How to get text between two words/characters

Question

How can I get some text between two known words?

Sub-select example:

(SELECT TOP(1)Note
FROM Clients
WHERE (ID=@ID) AND (Note IS NOT NULL) AND (Note NOT         


        

Answer 1

CREATE FUNCTION [dbo].[fnSplitString] 
( 
    @string NVARCHAR(MAX),
    @id int , ,
    @sword varchar(30) ,
    @delimiter varchar(30) 
) 
RETURNS @output TABLE(splitdata NVARCHAR(MAX) , id int
) 
BEGIN 
    DECLARE @start INT, @end INT 
    declare @start1 varchar(100)
    set @start1 = @sword

    declare @end1 varchar(100)
    set @end1 =  @delimiter
    SELECT @start = 1, @end = CHARINDEX(@delimiter, @string) + Len(@delimiter) - 1
    WHILE @start < LEN(@string) + 1 BEGIN 
        IF @end = 0  
            SET @end = LEN(@string) + 1

        INSERT INTO @output (splitdata , id)  
        VALUES(substring( LEFT((SUBSTRING(@string, @start, @end - @start)), charindex(@end1, (SUBSTRING(@string, @start, @end - @start)), CHARINDEX(@start1, (SUBSTRING(@string, @start, @end - @start))))-1), CHARINDEX(@start1, (SUBSTRING(@string, @start, @end - @start))) + len(@start1), LEN((SUBSTRING(@string, @start, @end - @start))))  ,@id ) 

        SET @start = @end + 1 
        SET @end = CHARINDEX(@delimiter, @string, @start)+ Len(@delimiter) - 1
         END 
    RETURN 
END

Select note , fnSplitString (note , id ,'startword', 'endword') from clients 

Answer 2

Guys I found my own solution:

DECLARE @S VARCHAR(MAX) = 'Blablabla bla bla blabla bla bla bla (21.08.2015) "some text" word bla bla Blablabla bla bla blabla bla bla bla (25.08.2015) "another text" word bla bla Blablabla bla bla blabla bla bla bla (28.08.2015) "TEXT I NEED TO GET" word bla bla blabla bla bla.'

SELECT LEFT(REVERSE(LEFT(REVERSE(@S), CHARINDEX(')', REVERSE(@S)) - 2))+'word', CHARINDEX('word',REVERSE(LEFT(REVERSE(@S), CHARINDEX(')', REVERSE(@S)) - 2))+'word')-1) AS respond

This returns exactly:

"TEXT I NEED TO GET"

Thank you all for leading me to this point.

Answer 3

Maybe something like this:

You try to find the last occurence. This is the first in a reverted string...

--This is a mockup of your table to hold your strings:

DECLARE @tbl TABLE(YourLine VARCHAR(100));
INSERT INTO @tbl VALUES
 ('Blablabla bla bla blabla bla bla bla (21.08.2015) "some text" word bla bla')
,('Blablabla bla bla blabla bla bla bla (25.08.2015) "another text" word bla bla') 
,('Blablabla bla bla blabla bla bla bla (28.08.2015) "TEXT I NEED TO GET" word bla bla blabla bla bla.');

--Here I specify the word you are searching for

DECLARE @SearchWord VARCHAR(100)='word';

--This is the query

SELECT REVERSE(C.CutOut) AS YourSnippet
FROM @tbl AS tbl

--CROSS APPLY works row-wise: Take the current line and revert it!

CROSS APPLY(SELECT REVERSE(tbl.YourLine) AS RevLine) AS A

--Find the first occurence of ) and the first occurence of the reverted searchword!

CROSS APPLY(SELECT CHARINDEX(')',A.RevLine) AS PosParanthesis
                  ,CHARINDEX(REVERSE(@SearchWord),A.RevLine) AS PosWord) AS B 

--Cut the string at the position found to the length calculated

CROSS APPLY(SELECT SUBSTRING(A.RevLine,B.PosWord + LEN(@SearchWord)+1,B.PosParanthesis-B.PosWord-LEN(@SearchWord)-1) AS CutOut) AS C

The result of the las CROSS APPLY is your target string, but reverse. Therefore a last reverse in the main SELECT...

Answer 4

declare @start varchar(100)
set @start = 'startword'

declare @end varchar(100)
set @end = 'endword'

SELECT TOP(1) Note , substring( LEFT(note, charindex(@end, note, CHARINDEX(@start, note))-1), CHARINDEX(@start, note) + len(@start), LEN(note))   result
FROM Clients
WHERE (ID=@ID) AND (Note IS NOT NULL)