我有一张这样的桌子
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
问问题
1123770 次
38 回答
147
您的目的可以使用以下查询来解决 -
Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name,
Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as Surname
from Table1
sql server中没有现成的Split函数，所以我们需要创建自定义函数。
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
于 2012-05-14T10:43:39.273 回答
58
;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
并查看下面的链接以供参考
http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html
于 2013-02-27T09:30:02.983 回答
47
基于xml的答案简单而干净
参考这个
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)
于 2013-06-19T07:32:22.557 回答
42
我觉得这很酷
SELECT value,
PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
PARSENAME(REPLACE(String,',','.'),1) 'Surname'
FROM table WITH (NOLOCK)
于 2014-06-20T05:39:27.113 回答
30
交叉申请
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
于 2015-05-13T17:23:19.570 回答
23
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。
样本数据
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，如LEFT
SELECT
Value,
LEFT(String,CHARINDEX(',',String)-1) as Fname,
LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname
FROM @tbl1
如果字符串中有更多 2 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和使用.nodes来获取字符串项。XMLaads 和 bvr 在他们的解决方案中详细说明了基于解决方案的解决方案。
这个问题的答案使用拆分器，所有使用WHILE拆分效率低下。检查此性能比较。周围最好的分离器之一是DelimitedSplit8K由 Jeff Moden 创建的。你可以在这里阅读更多关于它的信息
分流器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
输出
Value Fname Lname
1 Cleo Smith
2 John Mathew
DelimitedSplit8K通过杰夫摩登
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
于 2015-06-11T09:18:52.690 回答
22
使用 SQL Server 2016，我们可以使用 string_split 来完成此操作：
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,',')
于 2016-09-07T18:20:38.327 回答
19
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
于 2019-02-28T13:34:01.397 回答
15
SELECT id,
Substring(NAME, 0, Charindex(',', NAME)) AS firstname,
Substring(NAME, Charindex(',', NAME), Len(NAME) + 1) AS lastname
FROM spilt
于 2015-04-14T18:42:14.163 回答
15
试试这个（将 ' ' 的实例更改为 ',' 或您要使用的任何分隔符）
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
示例用法：
SELECT dbo.Wordparser(COLUMN, 1),
dbo.Wordparser(COLUMN, 2),
dbo.Wordparser(COLUMN, 3)
FROM TABLE
于 2016-12-27T21:16:57.063 回答
13
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 函数在逻辑上设计为解析四部分对象名称。PARSENAME 的好处是它不仅限于解析 SQL Server 的四部分对象名称——它会解析任何由点分隔的函数或字符串数据。
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。这篇文章正在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。
例子：
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) 来针对分隔符替换的值运行 PARSENAME。CTE 对于返回临时视图或结果集很有用。
之后，使用 UNPIVOT 函数将一些列转换为行；SUBSTRING 和 CHARINDEX 函数用于清理数据中的不一致，最后使用了 LAG 函数（SQL Server 2012 的新功能），因为它允许引用以前的记录。
于 2015-12-23T09:51:14.277 回答
10
我们可以像这样创建一个函数
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 值分隔到我们各自的列中
于 2014-03-26T10:42:15.433 回答
9
我认为以下功能对您有用：
您必须首先在 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',',')
执行：
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：
于 2017-08-28T07:32:48.387 回答
8
您可以使用表值函数STRING_SPLIT，该函数仅在兼容级别 130 下可用。如果您的数据库兼容级别低于 130，SQL Server 将无法找到并执行该STRING_SPLIT函数。您可以使用以下命令更改数据库的兼容性级别：
ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130
句法
SELECT * FROM STRING_SPLIT ( string, separator )
于 2017-07-09T08:54:06.287 回答
7
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
于 2015-03-18T04:36:22.143 回答
7
尝试这个：
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
于 2016-03-14T11:47:23.717 回答
6
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
示例用法：
SELECT dbo.F_ExtractSubString(COLUMN, 1, ', '),
dbo.F_ExtractSubString(COLUMN, 2, ', '),
dbo.F_ExtractSubString(COLUMN, 3, ', ')
FROM TABLE
于 2017-12-19T14:55:21.320 回答
5
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该问题的第一个线程，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，但我希望我可以帮助其他去这个线程寻找更复杂解决方案的人。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
解决方案 1：使用 XML 函数：此解决方案是迄今为止最慢的解决方案
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函数和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 - 比解决方案 2 快一点
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，所以我需要对第一个字符串使用子字符串函数。
我希望我的解决方案能帮助其他进入该线程的人寻找更复杂的拆分列方法
于 2015-07-14T09:42:52.287 回答
5
使用字符串函数:)
select Value,
substring(String,1,instr(String," ") -1) Fname,
substring(String,instr(String,",") +1) Sname
from tablename;
使用了两个函数，
1. substring(string, position, length) ==> 返回从位置到长度的字符串
2. instr(string,pattern)==> 返回模式的位置。
如果我们不在子字符串中提供长度参数，它会返回到字符串结尾
于 2015-09-30T08:36:00.330 回答
4
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
于 2017-01-07T14:31:58.880 回答
3
表：
Value ColOne
--------------------
1 Cleo, Smith
如果没有太多列，以下应该可以工作
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
于 2014-12-29T22:30:56.443 回答
3
您可能会在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
于 2012-05-14T10:43:56.293 回答
3
这很简单，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)))
于 2016-03-16T07:40:44.247 回答
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
于 2016-08-10T13:10:47.207 回答
3
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
于 2018-05-22T09:17:21.323 回答
2
我发现使用上面的 PARSENAME 会导致任何带有句点的名称都为空。
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
于 2015-04-28T03:17:23.917 回答
2
问题很简单，但问题很热门:)
所以我为string_split()创建了一些包装器，它以更通用的方式旋转结果。它是返回值 (nn, value1, value2, ... , value50) 的表函数 - 对于大多数 CSV 行来说已经足够了。如果有更多值，它们将换行到下一行 - nn表示行号。设置第三个参数@columnCnt = [yourNumber]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，它会有所帮助:)
于 2019-11-22T21:45:40.790 回答
2
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
于 2017-03-08T13:28:34.090 回答
2
我重写了上面的答案并使它变得更好：
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
示例用法：
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
于 2019-04-15T19:36:10.073 回答
2
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
于 2016-12-08T18:51:52.040 回答
0
您可以使用 SQL Server 的“STRING_SPLIT”函数：
STRING_SPLIT（字符串，分隔符）
于 2019-11-27T13:02:05.757 回答
0
这是一个老问题，但如果升级到 SQL Server 2017+ 是可能的，那么基于 JSON 的方法也是一种选择。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, Smithinto["Cleo"," Smith"]) 并使用JSON_VALUE().将存储在
String列中的文本转换为有效的嵌套 JSON 数组 (Cleo, Smithinto ) 并使用显式模式（列定义）[["Cleo"," Smith"]]解析该数组。OPENJSON()
桌子：
SELECT [Value], [String]
INTO Data
FROM (VALUES
(1, 'Cleo, Smith'),
(2, 'John, Smith'),
(3, 'Marian')
) v ([Value], [String])
声明使用JSON_VALUE()：
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
声明使用OPENJSON()：
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
作为附加说明，使用此技术，您可以通过添加适当的 JSON 轻松解析具有两列以上的文本path。
于 2021-04-07T06:36:15.387 回答
0
您可以使用拆分功能。
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
于 2017-01-09T12:30:58.517 回答
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
于 2015-10-16T09:29:30.317 回答
-1
试试下面：
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')
询问：
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
如果您对代码有任何疑问，请告诉我
于 2018-05-11T10:59:56.883 回答
-1
试试这个
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
于 2020-03-27T10:59:52.837 回答
-2
选择字符串。LEFT(STRING,Charindex(",",STRING)-1 as SURE Name RIGHT(STRING,LEN(STRING)-CHARINDEX(","STRING)) AS ]NAME FROM TABLE NAME
于 2021-03-06T05:34:29.963 回答
-2
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;
于 2016-06-12T06:59:08.230 回答