How can find special row and replace special strings

Question

I wrote a todo list script with "add","clear" and "find" argument. I need to pass a "done" argument and script find related row and replace its first column with another value. How can I do that?

Result of script shows in the follow :

0,H,First Task
0,L,Second Task
0,M,Third Task

and now i want to execute something like ./todo done 2, and obtain the following result:

0,H,First Task
1,L,Second Task
0,M,Third Task

Answer 1

file=todo.list
line=2
perl -F, -pi -lapse '
  $F[0] = 1 if $. == $line;
  $_ = join ",", @F' -- -line="$line" -- "$file"

Or with gawk:

file=todo.list
line=2
gawk -i /usr/share/awk/inplace.awk -v line="$line" -F, -v OFS=, -e '
  NR == line {$1 = 1}
  {print}' -E /dev/null "$file"

With GNU sed, you could do:

sed -i -e "${line}s/^[^,]*/1/" -- "$file"

But you'd have to sanitise $line (make sure it contains a sequence of [0123456789] characters) beforehand or otherwise, that would introduce an arbitrary command execution vulnerability.

With gawk, do not use -i inplace as gawk tries to load the inplace extension (as inplace or inplace.awk) from the current working directory first, where someone could have planted malware. The path of the inplace extension supplied with gawk may vary with the system, see the output of gawk 'BEGIN{print ENVIRON["AWKPATH"]}'

Answer 2

#!/bin/bash
fis=$1
col1=$2
if [[ $fis == "done" ]]
then
awk -v col1="$col1" -F "," 'BEGIN{OFS=","}NR==col1{gsub(".*","PRAVEEN",$1)}1' ht.txt
fi

output

0,H,First Task
PRAVEEN,L,Second Task
0,M,Third Task