如何将逗号分隔值拆分为列
我有一张这样的表
Value String
-------------------
1 Cleo, Smith
我想将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
38 个答案:
答案 0 :(得分：118)
您的目的可以使用以下查询解决 -
Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name,
Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as Surname
from Table1
在sql server中没有现成的Split函数，所以我们需要创建用户定义的函数。
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
答案 1 :(得分：41)
;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
并查看以下链接以供参考
http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html
答案 2 :(得分：38)
xml基本答案简单明了
参考this
DECLARE @S varchar(max),
@Split char(1),
@X xml
SELECT @S = 'ab,cd,ef,gh,ij',
@Split = ','
SELECT @X = CONVERT(xml,' <root> <myvalue>' +
REPLACE(@S,@Split,'</myvalue> <myvalue>') + '</myvalue> </root> ')
SELECT T.c.value('.','varchar(20)'), --retrieve ALL values at once
T.c.value('(/root/myvalue)[1]','VARCHAR(20)') , --retrieve index 1 only, which is the 'ab'
T.c.value('(/root/myvalue)[2]','VARCHAR(20)'),
T.c.value('(/root/myvalue)[3]','VARCHAR(20)')
FROM @X.nodes('/root/myvalue') T(c)
答案 3 :(得分：25)
我认为这很酷
SELECT value,
PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
PARSENAME(REPLACE(String,',','.'),1) 'Sur Name'
FROM table WITH (NOLOCK)
答案 4 :(得分：23)
使用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
答案 5 :(得分：15)
尝试此操作（更改''to'，'或您想要使用的分隔符的实例）
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
使用示例：
SELECT dbo.Wordparser(COLUMN, 1),
dbo.Wordparser(COLUMN, 2),
dbo.Wordparser(COLUMN, 3)
FROM TABLE
答案 6 :(得分：15)
有多种方法可以解决这个问题，并且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。
示例数据
DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))
INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');
使用字符串函数，例如LEFT
SELECT
Value,
LEFT(String,CHARINDEX(',',String)-1) as Fname,
LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname
FROM @tbl1
如果String中有2个项目，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并使用.nodes来获取字符串项。基于XML的解决方案已由aads和bvr在其解决方案中详细说明。
使用拆分器的这个问题的答案都使用WHILE，这对于拆分来说是低效的。检查此performance comparison。最好的分裂者之一是DelimitedSplit8K，由杰夫莫登创建。您可以阅读更多相关信息here
包含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
<强>输出
Value Fname Lname
1 Cleo Smith
2 John Mathew
DelimitedSplit8K
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
答案 7 :(得分：10)
我认为PARSENAME是用于此示例的简洁函数，如本文所述：http://www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/
PARSENAME函数在逻辑上设计用于解析四部分对象名称。 PARSENAME的优点在于它不仅仅解析SQL Server的四部分对象名称 - 它将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块的整数值。本文讨论解析和旋转分隔数据 - 公司电话号码，但它也可用于解析姓名/姓氏数据。
示例：
USE COMPANY;
SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue';
本文还介绍了使用名为“replaceChars”的公用表表达式（CTE）来对分隔符替换值运行PARSENAME。 CTE对于返回临时视图或结果集很有用。
之后，UNPIVOT函数用于将一些列转换为行; SUBSTRING和CHARINDEX函数已用于清除数据中的不一致性，最终使用了LAG函数（SQL Server 2012的新功能），因为它允许引用以前的记录。
答案 8 :(得分：10)
使用SQL Server 2016，我们可以使用string_split来完成此任务：
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,',')
答案 9 :(得分：9)
我们可以创建一个函数
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
然后我们可以使用SELECT语句将CSV值分成各自的列
答案 10 :(得分：7)
我认为以下功能对您有用：
您必须先在SQL中创建一个函数。像这样
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
您可以调用此函数，如下所示：
select * from fn_split('1,24,5',',')
实现：
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
结果将是这样的：
答案 11 :(得分：6)
使用Parsename（）函数
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
结果
Name Family
----- ------
Aria Karimi
Bab Karimi
Joe Karimi
答案 12 :(得分：6)
SELECT id,
Substring(NAME, 0, Charindex(',', NAME)) AS firstname,
Substring(NAME, Charindex(',', NAME), Len(NAME) + 1) AS lastname
FROM spilt
答案 13 :(得分：5)
试试这个：
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
答案 14 :(得分：4)
使用instring函数:)
select Value,
substring(String,1,instr(String," ") -1) Fname,
substring(String,instr(String,",") +1) Sname
from tablename;
使用了两个功能，
1. substring(string, position, length) ==＆gt;返回从positon到长度的字符串
2. instr(string,pattern) ==＆gt;返回模式的位置。
如果我们不在子字符串中提供长度参数，则返回直到字符串结尾
答案 15 :(得分：3)
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
使用示例：
SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
dbo.F_ExtractSubString(COLUMN, 2, ', '),
dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM TABLE
答案 16 :(得分：3)
我遇到了类似的问题，但是复杂的问题，因为这是我发现的关于该问题的第一个线程，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，但我希望我可以帮助其他人去寻找更复杂的解决方案。我不得不拆分一个包含5个数字的字符串（列名：levelsFeed），并在一个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
解决方案1：使用XML函数： 迄今为止最慢解决方案的解决方案
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)
解决方案2：使用Split功能和pivot。 （split函数将字符串拆分为列名为Data的行）
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
解决方案3：使用字符串操作函数 - 在解决方案2上以最小的余量最快
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
因为levelsFeed包含5个字符串值，我需要使用子字符串函数作为第一个字符串。
我希望我的解决方案可以帮助其他人找到更复杂的分割方法
答案 17 :(得分：3)
您可以使用仅在兼容级别130下可用的内置STRING_SPLIT函数。如果您的数据库兼容级别低于130，则SQL Server将无法找到并执行STRING_SPLIT函数。您可以使用以下命令更改数据库的兼容级别：
ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130
语法
STRING_SPLIT ( string , separator )
答案 18 :(得分：3)
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
答案 19 :(得分：2)
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
答案 20 :(得分：2)
这对我有用
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
答案 21 :(得分：2)
您可以在 SQL User Defined Function to Parse a Delimited String 中找到有用的解决方案（来自The Code Project）。
这是此页面的代码部分：
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
答案 22 :(得分：2)
这很容易，您可以通过以下查询来获取它：
DECLARE @str NVARCHAR(MAX)='ControlID_05436b78-04ba-9667-fa01-9ff8c1b7c235,3'
SELECT LEFT(@str, CHARINDEX(',',@str)-1),RIGHT(@str,LEN(@str)-(CHARINDEX(',',@str)))
答案 23 :(得分：2)
mytable的：
Value ColOne
--------------------
1 Cleo, Smith
如果没有太多列
，以下情况应该有效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', '')
结果：
Value ColOne ColTwo
--------------------
1 Cleo Smith
答案 24 :(得分：1)
create temp tablex as select table1.column1, table1.column2, table2.column3, table2.column4 from table1, table2 where table1.PrrimaryKey = table2.ForiegnKey;
答案 25 :(得分：1)
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
答案 26 :(得分：1)
我发现如上所述使用PARSENAME会导致任何带有句点的名称被取消。
因此，如果名称中有一个首字母或一个标题后跟一个点，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
答案 27 :(得分：1)
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
答案 28 :(得分：0)
您可以使用分割功能。
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
答案 29 :(得分：0)
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
答案 30 :(得分：0)
我重新编写了上面的答案，并将其做得更好：
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
用法示例：
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
答案 31 :(得分：0)
问题很简单，但问题很严重：）
因此，我为 string_split（）创建了一些包装器，该包装器以更通用的方式产生了数据透视。它是表函数，可返回值（nn，value1，value2，...，value50）-对于大多数CSV行而言已经足够。如果有更多值，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
)
使用示例：
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)
希望，这会有所帮助：）
答案 32 :(得分：0)
这是一个老问题，但如果可以升级到 SQL Server 2017+，也可以选择基于 JSON 的方法。这个想法是进行适当的转换：
将存储在
String列中的文本转换为有效的 JSON 数组（Cleo, Smith为["Cleo"," Smith"]）并使用JSON_VALUE()解析此数组。将存储在
String列中的文本转换为有效的嵌套 JSON 数组（Cleo, Smith为[["Cleo"," Smith"]]）并使用OPENJSON()和显式架构解析此数组（列定义）。
表格：
SELECT [Value], [String]
INTO Data
FROM (VALUES
(1, 'Cleo, Smith'),
(2, 'John, Smith'),
(3, 'Marian')
) v ([Value], [String])
使用 JSON_VALUE() 的语句：
SELECT
[Value],
TRIM(JSON_VALUE(CONCAT('["', REPLACE(STRING_ESCAPE([String], 'json'), ',', '","'), '"]'), 'lax $[0]')) AS Name,
TRIM(JSON_VALUE(CONCAT('["', REPLACE(STRING_ESCAPE([String], 'json'), ',', '","'), '"]'), 'lax $[1]')) AS Surname
FROM Data
使用 OPENJSON() 的语句：
SELECT d.[Value], TRIM(j.[Name]) AS [Name], TRIM(j.[Surname]) AS [Surname]
FROM Data d
OUTER APPLY OPENJSON(CONCAT('[["', REPLACE(STRING_ESCAPE(d.[String], 'json'), ',', '","'), '"]]')) WITH (
Name varchar(100) 'lax $[0]',
Surname varchar(100) 'lax $[1]'
) j
结果：
Value Name Surname
---------------------
1 Cleo Smith
2 John Smith
3 Marian
另外要注意的是，使用此技术，您可以通过添加适当的 JSON path 轻松解析两列以上的文本。
答案 33 :(得分：-1)
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
答案 34 :(得分：-1)
您可以使用SQL Server的“ STRING_SPLIT”功能：
STRING_SPLIT（字符串，分隔符）
答案 35 :(得分：-1)
尝试一下
CREATE FUNCTION [dbo].[Split]
(
@ListOfValues varchar(max),
@ValueSeparator varchar(10)
)
RETURNS @ListOfValuesInRows TABLE
(
Value varchar(max)
)
AS
BEGIN
IF Len(@ListOfValues) = 0
RETURN
if @ValueSeparator <> ' '
Begin
WHILE CHARINDEX(@ValueSeparator, @ListOfValues) > 0
BEGIN
INSERT INTO @ListOfValuesInRows
SELECT LTRIM(RTRIM(SUBSTRING(@ListOfValues, 1, CHARINDEX(@ValueSeparator, @ListOfValues)-1)))
SET @ListOfValues = SubString(@ListOfValues, CharIndex(@ValueSeparator, @ListOfValues)+Len(@ValueSeparator), Len(@ListOfValues))
END
INSERT INTO @ListOfValuesInRows
SELECT LTRIM(RTRIM(@ListOfValues))
End
Else
BEGIN
DECLARE @xml XML;
SET @xml = N'<t>' + REPLACE(@ListOfValues, @ValueSeparator, '</t><t>') + '</t>';
INSERT INTO @ListOfValuesInRows (Value)
SELECT LTRIM(RTRIM(r.value( '.', 'varchar(MAX)' ))) AS item
FROM @xml.nodes( '/t' ) AS records( r )
END
RETURN
END
答案 36 :(得分：-1)
选择 STRING 。 LEFT(STRING,Charindex(",",STRING)-1 as SURE Name RIGHT(STRING,LEN(STRING)-CHARINDEX(","STRING)) AS ]NAME FROM TABLE NAME
答案 37 :(得分：-3)
ALTER FUNCTION [dbo].[StringListTo] (@StringList Nvarchar(max),@Separators char(1),@start int, @index int )
RETURNS nvarchar(max)
AS
BEGIN
declare @out Nvarchar(max)
declare @i int
declare @start_old int
set @start=@start+1
set @i=1
while(@i<=@index)
begin
set @start_old=@start
set @start=CHARINDEX('.',@StringList,@start+1)
if (@start>0)
begin
set @out=Substring(@StringList,@start_old+1,@start-@start_old-1)
end
else
begin
set @out=Substring(@StringList,@start_old+1,len(@StringList)-1)
end
set @i=@i+1
end
RETURN @out
END;
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