I got the following entry in my database:
images/test.jpg I want to trim the entry so I get: test
So basically, I want everything after / and before .
How can I solve it?
620 Answers
use the following function
left(@test, charindex('/', @test) - 1) If you want to get this out of your table using SQL, take a look at the following functions that will help you: SUBSTRING and CHARINDEX. You can use those to trim your entries.
A possible query will look like this (where col is the name of the column that contains your image directories:
SELECT SUBSTRING(col, LEN(SUBSTRING(col, 0, LEN(col) - CHARINDEX ('/', col))) + 1, LEN(col) - LEN(SUBSTRING(col, 0, LEN(col) - CHARINDEX ('/', col))) - LEN(SUBSTRING( col, CHARINDEX ('.', col), LEN(col)))); Bit of an ugly beast. It also depends on the standard format of 'dir/name.ext'.
Edit:
This one (inspired by praveen) is more generic and deals with extensions of different length:
SELECT SUBSTRING(col, LEN(LEFT(col, CHARINDEX ('/', col))) + 1, LEN(col) - LEN(LEFT(col, CHARINDEX ('/', col))) - LEN(RIGHT(col, LEN(col) - CHARINDEX ('.', col))) - 1); Before
SELECT SUBSTRING(ParentBGBU,0,CHARINDEX('/',ParentBGBU,0)) FROM dbo.tblHCMMaster; After
SELECT SUBSTRING(ParentBGBU,CHARINDEX('-',ParentBGBU)+1,LEN(ParentBGBU)) FROM dbo.tblHCMMaster ----select characters before / including /
select SUBSTRING ('abcde/wxyz',0,CHARINDEX('/','abcde/wxyz')+1) --select characters after / including /
select SUBSTRING('abcde/wxyz',CHARINDEX('/','abcde/wxyz'),LEN('abcde/wxyz')) declare @T table ( Col varchar(20) ) insert into @T Select 'images/test1.jpg' union all Select 'images/test2.png' union all Select 'images/test3.jpg' union all Select 'images/test4.jpeg' union all Select 'images/test5.jpeg' Select substring( LEFT(Col,charindex('.',Col)-1),charindex('/',Col)+1,len(LEFT(Col,charindex('.',Col)-1))-1 ) from @T I have made a method which is much more general :
so :
DECLARE @a NVARCHAR(MAX)='images/test.jpg'; --Touch here DECLARE @keysValueToSearch NVARCHAR(4000) = '/' DECLARE @untilThisCharAppears NVARCHAR(4000) = '.' DECLARE @keysValueToSearchPattern NVARCHAR(4000) = '%' + @keysValueToSearch + '%' --Nothing to touch here SELECT SUBSTRING( @a, PATINDEX(@keysValueToSearchPattern, @a) + LEN(@keysValueToSearch), CHARINDEX( @untilThisCharAppears, @a, PATINDEX(@keysValueToSearchPattern, @a) + LEN(@keysValueToSearch) ) -(PATINDEX(@keysValueToSearchPattern, @a) + LEN(@keysValueToSearch)) ) I just did this in one of my reports and it was very simple.
Try this:
=MID(Fields!.Value,8,4) Note: This worked for me because the value I was trying to get was a constant not sure it what you are trying to get is a constant as well.
1I know this has been a while.. but here is an idea
declare @test varchar(25) = 'images/test.jpg' select @test as column_name , parsename(replace(@test,'/','.'),1) as jpg ,parsename(replace(@test,'/','.'),2) as test ,parsename(replace(@test,'/','.'),3) as images SELECT Substring('', 1, ( Charindex('@', '') - 1 )) Before, RIGHT('', ( Charindex('@', '') + 1 )) After I found Royi Namir's answer useful but expanded upon it to create it as a function. I renamed the variables to what made sense to me but you can translate them back easily enough, if desired.
Also, the code in Royi's answer already handled the case where the character being searched from does not exist (it starts from the beginning of the string), but I wanted to also handle cases where the character that is being searched to does not exist.
In that case it acts in a similar manner by starting from the searched from character and returning the rest of the characters to the end of the string.
CREATE FUNCTION [dbo].[getValueBetweenTwoStrings](@inputString NVARCHAR(4000), @stringToSearchFrom NVARCHAR(4000), @stringToSearchTo NVARCHAR(4000)) RETURNS NVARCHAR(4000) AS BEGIN DECLARE @retVal NVARCHAR(4000) DECLARE @stringToSearchFromSearchPattern NVARCHAR(4000) = '%' + @stringToSearchFrom + '%' SELECT @retVal = SUBSTRING ( @inputString, PATINDEX(@stringToSearchFromSearchPattern, @inputString) + LEN(@stringToSearchFrom), (CASE CHARINDEX( @stringToSearchTo, @inputString, PATINDEX(@stringToSearchFromSearchPattern, @inputString) + LEN(@stringToSearchFrom)) WHEN 0 THEN LEN(@inputString) + 1 ELSE CHARINDEX( @stringToSearchTo, @inputString, PATINDEX(@stringToSearchFromSearchPattern, @inputString) + LEN(@stringToSearchFrom)) END) - (PATINDEX(@stringToSearchFromSearchPattern, @inputString) + LEN(@stringToSearchFrom)) ) RETURN @retVal END Usage:
SELECT dbo.getValueBetweenTwoStrings('images/test.jpg','/','.') AS MyResult You can try this:
Declare @test varchar(100)='images/test.jpg' Select REPLACE(RIGHT(@test,charindex('/',reverse(@test))-1),'.jpg','') Below query gives you data before '-' Ex- W12345A-4S
SELECT SUBSTRING(Column_Name,0, CHARINDEX('-',Column_Name)) as 'new_name' from [abc]. Output - W12345A
Inspired by the work of Josien, I wondered about a simplification.
Would this also work? Much shorter:
SELECT SUBSTRING(col, CHARINDEX ('/', col) + 1, CHARINDEX ('.', col) - CHARINDEX ('/', col) - 1); (I can't test right now because of right issues at my company SQL server, which is a problem in its own right)
Simply Try With LEFT ,RIGHT ,CHARINDEX
select LEFT((RIGHT(a.name,((CHARINDEX('/', name))+1))),((CHARINDEX('.', (RIGHT(a.name, ((CHARINDEX('/', name))+1)))))-1)) splitstring, a.name from (select 'images/test.jpg' as name)a I got some invalid length errors. So i made this function, this should not give any length problems. Also when you do not find the searched text it will return a NULL.
CREATE FUNCTION [FN].[SearchTextGetBetweenStartAndStop](@string varchar(max),@SearchStringToStart varchar(max),@SearchStringToStop varchar(max)) RETURNS varchar(max) BEGIN SET @string = CASE WHEN CHARINDEX(@SearchStringToStart,@string) = 0 OR CHARINDEX(@SearchStringToStop,RIGHT(@string,LEN(@string) - CHARINDEX(@SearchStringToStart,@string) + 1 - LEN(@SearchStringToStart))) = 0 THEN NULL ELSE SUBSTRING(@string ,CHARINDEX(@SearchStringToStart,@string) + LEN(@SearchStringToStart) + 1 ,(CHARINDEX(@SearchStringToStop,RIGHT(@string,LEN(@string) - CHARINDEX(@SearchStringToStart,@string) + 1 - LEN(@SearchStringToStart)))-2) ) END RETURN @string END declare @searchStart nvarchar(100) = 'search '; declare @searchEnd nvarchar(100) = ' '; declare @string nvarchar(4000) = 'This is a string to search (hello) in this text '; declare @startIndex int = CHARINDEX(@searchStart, @string,0) + LEN(@searchStart); declare @endIndex int = CHARINDEX(@searchEnd, @string, @startIndex + 1); declare @length int = @endIndex - @startIndex; declare @sub nvarchar(4000) = SUBSTRING(@string, @startIndex, @length) select @startIndex, @endIndex, @length, @sub This is a little more legible than the one-liners in this answer which specifically answer the question, but not in a generic way that would benefit all readers. This could easily be made into a function as well with a slight modification.
If there are more than one or none occurences of given character use this:
DECLARE @rightidx int = CASE WHEN 'images/images/test.jpg' IS NULL OR (CHARINDEX('.', 'images/images/test.jpg')) <= 0 THEN LEN('images/images/test.jpg') ELSE (CHARINDEX('.', REVERSE('images/images/test.jpg')) - 1) END SELECT RIGHT('images/images/test.jpg', @rightidx) This was the approach I took.
CREATE FUNCTION dbo.get_text_before_char(@my_string nvarchar(255),@my_char char(1)) RETURNS nvarchar(255) AS BEGIN; return IIF(@my_string LIKE '%' + @my_char + '%',left (@my_string, IIF(charindex(@my_char, @my_string) - 1<1,1,charindex(@my_char, @my_string) - 1)),''); END; CREATE FUNCTION dbo.get_text_after_char(@my_string nvarchar(255),@my_char char(1)) RETURNS nvarchar(255) AS BEGIN; return IIF ( @my_string LIKE '%' + @my_char + '%' ,RIGHT ( @my_string , IIF ( charindex ( @my_char ,reverse(@my_string) )-1 < 1 ,1 ,charindex ( @my_char ,reverse(@my_string) )-1 ) ) , '' ) END; SELECT dbo.get_text_before_char('foo-bar','-') , dbo.get_text_after_char('foo-bar','-') if Input= pg102a-wlc01s.png.intel.com and Output should be pg102a-wlc01s
we can use below query :
select Substring(pc.name,0,charindex('.',pc.name,0)),pc.name from tbl_name pc declare @test varchar(100)='images/test.jpg' select right(left(@test, charindex('.', @test) - 1),4) 
