Shell Programming and Scripting

I have a string as follows
IS*blahblah
TED~blahblah
etc.
I want to list down only IS and TED

Can someone help me?

How about this ?

Code:

#! /bin/sh

while read line
do
tmp=`echo $line | awk -F"[*~]" '{ printf $1 }'`
echo $tmp
done < input.txt

You can use tmp if you need to retain the value.

If you need to avoid tmp then this should work

Code:

#! /bin/sh

while read line
do
echo $line | awk -F"[*~]" '{ printf $1 "\n" }'
done < input.txt

From the prompt this would work,

Code:

awk -F"[*~]" '{ printf $1 "\n" }' < input.txt

You can introduce all your required delimiters within the "[ ]" of -F

Vino

Try This :--

sed -e 's/^IS.[a-z]*/IS/' -e 's/^TED.[a-z]*/TED/' < filename


Thnx
Rahul

Can be simplified further....

awk -F"[*~]" '{print $1}' input.txt

Cheers
ZB

Hi folks,

None of the solutions did not work. Can someone please help me.

My requirement was
The file has many lines. In each line, after 3 or 4 characters, there is a special character like *, ~ or + etc. The first 3 or 4 characters can be Alphabet, or Numeric. I wanted to extract the first 3 or 4 characters from each line with out the special character.

Thanks,
AK.

And..

Code:

awk -F"[*~]" '{print $1}' input.txt

didnt work ??

If it didnt work, post the results of the above awk script for a sample section of your input file. Best would be the part of the input file, where the script fails.

Vino

Thanks Vino for your help

My input file is as below

ISA~00~ ~00~ ~ZZ~12345 ~ZZ~54321 GS~SH~12345 ~54321
ST~86~000000007
BSN~00~0958
DTM~011~ET

And, when I ran the code above, the below results appeared

ISA~00~ ~00~ ~ZZ~12345 ~ZZ~54321 GS~SH~12345 ~54321
ST~86~000000007
BSN~00~0958
DTM~011~ET


My requirement is to get the first 2 or 3 characters like ISA, GS, ST, BSN and DTM. In this file, the special character is '~', but in other files it may vary.