在 SQLServer 中有 LastIndexOf 吗?

我试图从一个字符串中解析出一个值,这个字符串包含了 绳子最后 索引。目前,我正在做一个可怕的黑客攻击,包括反转一个字符串:

SELECT REVERSE(SUBSTRING(REVERSE(DB_NAME()), 1,
CHARINDEX('_', REVERSE(DB_NAME()), 1) - 1))

对我来说,这些代码几乎是不可读的。我刚刚升级到 SQLServer2016,我希望有一个更好的方法。 有吗?

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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()) + '_'))

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

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 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

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
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!

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

Try:

select LEN('tran van abc') + 1 - CHARINDEX(' ', REVERSE('tran van abc'))

So, the last index of ' ' is : 9