如何限制子字符串以在SQL Server中的两个空格之间提取文本？
列
2017-03-09 Today ABC.XYZ Work_In_Progress
输出应为：
ABC.XYZ
我可以从第二个空间拉出但不能限制到第三个空间：
SELECT ID, SUBSTRING(HISTORY, CHARINDEX(' ', HISTORY, CHARINDEX(' ', HISTORY) +1)+1,LEN(HISTORY))
from Test;
答案 0 :(得分：5)
如果您知道＆＃34;位置＆＃34;还有另一种选择。你要去。在这种情况下，第3个字符串
mpl::push_back<强>返回
Declare @YourTable table (ID int,SomeColumn varchar(max))
Insert Into @YourTable values
(1,'2017-03-09 Today ABC.XYZ Work_In_Progress')
Select ID
,SomeValue = Cast('<x>' + replace(SomeColumn,' ','</x><x>')+'</x>' as xml).value('/x[3]','varchar(max)')
From @YourTable
XML安全版
ID SomeValue
1 ABC.XYZ
答案 1 :(得分：3)
以下是使用Charindex和Substring的一种方法。
DECLARE @str VARCHAR(100) = '2017-03-09 Today ABC.XYZ Work_In_Progress'
SELECT Substring(string, scd+1, thrd - scd) AS Result
FROM (SELECT @str string,
Charindex(' ', @str) AS fst) A
CROSS apply(VALUES (Charindex(' ', string, fst + 1))) cs (scd)
CROSS apply(VALUES (Charindex(' ', string, scd + 1))) cs1 (thrd)
答案 2 :(得分：2)
如果您获取NGrams8K的副本，则可以创建名为SubstringBetween8K的此函数：
CREATE FUNCTION dbo.substringBetween8K
(
@string varchar(8000),
@start tinyint,
@stop tinyint,
@delimiter char(1)
)
/*****************************************************************************************
Purpose:
Takes in input string (@string) and returns the text between two instances of a delimiter
(@delimiter); the location of the delimiters is defined by @start and @stop.
For example: if @string = 'xx.yy.zz.abc', @start=1, @stop=3, and @delimiter = '.' the
function will return the text: yy.zz; this is the text between the first and third
instance of "." in the string "xx.yy.zz.abc".
Compatibility:
SQL Server 2008+
Syntax:
--===== Autonomous use
SELECT sb.token, sb.position, sb.tokenLength
FROM dbo.substringBetween8K(@string, @start, @stop, @delimiter); sb;
--===== Use against a table
SELECT sb.token, sb.position, sb.tokenLength
FROM SomeTable st
CROSS APPLY dbo.substringBetween8K(st.SomeColumn1, 1, 2, '.') sb;
Parameters:
@string = varchar(8000); Input string to parse
@delimiter = char(1); this is the delimiter use to determine where the output starts/ends
@start = tinyint; the first instance of @delimiter to search for; this is where the
output should start. When @start is 0 then the function will return
everything from the beginning of @string until @end.
@stop = tinyint; the last instance of @delimiter to search for; this is where the
output should end. When @end is 0 then the function will return everything
from @start until the end of the string.
Return Types:
Inline Table Valued Function returns:
token = varchar(8000); the substring between the two instances of @delimiter defined by
@start and @stop
position = smallint; the location of where the substring begins
tokenlength = length of the return token
---------------------------------------------------------------------------------------
Developer Notes:
1. Requires NGrams8K. The code for NGrams8K can be found here:
http://www.sqlservercentral.com/articles/Tally+Table/142316/
2. This function is what is referred to as an "inline" scalar UDF." Technically it's an
inline table valued function (iTVF) but performs the same task as a scalar valued user
defined function (UDF); the difference is that it requires the APPLY table operator
to accept column values as a parameter. For more about "inline" scalar UDFs see this
article by SQL MVP Jeff Moden: http://www.sqlservercentral.com/articles/T-SQL/91724/
and for more about how to use APPLY see the this article by SQL MVP Paul White:
http://www.sqlservercentral.com/articles/APPLY/69953/.
Note the above syntax example and usage examples below to better understand how to
use the function. Although the function is slightly more complicated to use than a
scalar UDF it will yield notably better performance for many reasons. For example,
unlike a scalar UDFs or multi-line table valued functions, the inline scalar UDF does
not restrict the query optimizer's ability generate a parallel query execution plan.
3. dbo.substringBetween8K is deterministic; for more about deterministic and
nondeterministic functions see https://msdn.microsoft.com/en-us/library/ms178091.aspx
Examples:
DECLARE @string varchar(8000) = '123.ABC456.333.222.3333XXX.$$$'
-- beginning of string to 2nd delimiter, 2nd delimiter to end of the string
SELECT '0, 2', * FROM dbo.substringBetween8K(@string,0,2, '.') UNION ALL
SELECT '2, 0', * FROM dbo.substringBetween8K(@string,2,0, '.') UNION ALL
-- Between the 1st & 2nd, then 2nd & 5th delimiters
SELECT '1, 2', * FROM dbo.substringBetween8K(@string,1,2, '.') UNION ALL
SELECT '2, 5', * FROM dbo.substringBetween8K(@string,2,5, '.') UNION ALL
-- dealing with NULLS, delimiters that don't exist and when @first = @last
SELECT '2, 10', * FROM dbo.substringBetween8K(@string,2,10,'.') UNION ALL
SELECT '1, NULL',* FROM dbo.substringBetween8K(@string,1,NULL,'.') UNION ALL
SELECT '1, 1', * FROM dbo.substringBetween8K(@string,1,NULL,'.');
---------------------------------------------------------------------------------------
Revision History:
Rev 00 - 20160720 - Initial Creation - Alan Burstein
****************************************************************************************/
RETURNS TABLE WITH SCHEMABINDING AS RETURN
WITH
chars AS
(
SELECT instance = 0, position = 0 WHERE @start = 0
UNION ALL
SELECT ROW_NUMBER() OVER (ORDER BY position), position
FROM dbo.NGrams8k(@string,1)
WHERE token = @delimiter
UNION ALL
SELECT -1, DATALENGTH(@string)+1 WHERE @stop = 0
)
SELECT token =
SUBSTRING
(
@string,
MIN(position)+1,
NULLIF(MAX(position),MIN(position)) - MIN(position)-1
),
position = CAST(
CASE WHEN NULLIF(MAX(position),MIN(position)) - MIN(position)-1 > 0
THEN MIN(position)+1 END AS smallint),
tokenLength = CAST(NULLIF(MAX(position),MIN(position)) - MIN(position)-1 AS smallint)
FROM chars
WHERE instance IN (@start, NULLIF(@stop,0), -1);
要获得第3和第4空间之间的文本，您可以这样做：
DECLARE @txt varchar(100) = '2017-03-09 Today ABC.XYZ Work_In_Progress';
SELECT token
FROM dbo.substringBetween8K(@txt, 2, 3, ' ');
退货： ABC.XYZ
要对表使用，您可以这样做：
DECLARE @table TABLE(txt varchar(100));
INSERT @table VALUES ('2017-03-09 Today ABC.XYZ Work_In_Progress'),
('2011-05-09 Today 123.999 Work_NotIn_Progress');
SELECT txt, token
FROM @table
CROSS APPLY dbo.substringBetween8K(txt, 2, 3, ' ');
<强>返回：
txt token
-------------------------------------------------- -------
2017-03-09 Today ABC.XYZ Work_In_Progress ABC.XYZ
2011-05-09 Today 123.999 Work_NotIn_Progress 123.999
答案 3 :(得分：1)
您可以轻松使用LEFT和您已创建的逻辑。
DECLARE @HISTORY VARCHAR(50) = '2017-03-09 Today ABC.XYZ Work_In_Progress'
SELECT LEFT(SUBSTRING(@HISTORY, CHARINDEX(' ', @HISTORY, CHARINDEX(' ', @HISTORY) +1)+1,LEN(@HISTORY))
, CHARINDEX(' ' , SUBSTRING(@HISTORY, CHARINDEX(' ', @HISTORY, CHARINDEX(' ', @HISTORY) +1)+1,LEN(@HISTORY))) - 1)