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So - you want to replace a space with the 1 - not adding one. Arrange for the replacement:
sed 's/\(.\{13\}\) /\11/' file

Yes, there is a way to do it. Do it exactly as I showed you in post #2 in this thread.

Hi,
In bash with pipeline work under many conditions, just a silly example to present bash pipeline:
In general, not work by default:
$ while read line; do echo $line | IFS="," read dbt_acct...

This works because the fourth field of the first line is empty; that loop prints the empty string "" immediatedly followed by "home". With any string in he first line's fourth field it would print...

Try:
while IFS=, read -r dbt_acct val_dt crncy AMT dbt_GOP
do
print "$AMT"
done <abc.txt

How aboutfind . ! \( -perm -u=w -o -perm -g=w \) to print files that have either user's or group's write bit missing?

I got a request to explain the sed statement in post #4:


-e :a # define label a
-e '/|$/{ # if the last character is a vertical bar, then
$!N # if it is not the last line...

Please use code tags as required by forum rules!

Does your cut provide the --complement option?cut -c$a --complement one.txt
xyz 33 skfo 4r443
sfs bcd sd fsdf sdfd
abc c 33 abc dsaf

That depends on the system and shell you use. With a recent bash, shopt -s extglob
ls -la !(MIS*)
might work.

Just using standard shell features, you can get what you want with:
ls -l [^M][^I][^S]* M[^I][^S]* [^M]I{^S]* [^M][^I]S* MI[^S]* M[^I]S* [^M]IS* 2>/dev/null
With some shells with some non-standard...