SQL Server 2005:charindex starting from the end

问题

I have a string 'some.file.name',I want to grab 'some.file'.

To do that,I need to find the last occurrence of '.' in a string.

My solution is :

 declare @someStr varchar(20)

 declare @reversedStr varchar(20)

 declare @index int

 set @someStr = '001.002.003'

 set @reversedStr = reverse(@someStr)

 set @index = len(@someStr) - charindex('.',@reversedStr)

 select left(@someStr,@index)

Well,isn't it too complicated?I was just intented to using 'some.file' in a where-clause.

Anyone has a good idea?

回答1:

What do you need to do with it?? Do you need to grab the characters after the last occurence of a given delimiter?

If so: reverse the string and search using the normal CHARINDEX:

declare @test varchar(100)
set @test = 'some.file.name'

declare @reversed varchar(100)
set @reversed = REVERSE(@test)

select 
    REVERSE(SUBSTRING(@reversed, CHARINDEX('.', @reversed)+1, 100))

You'll get back "some.file" - the characters up to the last "." in the original file name.

There's no "LASTCHARINDEX" or anything like that in SQL Server directly. What you might consider doing in SQL Server 2005 and up is great a .NET extension library and deploy it as an assembly into SQL Server - T-SQL is not very strong with string manipulation, whereas .NET really is.

回答2:

This will also work:

DECLARE
     @test     VARCHAR(100)

SET @test = 'some.file.name'

SELECT
     LEFT(@test, LEN(@test) - CHARINDEX('.', REVERSE(@test)))

回答3:

A very simple way is:

select right(@str,charindex('.',reverse(@str))-1)

回答4:

Take one ')'

declare @test varchar(100)
set @test = 'some.file.name'
select left(@test,charindex('.',@test)+charindex('.',@test)-1)

回答5:

CREATE FUNCTION [dbo].[Instr] (
  -------------------------------------------------------------------------------------------------
  -- Name:     [dbo].[Instr]
  -- Purpose:  Find The Nth Value Within A String
  -------------------------------------------------------------------------------------------------
  -- Revisions:
  --   25-FEB-2011 - HESSR - Initial Revision
  -------------------------------------------------------------------------------------------------
  -- Parameters:
  --    1) @in_FindString - NVARCHAR(MAX) - INPUT - Input Find String
  --    2) @in_String - NVARCHAR(MAX) - INPUT - Input String
  --    3) @in_StartPos - SMALLINT - INPUT - Position In The String To Start Looking From
  --          (If Start Position Is Negative, Search Begins At The End Of The String)
  --          (Negative 1 Starts At End Position 1, Negative 3 Starts At End Position Minus 2)
  --    4) @in_Nth - SMALLINT - INPUT - Nth Occurrence To Find The Location For
  -------------------------------------------------------------------------------------------------
  -- Returns: SMALLINT - Position Of String Segment (Not Found = 0)
  -------------------------------------------------------------------------------------------------
  @in_FindString             NVARCHAR(MAX),
  @in_String                 NVARCHAR(MAX),
  @in_StartPos               SMALLINT            = NULL,
  @in_Nth                    SMALLINT            = NULL
  ) 
  RETURNS                    SMALLINT
AS
BEGIN

  DECLARE @loc_FindString NVARCHAR(MAX);
  DECLARE @loc_String NVARCHAR(MAX);
  DECLARE @loc_Position SMALLINT;
  DECLARE @loc_StartPos SMALLINT;
  DECLARE @loc_Nth SMALLINT;
  DECLARE @loc_Idx SMALLINT;
  DECLARE @loc_FindLength SMALLINT;
  DECLARE @loc_Length SMALLINT;

  SET @loc_FindString = @in_FindString;
  SET @loc_String = @in_String;

  SET @loc_Nth = ISNULL(ABS(@in_Nth), 1);
  SET @loc_FindLength = LEN(@loc_FindString+N'.') - 1;
  SET @loc_Length = LEN(@loc_String+N'.') - 1;

  SET @loc_StartPos = ISNULL(@in_StartPos, 1);
  SET @loc_Idx = 0;

  IF (@loc_StartPos = ABS(@loc_StartPos))
    BEGIN
      WHILE (@loc_Idx < @loc_Nth)
        BEGIN
          SET @loc_Position = CHARINDEX(@loc_FindString,@loc_String,@loc_StartPos);
          IF (@loc_Position > 0)
            SET @loc_StartPos = @loc_Position + @loc_FindLength
          ELSE
            SET @loc_Idx = @loc_Nth;
          SET @loc_Idx = @loc_Idx + 1;
        END;
    END
  ELSE
    BEGIN
      SET @loc_StartPos = ABS(@loc_StartPos);
      SET @loc_FindString = REVERSE(@in_FindString);
      SET @loc_String = REVERSE(@in_String);
      WHILE (@loc_Idx < @loc_Nth)
        BEGIN
          SET @loc_Position = CHARINDEX(@loc_FindString,@loc_String,@loc_StartPos);
          IF (@loc_Position > 0)
            SET @loc_StartPos = @loc_Position + @loc_FindLength
          ELSE
            SET @loc_Idx = @loc_Nth;
          SET @loc_Idx = @loc_Idx + 1;
        END;
      IF (@loc_Position > 0)
        SET @loc_Position = @loc_Length - @loc_Position + (1 - @loc_FindLength) + 1;
    END;

  RETURN (@loc_Position);

END;
GO

回答6:

Here is a shorter version

DECLARE @someStr varchar(20)
set @someStr = '001.002.003'

SELECT REVERSE(Substring(REVERSE(@someStr),CHARINDEX('.', REVERSE(@someStr))+1,20))

来源：https://stackoverflow.com/questions/1851650/sql-server-2005charindex-starting-from-the-end

标签

sql

sql-server

string

sql-server-2005