Is there a LastIndexOf in SQL Server? - sql

Question

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.
Is there?

Answer 1

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 left():
select left(db_name(), len(db_name()) - charindex('_', reverse(db_name()) + '_'))

Answer 2

Wrote 2 functions, 1 to return LastIndexOf for the selected character.
CREATE FUNCTION dbo.LastIndexOf(#source nvarchar(80), #pattern char)
RETURNS int
BEGIN
RETURN (LEN(#source)) - CHARINDEX(#pattern, REVERSE(#source))
END;
GO
and 1 to return a string before this LastIndexOf. Maybe it will be useful to someone.
CREATE FUNCTION dbo.StringBeforeLastIndex(#source nvarchar(80), #pattern char)
RETURNS nvarchar(80)
BEGIN
DECLARE #lastIndex int
SET #lastIndex = (LEN(#source)) - CHARINDEX(#pattern, REVERSE(#source))
RETURN SUBSTRING(#source, 0, #lastindex + 1)
-- +1 because index starts at 0, but length at 1, so to get up to 11th index, we need LENGTH 11+1=12
END;
GO

Answer 3

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

Answer 4

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

Answer 5

CREATE FUNCTION dbo.LastIndexOf(#text NTEXT, #delimiter NTEXT)
RETURNS INT
AS
BEGIN
IF (#text IS NULL) RETURN NULL;
IF (#delimiter IS NULL) RETURN NULL;
DECLARE #Text2 AS NVARCHAR(MAX) = #text;
DECLARE #Delimiter2 AS NVARCHAR(MAX) = #delimiter;
DECLARE #Index AS INT = CHARINDEX(REVERSE(#Delimiter2), REVERSE(#Text2));
IF (#Index < 1) RETURN 0;
DECLARE #ContentLength AS INT = (LEN('|' + #Text2 + '|') - 2);
DECLARE #DelimiterLength AS INT = (LEN('|' + #Delimiter2 + '|') - 2);
DECLARE #Result AS INT = (#ContentLength - #Index - #DelimiterLength + 2);
RETURN #Result;
END
Allows for multi-character delimiters like ", " (comma space).
Returns 0 if the delimiter is not found.
Takes a NTEXT for comfort reasons as NVARCHAR(MAX)s are implicitely cast into NTEXT but not vice-versa.
Handles delimiters with leading or tailing space correctly!

Answer 6

Try:
select LEN('tran van abc') + 1 - CHARINDEX(' ', REVERSE('tran van abc'))
So, the last index of ' ' is : 9

Answer 7

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 REVERSE function.
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 Occurence property.
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

Answer 8

Try this.
drop table #temp
declare #brokername1 nvarchar(max)='indiabullssecurities,canmoney,indianivesh,acumencapitalmarket,sharekhan,edelweisscapital';
Create Table #temp
(
ID int identity(1,1) not null,
value varchar(100) not null
)
INSERT INTO #temp(value) SELECT value from STRING_SPLIT(#brokername1,',')
declare #id int;
set #id=(select max(id) from #temp)
--print #id
declare #results varchar(500)
select #results = coalesce(#results + ',', '') + convert(varchar(12),value)
from #temp where id<#id
order by id
print #results