moden创建的DelimitedSplit8K。你可以阅读更多关于它的内容 带枢轴的分离器 输出 Jeff Moden设计的DSPLIT8K
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
我遇到了一个类似的问题，但很复杂，因为这是我发现的关于这个问题的第一个线程，所以我决定发布我的发现。我知道这是一个简单问题的复杂解决方案，但我希望我能帮助其他人谁去这个线程寻找更复杂的解决方案。我必须拆分一个包含5个数字的字符串列名：levelsFeed，并在单独的列中显示每个数字。 例如：8,1,2,2,2 应显示为：
1 2 3 4 5
-------------
8 1 2 2 2
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)
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
我希望我的解决方案能帮助其他人找到更复杂的拆分到列方法，使用instring函数：
select Value,
substring(String,1,instr(String," ") -1) Fname,
substring(String,instr(String,",") +1) Sname
from tablename;
如果我们在子字符串中不提供长度参数，它将返回到字符串末尾，我认为PARSENAME是用于此示例的整洁函数，如本文所述： PARSENAME函数在逻辑上设计为解析由四部分组成的对象名。PARSENAME的好处在于它不仅限于解析由SQL Server四部分组成的对象名，它还可以解析由点分隔的任何函数或字符串数据 第一个参数是要解析的对象，第二个参数是要返回的对象片段的整数值。本文讨论的是解析和旋转分隔数据-公司电话号码，但它也可以用于解析姓名数据 例如：
USE COMPANY;
SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue';
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
这很简单，您可以通过以下查询获得：
DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3'
SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))
使用SQL Server 2016，我们可以使用字符串分割来实现这一点：
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,',')
请尝试将“”的实例更改为“”，“”或任何要使用的分隔符
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
SELECT dbo.Wordparser(COLUMN, 1),
dbo.Wordparser(COLUMN, 2),
dbo.Wordparser(COLUMN, 3)
FROM TABLE
SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
dbo.F_ExtractSubString(COLUMN, 2, ', '),
dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM TABLE
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130
SELECT * FROM STRING_SPLIT ( string, separator )
我认为以下功能适用于您： 您必须首先在SQL中创建一个函数。像 是 您可以调用此函数，如下所示：
select * from fn_split('1,24,5',',')
此功能最快：
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
SELECT dbo.Wordparser(COLUMN, 1),
dbo.Wordparser(COLUMN, 2),
dbo.Wordparser(COLUMN, 3)
FROM TABLE
SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
dbo.F_ExtractSubString(COLUMN, 2, ', '),
dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM TABLE
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
我在上面重新写了一个答案，让它变得更好：
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
SELECT dbo.Wordparser(COLUMN, 1),
dbo.Wordparser(COLUMN, 2),
dbo.Wordparser(COLUMN, 3)
FROM TABLE
SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
dbo.F_ExtractSubString(COLUMN, 2, ', '),
dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM TABLE
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
问题很简单，但问题很棘手： 因此，我为string_split创建了一些包装器，这将以更通用的方式实现pivot。它是一个表函数，返回值nn，value1，value2，值50-足够用于大多数CSV行。如果有更多的值，它们将换行到下一行-nn表示行号。将第三个参数@columnCnt=[yourNumber]设置为在特定位置换行：
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
)
select * from dbo.fn_split50('zz1,aa2,ss3,dd4,ff5', ',', DEFAULT)
希望，这会有帮助：可能重复的From：输出行可能是任意顺序的。顺序不能保证与输入字符串中子字符串的顺序匹配。它是在SQL Server 2016中添加的。@HABO，因此，在SQL Server 2008中询问如何使用它是没有用的：-您是否有可能在此处总结解决方案，以确保链接一旦消失，答案不会过时？您的要求仅针对姓名和姓氏。您还需要注意，对于超过128个字符的物品。不错。也适用于我的数据集！这更好。。它简单又短。我真的很喜欢这种方式。如果要拆分的值超过2个，例如1、2、3，则CHARINDEX和SUBSTRING会变得一团糟。谢谢，好主意。至少对我来说，它的速度是CHARINDEX plus子字符串混乱的三倍-很好的解决方案，但是XML中的一些字符是非法的，例如“&”，所以我必须将每个字段包装在CDATA标记中。。。将XML转换为xmlname@Tony需要将代码从Tony更新为CONVERTXML，因为xmlname缺少最后的s，如果您可以展开答案，并使用代码格式工具，这将非常有用。关闭，这将在lastname中包含逗号。在错误的位置得到+1。应该是SubstringNAME，Charindex'，'，NAME+1，LenNAME AS lastname我无法理解为什么需要在原始字符串的末尾添加2个逗号才能使其正常工作。为什么没有+'、'？@developer.ejay它就不能工作？是因为左/子字符串函数不能接受0值吗？太好了！您可以很容易地为每个需要的额外列复制/粘贴两行，然后只需增加数字，例如：select ParsedData.*从MyTable mt交叉应用选择str=mt.String+'，'f1交叉应用选择p1=charindex'，'，str ap1交叉应用选择p2=charindex'，'，str，p1+1 ap2交叉应用选择p3=charindex'，'str，p2+1 ap3交叉应用select FName=substring str，1，p1-1，LName=substring str，p1+1，p2-p1-1，Age=substring str，p2+1，p3-p2-1 parseddatan不确定您使用的是什么SQL方言，但在SQL Server中我们必须使用类似substring@str1.charindex@sep,@str-1后跟substring@str, charindex@sep，@str+1，len@str.I我正在使用SQL Server 2016，但它给出了一个错误无效的对象名“string\u split”。您可以检查数据库的兼容性级别吗？它必须是130，即sql server 2016。您可以使用此查询select*from sys.databasesright，我看到的是120，因此它必须是2016年的客户端Microsoft SQL Server Management Studio，而不是数据库服务器本身，因为如果我转到“帮助”->“关于”，我会看到SQL Server 2016 Management Studio v13.0.15000.23。如果db开发人员将级别设置为任何较低的值，以保持db兼容，则可能发生这种情况，即使实际安装的版本较高。使用此选项将级别设置为所需的高级别，只要db支持此选项：DECLARE@cl TINYINT；从[sys].[databases]中选择@cl=compatibility_level，其中name='mydb'；如果@cl<130 BEGIN ALTER DATABASE myDb SET COMPATIBILITY_LEVEL=130 END；除非您将其从行转回到列，否则这是无用的。我使用了您的代码，它很简单，但在Remouse_CHAT中存在拼写错误。它应该是Remouse_CHAR，脚本末尾的START应该是L_START。谢谢。虽然此代码可能会回答此问题，但提供有关此代码为什么和/或如何回答此问题的其他上下文将提高其长期价值。您知道如何处理xml特殊字符吗？感谢您提供此代码片段，它可能会提供一些有限的即时帮助。通过说明为什么这是一个很好的问题解决方案来正确解释它的长期价值，并将使它对未来有其他类似问题的读者更有用。请您的答案添加一些解释，包括您所做的假设。请看Jeff Moden在下面@ughai answer中的数字表解决方案DelimitedSplit8K。SQL 2016现在提供了一个拆分函数SQL 2016及更高版本：SELECT*FROM STRING_split'John，Jeremy，Jack'，'使用循环拆分字符串效率极低。这里有几个更好的分割函数选项。美好的但在我的回答中，它不适用于2016年以下的SQL Server
我指出，它将只在兼容级别130及更高版本中可用。但是STRING_分割成多行，而不是每个分割都有多个列。这篇报道问的是关于分成多个栏目的问题，对吧？像个魔术师一样工作！这真的很酷。类似数组的功能非常有用，我对此一无所知。谢谢如果同一行中的值相同，则我失败。这不应是可接受的答案。。。多语句TVF非常糟糕！而一个更糟糕的循环将表现得非常糟糕。除此之外，这是一个只使用代码的答案，甚至不能解决问题—有更好的方法！对于SQL Server 2016+来说，查找字符串_SPLIT，它不包含片段的位置，这是一个巨大的失败！或者真正快速的JSON黑客。对于旧版本，请查看著名的XML黑客json和XML详细信息。或者寻找一个基于递归CTEs.SQL 2016及更高版本的may ITVF：从字符串中选择*“John，Jeremy，Jack”，“同意给定的解决方案”。但是，如果您是SQL Server 2016，则可以使用字符串分割函数。。你也可以在这里找到这个内置函数的用法，每个人都建议STRING_SPLIT，这个函数怎么能像预期的那样将字符串拆分成列而不是行呢？
select * from fn_split('1,24,5',',')
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
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
SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
dbo.F_ExtractSubString(COLUMN, 2, ', '),
dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM TABLE
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
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
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
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
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
)
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)