Sed replace pattern with line number

Question

I need to replace the pattern ### with the current line number.

I managed to Print in the next line with both AWK and SED.

sed -n \"/###/{

Answer 1

This might work for you (GNU sed):

sed = file | sed 'N;:a;s/\(\(.*\)\n.*\)###/\1\2/;ta;s/.*\n//'

Answer 2

Following on from https://stackoverflow.com/a/53519367/29924

If you try this on osx the version of sed is different and you need to do:

seq 1 `wc -l FILE | awk '{print $1}'` | xargs --verbose -IX sed -i bak "X s/__line__/X/" FILE

see https://markhneedham.com/blog/2011/01/14/sed-sed-1-invalid-command-code-r-on-mac-os-x/

Answer 3

Simple awk oneliner:

awk '{gsub("###",NR,$0);print}'

Answer 4

is this ok ?

kent$  echo "a
b
c
d
e"|awk '/d/{$0=$0" "NR}1'
a
b
c
d 4
e

if match pattern "d", append line number at the end of the line.

edit

oh, you want to replace the pattern not append the line number... take a look the new cmd:

kent$  echo "a
b
c
d
e"|awk '/d/{gsub(/d/,NR)}1'
a
b
c
4
e

and the line could be written like this as well: awk '1+gsub(/d/,NR)' file

Answer 5

Given the limitations of the = command, I think it's easier to divide the job in two (actually, three) parts. With GNU sed you can do:

$ sed -n '/###/=' test > lineno

and then something like

$ sed -e '/###/R lineno' test | sed '/###/{:r;N;s/###\([^\n]*\n\)\([^\n]*\)/\2\1/;tr;:c;s/\n\n/\n/;tc}'

I'm afraid there's no simple way with sed because, as well as the = command, the r and GNU extension R commands don't read files into the pattern space, but rather directly append the lines to the output, so the contents of the file cannot be modified in any way. Hence piping to another sed command.

If the contents of test are

fooo
bar ### aa
test
zz ### bar

the above will produce

fooo
bar 2 aa
test
zz 4 bar

Answer 6

one-liner to modify the FILE in place, replacing LINE with the corresponding line number:

seq 1 `wc -l FILE | awk '{print $1}'` | xargs -IX sed -i 'X s/LINE/X/' FILE