Comments on: SQL SERVER – Finding the Occurrence of Character in String

By: kalindi patel

kalindi patel — Thu, 07 Mar 2013 11:53:03 +0000

It is very much helpful to me and i used this site as i found perfect solution from here .

By: RFA

RFA — Thu, 21 Feb 2013 11:46:02 +0000

Simple and clever. Thank you so much!

By: Annonymous

Annonymous — Thu, 17 Jan 2013 04:55:16 +0000

Here’s what I did when searching in job steps for text (see Occurrences column)
———————————
———————————
———————————
DECLARE @SearchText VARCHAR(100)
SET @SearchText = ‘DBCC INPUT’
———————————
———————————
———————————

SELECT
job.job_id,
job.[name],
CASE job.enabled
WHEN 1 THEN ‘Yes’
ELSE ‘No’
END AS [Is Enabled],
jobstep.[step_name],
(LEN(jobstep.command) – LEN(REPLACE(jobstep.command COLLATE Latin1_General_CI_AS, @SearchText, ‘|’))) / LEN(@SearchText) + 1 AS [Occurrences],
‘…’ + SUBSTRING(jobstep.command, PATINDEX(‘%’ + @SearchText + ‘%’, jobstep.command COLLATE Latin1_General_CI_AS) – 20, 50) + ‘…’ AS [Found here]
FROM
msdb.dbo.sysjobs job
inner join msdb.dbo.sysjobsteps jobstep on job.job_id = jobstep.job_id
WHERE
jobstep.command COLLATE Latin1_General_CI_AS LIKE ‘%’ + @SearchText + ‘%’

By: Gary Melhaff

Gary Melhaff — Wed, 12 Dec 2012 22:30:45 +0000

I found method where programming isn’t necessary by using the new LIKE clause variations…you can just embed it within you query and no fancy CLR callouts to regex expressions are necessary…

Here is an example of replacing any character values in a string with null…(CASE WHEN money_field LIKE ‘%[a-zA-Z]%’ THEN NULL ELSE money_field END)

By: kunal

kunal — Wed, 28 Nov 2012 05:57:18 +0000

Here is i/p of a table:
Name Surname unique id
kunal & Raj Anand 1
Ram & k kelly & R Singh 2
Rahan & raj & kali kumar 3

How to get o/p like this:–

Name Surname unique id
kunal anand 1
Raj Anand 1
Ram Singh 2
k kelly Singh 2
R Singh 2
Rahan kumar 3
raj kumar 3
kali kumar 3

By: pgfiore

pgfiore — Mon, 26 Nov 2012 15:43:55 +0000

Yanick do you mean SELECT LEN(REPLACE(@test,’string’,’string+1char’) )– LEN(@test)?

By: Suvendu

Suvendu — Mon, 05 Nov 2012 07:48:39 +0000

Intelligent way, Sir :)

By: Yanick

Yanick — Mon, 29 Oct 2012 15:52:30 +0000

A simple and efficient way to count the occurrences in a string:
Here we are counting the number of As
DECLARE @test VARCHAR(150)
SET @test = ‘aaaaabbbbb’

SELECT REPLACE(@test,’a',’aa’) – LEN(@test)

By: mukhtar

mukhtar — Fri, 12 Oct 2012 13:20:12 +0000

i want to create a slang filter function.

want a query which returns no of slang in a string (select query).

i.e.
Table : Slans
slangs
—————–
aa
bb
cc
dd
ee

result required.
===================
string slang_cnt slangs
====================================================
aa cc bsf kk ff ee 3 aa, cc, ee

By: Tony Bromirski

Tony Bromirski — Thu, 06 Sep 2012 17:13:31 +0000

Thought I would share this handy (but simple) function for finding the Nth occurrence of a character in a string.

REF: http://www.sqlservercentral.com/scripts/Miscellaneous/30497/

Also, an example of how I used it…

–String value
–tblData.UserLogString = ‘John.Doe,4,Johnathan,Doe,Johnathan.Doe@domain.com,True,False|’

–Extract the UserID from string value.
SUBSTRING( tblData.UserLogString, 0, CHARINDEX( ‘,’, tblData.UserLogString, 0 ) )
Value returned: “John.Doe”

–Extract the User Name (First Last) from string value.
REPLACE(
SUBSTRING( tblData.UserLogString, CHARINDEX2( ‘,’, tblData.UserLogString, 2 ) + 1, ( CHARINDEX2( ‘,’, tblData.UserLogString, 4 ) – CHARINDEX2( ‘,’, tblData.UserLogString, 2 ) ) – 1 ),
‘,’ ,
‘ ‘ )
Value returned: “Johnathan Doe”

–Extract the email ID from string value.
SUBSTRING(tblData.UserLogString,
CHARINDEX2( ‘,’, tblData.UserLogString, 4 ) + 1 ,
( CHARINDEX2( ‘,’, tblData.UserLogString, 5 ) –
CHARINDEX2( ‘,’, tblData.UserLogString, 4 ) ) – 1 ),
Value returned: “Johnathan.Doe@domain.com”

I’m sure this can be done many other ways… Also I disagreed with the author in how I created the SPROC – I created it with the command:
CREATE FUNCTION [dbo].[CharIndex2]

Cheers,

(p.s. Pinal – I enjoy your column!)

By: Velu

Velu — Mon, 30 Jul 2012 13:22:04 +0000

thanks for u r cmnt….

By: Aarti

Aarti — Sat, 28 Jul 2012 12:46:57 +0000

can u plz tell how to fing a name which contains only 2 time “a”

By: ONjefu

ONjefu — Wed, 06 Jun 2012 06:27:32 +0000

Very handy. Thank you!

By: Sahhnawaz

Sahhnawaz — Tue, 06 Mar 2012 08:21:14 +0000

Hello Pinal,
It is very interesting example, and your example is marvelous..
salute to you

By: ashish

ashish — Mon, 21 Nov 2011 10:59:33 +0000

Dear Pinal Sir,

Thanks a lot, your nots are always very helpful to me.

By: Sushant Dalvi

Sushant Dalvi — Thu, 11 Aug 2011 11:49:42 +0000

above table

Cityname operatorname prefix
Kolkata Unitech Wireless 90620-90629 82960-82969
Kolkata Datacom Solutions 90730-90739
Kolkata BSNL 94330-94339 94320-94325 94770-94779
Kolkata Loop 91100-91109
Kolkata Reliance Telecom 98830-98839 96810-96819 88200-88209
Mumbai BPL Mobile 98210-98219 96640-96649 97730-97739
Mumbai Bharti Airtel 98670-98679 98920-98929 99670-99674

Wanted output like mentione in below table

Cityname operatorname prefix
Kolkata Unitech Wireless 90620
Kolkata Unitech Wireless 90621
Kolkata Unitech Wireless 90622
Kolkata Unitech Wireless 90623
Kolkata Unitech Wireless 90624
Kolkata Unitech Wireless 90625
Kolkata Unitech Wireless 90626
Kolkata Unitech Wireless 90627
Kolkata Unitech Wireless 90628
Kolkata Unitech Wireless 90629
Kolkata Unitech Wireless 82960
…..
…..
…..
Kolkata Unitech Wireless 82969
Kolkata Datacom Solutions 90730
Kolkata Datacom Solutions 90731
So on…..

pls suggest

By: Ajit Pratap

Ajit Pratap — Wed, 25 May 2011 08:58:01 +0000

good

thanks a lot

Regards

Ajit

By: Hien Le

Hien Le — Wed, 23 Feb 2011 00:31:19 +0000

Hi Pinal,

Great stuff, but in regards to this comment of yours –

“… Additionally, make sure that your strings does not have leading or trailing empty spaces. If you have it, you may want to use LTRIM or RTRIM functions.”

Is there a reason for not using DATALENGTH instead of LEN?

With the use of DATALENGTH you won’t have to LTRIM and RTRIM. Plus you can also search for space(s) as well.

I have only carried out limit testing and have not done any searching for issue(s) with DATALENGTH.

DECLARE
@varString VARCHAR(4000),
@varCharPattern1 VARCHAR(20),
@varCharPattern2 VARCHAR(20),
@varCharPattern3 VARCHAR(20)

SET @varString = ‘ Some random text aslkja aslkwe ld axls ‘
SET @varCharPattern1 = ‘ ‘
SET @varCharPattern2 = ‘as’
SET @varCharPattern3 = ‘ ‘

SELECT (DATALENGTH(@varString) – DATALENGTH(REPLACE(@varString, @varCharPattern1, ”)))/DATALENGTH(@varCharPattern1)
SELECT (DATALENGTH(@varString) – DATALENGTH(REPLACE(@varString, @varCharPattern2, ”)))/DATALENGTH(@varCharPattern2)
SELECT (DATALENGTH(@varString) – DATALENGTH(REPLACE(@varString, @varCharPattern3, ”)))/DATALENGTH(@varCharPattern3)

Thanks,

Hien

By: Hamid

Hamid — Sat, 02 Oct 2010 11:50:25 +0000

ALTER FUNCTION [dbo].[CountWord]
(
– Add the parameters for the function here
@RowStr nvarchar(1000),
@WordStr nvarchar(255)
)
RETURNS int
AS
BEGIN

declare @WordCount int
declare @index int

set @index =0
set @WordCount = 0

while CHARINDEX(@WordStr, @RowStr,@index)>0
begin

SET @index = CHARINDEX(@WordStr,@RowStr,@index)
SET @WordCount = @WordCount + 1
if @index =0
break
else
SET @index = @index + 1
end

RETURN @WordCount

END

By: Shashi Sekhar

Shashi Sekhar — Sun, 26 Sep 2010 17:48:26 +0000

Sir,
I am studing your blog. It is very good for everyone who wants to learn SQL.
Here in your blog based on Finding the occurrence of character in a string you gave two example. One is for caracter and other for word. Here i want to tell you that in the first example there is some problem. When i run the example i found that the result is different while in your case the result are same.

CntReplaceChars CntOccurrenceChars.
8 2

By: Jagan

Jagan — Wed, 25 Aug 2010 15:55:47 +0000

Hi Pinal,

Thanks for the SQL. I love your articles. Very good. Continue this effort.

Take care,
Jagan.

By: Animesh Upadhyay

Animesh Upadhyay — Fri, 20 Aug 2010 22:43:08 +0000

Hi Pinal,

I am a regular reader of your blog, kindly accept my heartfelt thanks for sharing your wisdom. Thanks again for the great post and the t-sql examples. It seems there is a typo error in the output for the first example, it should be:

CntReplacedChars CntOccuranceChars
——————– ——————–
8 2

Just to avoid any doubt for the readers.

Thanks & Regards,
Animesh

By: amigasger

amigasger — Mon, 16 Aug 2010 22:51:31 +0000

Thanks again, Pinal.
I really love these T-SQL-examples. Especially string-manipulation is hard in T-SQL compared to other languages, so thumbs up for sharing your knowledge.

By: Ramdas

Ramdas — Mon, 16 Aug 2010 13:34:14 +0000

Cool example.
Hi Fairoze,
LEN – Returns the lenght of the string.
REPLACE – allows one to replace characters in a given string.
Check out books online for more examples and detailed explanation.

By: tobi

tobi — Mon, 16 Aug 2010 10:14:55 +0000

I am sorry for all of you who have to do string manipulation in SQL during their day jobs because they are stuck in a legacy codebase.