May 18, 2017 · I'm trying to extract a code that varies in length that exists after the first two underscores and before the third underscore in a field in a table.

Jan 2, 2020 · I have a column of data that contains varchar data and I've been trying to return only the section contained with brackets using CHARINDEX and SUBSTRING : column data: 'this is an (example)' I use the following code to return 'example' from the above data:

"Access Database \ Server \ Windows" i want to select whatever is there after the first '\' in one query and in second query i want to select the text after second '\' between the '\' the length of the text keeps changing. I have tried: SELECT Somecolumn=Substring(column1, 0, Charindex('\', column1, 1)) FROM dbo.sometable The result:

Jun 13, 2013 · substring in sql server 2008. 2. Substring and Charindex. 2. SQL Charindex and SUbstring. 0.

Jun 22, 2009 · This answer uses MS SQL Server 2008 (I don't have access to MS SQL Server 2000), but the way I see it according to the OP are 3 situations to take into consideration. From what I've tried no answer here covers all 3 of them: Return the …

Oct 13, 2015 · SELECT * , SUBSTRING(name, 1, ( CHARINDEX('z_a', NAME) - 1 )) AS nameNew FROM myTable This is for the first step, trimming the string, for the 2nd step (making it easier to read) I don't know how to target the digit and place an _. Any help would be appreciated. Using sql server 2012. edit: First of all thank you for your time and solutions.

Once you have one of the split strings from here,you can do it in a set based way like this... declare @string varchar(max) set @string='C:\Program Files\Microsoft SQL Server\MSSQL\DATA\AdventureWorks_Data.mdf' ;with cte as (select *,row_number() over (order by (select null)) as rownum from [dbo].[SplitStrings_Numbers](@string,'\') ) select top 1 item from cte order by rownum desc **Output ...

Nov 20, 2012 · You can use a combiation of LEFT and CHARINDEX to find the index of the first space, and then grab everything to the left of that. SELECT LEFT(YourColumn, charindex(' ', YourColumn) - 1) And in case any of your columns don't have a space in them:

Apr 15, 2014 · You might want to create split function. CREATE FUNCTION [dbo].[fnSplitString] ( @string NVARCHAR(MAX), @delimiter CHAR(1) ) RETURNS @output TABLE(splitdata NVARCHAR(MAX) ) BEGIN DECLARE @start INT, @end INT SELECT @start = 1, @end = CHARINDEX(@delimiter, @string) WHILE @start < LEN(@string) + 1 BEGIN IF @end = 0 SET @end = LEN(@string) + 1 INSERT INTO @output (splitdata) VALUES(SUBSTRING ...

Jan 4, 2012 · there is a bug in this method. this method only works if you are sure you have n occurrence. if you try to find 3rd occurrence and you have 1 occurrence in first charindex you get x but then when you search `x+1' you get 0 but now you search from 1 and again you get the x. so if you have just one occurrence you get its location or 1 as output.