Is there a LastIndexOf in SQL Server?
I am trying to parse out a value from a string that involves getting the last index of a string. Currently, I am doing a horrible hack that involves reversing a string:
SELECT REVERSE(SUBSTRING(REVERSE(DB_NAME()), 1, CHARINDEX('_', REVERSE(DB_NAME()), 1) - 1))
To me this code is nearly unreadable. I just upgraded to SQL Server 2016 and I hoping there is a better way.
4 Solutions collect form web for “Is there a LastIndexOf in SQL Server?”
If you want everything after the last
_, then use:
select right(db_name(), charindex('_', reverse(db_name()) + '_') - 1)
If you want everything before, then use
select left(db_name(), len(db_name()) - charindex('_', reverse(db_name()) + '_'))
No, SQL server doesnt have LastIndexOf.
This are the available string functions
But you can always can create your own function
CREATE FUNCTION dbo.LastIndexOf(@source text, @pattern char) RETURNS AS BEGIN DECLARE @ret text; SELECT into @ret REVERSE(SUBSTRING(REVERSE(@source), 1, CHARINDEX(@pattern, REVERSE(@source), 1) - 1)) RETURN @ret; END; GO
I came across this thread while searching for a solution to my similar problem which had the exact same requirement but was for a different kind of database that was lacking the
In my case this was for a OpenEdge (Progress) database, which has a slightly different syntax. This made the
INSTR function available to me that most Oracle typed databases offer.
So I came up with the following code:
SELECT INSTR(foo.filepath, '/',1, LENGTH(foo.filepath) - LENGTH( REPLACE( foo.filepath, '/', ''))) AS IndexOfLastSlash FROM foo
However, for my specific situation (being the OpenEdge (Progress) database) this did not result into the desired behaviour because replacing the character with an empty char gave the same length as the original string. This doesn’t make much sense to me but I was able to bypass the problem with the code below:
SELECT INSTR(foo.filepath, '/',1, LENGTH( REPLACE( foo.filepath, '/', 'XX')) - LENGTH(foo.filepath)) AS IndexOfLastSlash FROM foo
Now I understand that this code won’t solve the problem for T-SQL because there is no alternative to the
INSTR function that offers the
Just to be thorough I’ll add the code needed to create this scalar function so it can be used the same way like I did in the above examples. And will do exactly what the OP wanted, serve as a LastIndexOf method for SQL Server.
-- Drop the function if it already exists IF OBJECT_ID('INSTR', 'FN') IS NOT NULL DROP FUNCTION INSTR GO -- User-defined function to implement Oracle INSTR in SQL Server CREATE FUNCTION INSTR (@str VARCHAR(8000), @substr VARCHAR(255), @start INT, @occurrence INT) RETURNS INT AS BEGIN DECLARE @found INT = @occurrence, @pos INT = @start; WHILE 1=1 BEGIN -- Find the next occurrence SET @pos = CHARINDEX(@substr, @str, @pos); -- Nothing found IF @pos IS NULL OR @pos = 0 RETURN @pos; -- The required occurrence found IF @found = 1 BREAK; -- Prepare to find another one occurrence SET @found = @found - 1; SET @pos = @pos + 1; END RETURN @pos; END GO
To avoid the obvious, when the
REVERSE function is available you do not need to create this scalar function and you can just get the required result like this:
SELECT LEN(foo.filepath) - CHARINDEX('\', REVERSE(foo.filepath))+1 AS LastIndexOfSlash FROM foo
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:** AdventureWorks_Data.mdf