Shell Programming and Scripting

As per the OP
He has already have code to separate the numbers here so, He just want to remove leading zeros.

In case you are interested in a much more efficient string processing approach which does not need to create an entirely new shell just to strip the zeroes, here's a modified version of pamu's suggestion:

Code:

while
    read line
do
    wardevent=${line#"${line%%[!0]*}"}
    echo $wardevent
done

That assumes (like pamu's suggestion) that the value of $line consists of nothing but digits. If that is not the case, then the read statement needs to be fortified and the echo statement should be abandoned in favor of printf and $wardevent double-quoted.

Regards,
Alister

Code:

while read v junk
do
    wardevent=$((10#$v))
    echo $wardevent
done < file

Slight modification of pamu's code:

Code:

while read No REST; do   printf "%2d %s\n" "$No" "$REST"; done<file
 1 
 2 123
 3 Hello

Adapt printf's format string to your needs.

Like the shell built-in numerics, printf "%d" treats a value with a leading 0 as an octal value, so 08 and 09 give error.
A workaround is printf "%.f".

... or give it the decimal prefix:

Code:

while read No REST; do   printf "%02d %s\n" $((10#$No)) "$REST"; done<file
01 
02 123
03 Hello
08 sklöfj
09 dfdsf

printf is a poor approach. Post #3 states that the result of stripping the leading zeroes will be saved for further use. To save printf's output, it's necessary to use command substitution, which does not improve upon the already accepted solution, $(expr ...).

If the shell (or shells) which must run this code support the base# notation, I recommend MadeInGermany's arithmetic expansion approach, $((10#$v)). It's efficient and so simple that it's immediately understandable.

If the code must run on a shell which does not support base#, then my suggestion is probably the best choice.

Regards,
Alister