sed not substituting variable in while loop

Question

When i try to use line number from variable in sed it works individually but when same is done using while loop it not works.

eg.

cat file
den 
run
ten
run
fan

linenumber=2
replacetext=34
sed -i ""$linenumber"s/run/run\n"$replacetext"/" file

Output:
cat file
den
run
34
ten
run
fan

Above works and sed command is generated after variable expansion.

But when i use same in while loop it does work only for first instance and not all as i wanted to replace all instances with incrementing line number using pure sed solution. Awk or perl solution is not wanted.

eg.

replacetext=1
while read in; do replacetext=$((replacetext++)) && sed ""$in"s/run/run\n"$replacetext"/" file; done <<< "$(cat file | grep 'run' -n | cut -f1 -d ':')"

Expected Output: Match text will be replaced with itself followed by newline and a incrementing number.

den 
run
2
ten
run
3
fan

Obtained output:

den
run
2
ten
run
fan

Please only sed based solution required. And i wanted to know what unusual behaviour was occuring in while loop that it was not working.Because the loop variable is given the value of line number and then that line number is used in sed to replace macthed text on that line with itself folowed by newline and incremented number.

Sed seems to work when same variable line number is provided but not when used in while loop.

Answer 1

It fails because the stuff after <<< gets a list of the correct line numbers in the original file.

Then you edit the file multiple times, inserting extra lines.

So every sed except the first is being told the wrong line numbers for the s/// command.

One solution would be to reverse the list of line numbers, so edits did not affect earlier numbering.

This is a sed solution, in one pipeline, somewhat quirky.

The first sed gets the line numbers where the Pattern is found, using the = operator. The nl prefixes a sequence number to each line: like 3 10

The second sed outputs sed expressions: it converts each such line to your required sed edit, like: 10 s/$/\n3/.

The third sed collects and applies those edits from stdin, and does an in-situ edit of the file. In this case, it edits itself.

This is the script itself:

#! /bin/bash

Pattern="[Ss]ed"

RegEx='[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\).*'
DoSed='\2 s/\$/\\n\1/'

sed -n -e "/${Pattern}/=" badSed |
    nl -ba |
    sed -e "s:${RegEx}:${DoSed}:" |
    sed -i -f - badSed
 

This is what it does to itself:

#! /bin/bash

Pattern="[Ss]ed"

RegEx='[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\).*'
DoSed='\2 s/\$/\\n\1/'
1

sed -n -e "/${Pattern}/=" badSed |
2
    nl -ba |
    sed -e "s:${RegEx}:${DoSed}:" |
3
    sed -i -f - badSed
4

Answer 2

The line numbers are being messed up where the substitution needs to happen.

So we devise another strategy, we allow the while loop to first generate a valid set of sed commands, which are then applied back to the input file.

replacetext=1
while read in; do ((replacetext++)); echo "$in s/\$/\\n$replacetext/"; done <<<"$(grep -n run file | cut -d: -f1)" | sed -f - file > tmp && mv tmp file