UNIX for Dummies Questions & Answers

Hi All,
I have a string "TBM630300000000020080506094041000003818".I want to replace the last nine digits with another string stored in a variabe called "count".The variabe is also having nine digits.Could any one please help me on this how to accomplish.I need a detail syntax(not in the short form) since i am new to the unix scripting.

I tried with some flavour of "string" commands but those are not supported by my KSH environment.

Appreciate your help!!!!!!!!!!!!!!!

Thanks
Narasimharao.

Here is something to start with

Code:

echo "your_string" | sed -e "s/\(.*\)........./\1$count/"

A . represents any character. A .* means any number of characters. \(.*\) would collect the characters into a buffer. Since we need to replace the last nine numbers, you can represent them as nine consecutive dots or as [0-9]\{9\}. This will get replaced with the contents of $count.

Using ksh93

Code:

$ str="TBM630300000000020080506094041000003818"
$ print ${str%{9}([[:digit:]])}
TBM630300000000020080506094041
$ count="987654321"
$ print ${str%{9}([[:digit:]])}$count
TBM630300000000020080506094041987654321
$

Hi vino,
First of all Thanks for you.Its working fine for my requirement.Still i need a little bit clarification on the below code.What is the 1 before the "$count" variable signifies? Please explain me.with out 1 its not working fine.

Thanks in Advance.

Narasimharao.