UNIX for Dummies Questions & Answers

data contains:

1 ;1
500;2
10 ;6
111;9
100;0
20 ;1
2 ;5
and so on............

The program codes:

#!/usr/bin/ksh
echo "Please enter id1"
read n
echo "please enter id2"
read m
typeset -L3 n
grep "$n;$m" data > $n-x


The issues:

If I enter 500 for id1 and 2 for id2, I will get a Valid file "500-x" that contain 500;2.
If I enter 1 for id1 and 1 for id2, I will get an invalid file "1 -x" which I can not cat; I get these messages when I cat the file:
cannot open 1
cannot open -x

The questions:

Are there ways to get around this problem? Since I typeset -L3 n to get the data pattern that I need to grep. How can I un-typeset it when want to save it as a file?

When I enter 1 for id1 and 1 for id2; I need to have a file "1-x" as a valid file and when I cat it I should get: 1 ;1

The experts answers?


Many thanks!!!

I would use two variables:
typeset -L3 n2=$n
and then is n2 or n as required.

Thanks for the help! I see how it works but I don't understand the coding..Can you please explain? Are there other ways beside this coding?

Not sure what to explain...but
typeset -L3 n2
n2=$n
would be another way to do it.

Thanks! I see it now!