another special scenario for string searching.

Keanehelp · New User Joined: 27 May 2008 Posts: 71 Location: USA, CA.

Hi,

I have a different problem now. How can I search for a substring in a string. For ex what should I do for searching 'GHI' from 'ABCDEFGHIJKLMN'.

Thanks

dick scherrer · Posted: Tue Dec 30, 2008 4:21 am

Hello,

Similar to the "other" solution. . .

Loop thru the string using reference modification and each time thru compare the current location against "GHI". If you get to the "end", GHI is not in the string.

You might also look at INSPECT. At the top of the page is a link to "IBM Manuals" of which is the COBOL Language Reference. If you find something in the manual that is not clear, post what you found and your question about it and someone will be able to clarify.

Robert Sample · Posted: Tue Dec 30, 2008 4:22 am

If you just want to know if the substring exists in the string, INSPECT will work perfectly fine. If you want to know where the substring is in the string, use reference modification again checking substring against the right number of characters of the string. Your loop will run from 1 to (length of string - length of substring + 1).

SHAILESH OZA · New User Joined: 10 Jun 2005 Posts: 21 Location: Mumbai

Hi Keaan,

Either you can Inspect as follows

Inspect String1 for all 'GHI' tallying counter which will give how many times this verb is repeated.

and if you want to find out the location then you can do one thing as mentioned below.

Intialize I,

01 J OCCURS UPTO LENGTH OF STING LETS SAY 12
02 K PIC X(1)
perform varying i from 1 by 1 until Lengh of string

If (I:3) = 'GHI'
MOVE I TO J(K)
ADD 1 TO K
ELSE
CONTINUE
END-IF.

BECAUSE OF THIS WHAT WILL HAPPEN YOU COMES TO KNOW THE STARTING POSITION OF 'GHI' I.E G which is stored in the Array J.

lets check the positions in the J(K) what are the values theres and the

J(3) = 5 so if in the main string you check (5:3) you will get GHI. let's try this, if didn't get just reply back to me

Robert Sample · Posted: Tue Dec 30, 2008 10:05 pm

Shailesh: have you tested this code? You cannot run up to the length of the string because reference modification doesn't work past the end of the variable. Furthermore, I don't think checking one character to be 3 characters will work, and you're using J(K) as if K is numeric, which is very clearly will not be according to the sample given by the o/p.

dick scherrer · Posted: Wed Dec 31, 2008 1:13 am

Hello Shailesh,

Please do not post untested code. It is far too easy to test code like this and know that it works (rather than merely believing it could work).

Thank you,