私はこのようなテーブルを持っています
Value String
-------------------
1 Cleo, Smith
カンマ区切りの文字列を2つの列に分割したい
Value Name Surname
-------------------
1 Cleo Smith
固定された追加の列が2つだけ必要です
1123770 次
38 に答える
147
あなたの目的は次のクエリを使用して解決することができます-
Select Value , Substring(FullName, 1,Charindex(',', FullName)-1) as Name,
Substring(FullName, Charindex(',', FullName)+1, LEN(FullName)) as Surname
from Table1
SQL Serverには既製のSplit関数がないため、ユーザー定義関数を作成する必要があります。
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 に答える
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
また、以下のリンクを参照してください
http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html
于 2013-02-27T09:30:02.983 に答える
47
xmlベースの答えはシンプルでクリーンです
これを参照してください
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 に答える
42
かっこいいと思います
SELECT value,
PARSENAME(REPLACE(String,',','.'),2) 'Name' ,
PARSENAME(REPLACE(String,',','.'),1) 'Surname'
FROM table WITH (NOLOCK)
于 2014-06-20T05:39:27.113 に答える
30
CROSSAPPLYを使用
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 に答える
23
これを解決する方法は複数あり、すでに多くの異なる方法が提案されています。最も簡単なのは、 LEFT/SUBSTRINGおよびその他の文字列関数を使用して目的の結果を達成することです。
サンプルデータ
DECLARE @tbl1 TABLE (Value INT,String VARCHAR(MAX))
INSERT INTO @tbl1 VALUES(1,'Cleo, Smith');
INSERT INTO @tbl1 VALUES(2,'John, Mathew');
次のような文字列関数を使用するLEFT
SELECT
Value,
LEFT(String,CHARINDEX(',',String)-1) as Fname,
LTRIM(RIGHT(String,LEN(String) - CHARINDEX(',',String) )) AS Lname
FROM @tbl1
文字列に2つ以上のアイテムがある場合、このアプローチは失敗します。このようなシナリオでは、スプリッターを使用してから、文字列を使用またはPIVOT変換して文字列アイテムを取得できます。ベースのソリューションは、ソリューション内のaadsとbvrによって詳細に説明されています。XML.nodesXML
スプリッターを使用するこの質問への回答は、すべてWHILE分割には非効率的です。このパフォーマンス比較を確認してください。周りの最高のスプリッターの1つは、DelimitedSplit8KJeffModenによって作成されたものです。あなたはここでそれについてもっと読むことができます
スプリッター付き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
出力
Value Fname Lname
1 Cleo Smith
2 John Mathew
DelimitedSplit8Kジェフ・モデン
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 に答える
22
SQL Server 2016では、string_splitを使用してこれを実現できます。
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 に答える
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 に答える
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 に答える
15
これを試してください（''のインスタンスを'、または使用したい区切り文字に変更してください）
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
使用例：
SELECT dbo.Wordparser(COLUMN, 1),
dbo.Wordparser(COLUMN, 2),
dbo.Wordparser(COLUMN, 3)
FROM TABLE
于 2016-12-27T21:16:57.063 に答える
13
この記事で説明されているように、PARSENAMEはこの例で使用する優れた関数だと思います：http ：//www.sqlshack.com/parsing-and-rotating-delimited-data-in-sql-server-2012/
PARSENAME関数は、4つの部分からなるオブジェクト名を解析するように論理的に設計されています。PARSENAMEの優れている点は、SQL Serverの4つの部分からなるオブジェクト名だけを解析するだけでなく、ドットで区切られた関数または文字列データを解析することです。
最初のパラメーターは解析するオブジェクトであり、2番目のパラメーターは返すオブジェクトピースの整数値です。この記事では、区切られたデータ（会社の電話番号）の解析とローテーションについて説明していますが、名前/名前のデータの解析にも使用できます。
例：
USE COMPANY;
SELECT PARSENAME('Whatever.you.want.parsed',3) AS 'ReturnValue';
この記事では、「replaceChars」と呼ばれる共通テーブル式（CTE）を使用して、区切り文字で置き換えられた値に対してPARSENAMEを実行する方法についても説明しています。CTEは、一時的なビューまたは結果セットを返すのに役立ちます。
その後、UNPIVOT関数を使用して、一部の列を行に変換しました。データの不整合をクリーンアップするためにSUBSTRING関数とCHARINDEX関数が使用され、最後にLAG関数（SQL Server 2012の新機能）が使用され、以前のレコードを参照できるようになりました。
于 2015-12-23T09:51:14.277 に答える
10
このように関数を作成できます
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
次に、SELECTステートメントを使用してCSV値をそれぞれの列に分割できます
于 2014-03-26T10:42:15.433 に答える
9
私は次の機能があなたのために働くと思います：
最初にSQLで関数を作成する必要があります。このような
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
この関数は、次のように呼び出すことができます。
select * from fn_split('1,24,5',',')
実装：
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
結果は次のようになります。
于 2017-08-28T07:32:48.387 に答える
8
互換性レベル130でのみ使用できるテーブル値関数STRING_SPLITを使用できます。データベースの互換性レベルが130未満の場合、SQLServerはSTRING_SPLIT関数を見つけて実行できません。次のコマンドを使用して、データベースの互換性レベルを変更できます。
ALTER DATABASE DatabaseName SET COMPATIBILITY_LEVEL = 130
構文
SELECT * FROM STRING_SPLIT ( string, separator )
于 2017-07-09T08:54:06.287 に答える
7
Parsename（）関数を使用する
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
結果
Name Family
----- ------
Aria Karimi
Bab Karimi
Joe Karimi
于 2015-03-18T04:36:22.143 に答える
7
これを試して：
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 に答える
6
この関数は最も高速です。
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
使用例：
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 に答える
5
私は同様の問題に遭遇しましたが、複雑な問題でした。これがその問題に関して私が見つけた最初のスレッドであるため、私は自分の発見を投稿することにしました。私はそれが単純な問題の複雑な解決策であることを知っていますが、より複雑な解決策を探しているこのスレッドに行く他の人々を助けることができることを願っています。5つの数値（列名：levelsFeed）を含む文字列を分割し、各数値を別々の列に表示する必要がありました。例：8,1,2,2,2は次のように表示されます：
1 2 3 4 5
-------------
8 1 2 2 2
解決策1：XML関数を使用する：これまでで最も遅い解決策のためのこの解決策
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)
解決策2：分割機能とピボットを使用します。（split関数は、文字列を列名Dataの行に分割します）
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
解決策3：文字列操作関数を使用する-解決策2よりもわずかに速い
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
LevelsFeedには5つの文字列値が含まれているため、最初の文字列にsubstring関数を使用する必要がありました。
私のソリューションが、より複雑な列への分割メソッドを探しているこのスレッドに到達した他の人に役立つことを願っています
于 2015-07-14T09:42:52.287 に答える
5
ストリング関数の使用:)
select Value,
substring(String,1,instr(String," ") -1) Fname,
substring(String,instr(String,",") +1) Sname
from tablename;
2つの関数を使用しました
。1 。==>位置substring(string, position, length) から長さ
2までの文字列を返します。 instr(string,pattern)==>パターンの位置を返します。
サブ文字列に長さ引数を指定しない場合、文字列の終わりまで戻ります
于 2015-09-30T08:36:00.330 に答える
4
これは私のために働いた
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 に答える
3
区切り文字列を解析するSQLユーザー定義関数の解決策が役立つ場合があります（Code Projectから）。
これは、このページのコード部分です。
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 に答える
3
mytable：
Value ColOne
--------------------
1 Cleo, Smith
列が多すぎない場合は、次のように機能するはずです
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', '')
結果：
Value ColOne ColTwo
--------------------
1 Cleo Smith
于 2014-12-29T22:30:56.443 に答える
3
それはとても簡単です、あなたは以下のクエリによってそれを取ることができます：
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 に答える
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 に答える
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 に答える
2
上記のようにPARSENAMEを使用すると、ピリオドのある名前がnullになることがわかりました。
したがって、名前にイニシャルまたはタイトルの後にドットが続く場合、それらはNULLを返します。
私はこれが私のために働いたことを見つけました：
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 に答える
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 に答える
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 に答える
2
私は上記の答えを書き直して、それをより良くしました：
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
使用例：
SELECT dbo.CSVParser(COLUMN, 1),
dbo.CSVParser(COLUMN, 2),
dbo.CSVParser(COLUMN, 3)
FROM TABLE
于 2019-04-15T19:36:10.073 に答える
2
質問は簡単ですが、問題は熱いです:)
そこで、より一般的な方法で結果をピボットするstring_split（）のラッパーを作成します。これは、値（nn、value1、value2、...、value50）を返すテーブル関数です。ほとんどのCSV行には十分です。さらに値がある場合は、次の行に折り返されます-nnは行番号を示します。3番目のパラメーター@columnCnt=[yourNumber]を設定して、特定の位置でラップします。
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)
願わくば、それが役立つでしょう:)
于 2019-11-22T21:45:40.790 に答える
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 に答える
0
分割機能が利用できます。
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 に答える
0
SQLServerの「STRING_SPLIT」関数を使用できます。
STRING_SPLIT（文字列、区切り文字）
于 2019-11-27T13:02:05.757 に答える
0
これは古い質問ですが、SQL Server 2017以降へのアップグレードが可能な場合は、JSONベースのアプローチもオプションです。アイデアは、適切な変換を行うことです。
列に格納されているテキスト
Stringを有効なJSON配列（Cleo, Smithに["Cleo"," Smith"]）に変換し、この配列を。で解析しJSON_VALUE()ます。列に格納されているテキスト
Stringを有効なネストされたJSON配列（Cleo, Smithに）に変換し、この配列を明示的なスキーマ（列定義）[["Cleo"," Smith"]]で解析します。OPENJSON()
テーブル：
SELECT [Value], [String]
INTO Data
FROM (VALUES
(1, 'Cleo, Smith'),
(2, 'John, Smith'),
(3, 'Marian')
) v ([Value], [String])
を使用したステートメント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
を使用したステートメント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
結果：
Value Name Surname
---------------------
1 Cleo Smith
2 John Smith
3 Marian
補足として、この手法では、適切なJSONを追加することで、3つ以上の列を持つテキストを簡単に解析できますpath。
于 2021-04-07T06:36:15.387 に答える
-1
以下をお試しください：
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')
クエリ：
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
コードを理解する上で疑問がある場合はお知らせください
于 2018-05-11T10:59:56.883 に答える
-1
これを試してみてください
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 に答える
-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 に答える
-2
STRINGを選択します。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 に答える