I have the following test file
AAA BBB CCC
Using the following sed I can comment out the BBB line.
# sed -e '/BBB/s/^/#/g' -i file
I'd like to only comment out the line if it does not already has a # at the begining.
# sed -e '/^#/! /BBB/s/^/#/g' file sed: -e expression #1, char 7: unknown command: `/'
Any ideas how I can achieve this?
comment and uncomment a line with Shell Script Requirement is: 1. comment and uncomment the line with Shell Script: /opt/admin/fastpg/bin/fastpg.exe -c -=NET (using fastpg.exe as a search option) 2.
(quit) Exit sed without processing any more commands or input. (quit) This command is the same as q , but will not print the contents of pattern space.
Assuming you don't have any lines with multiple #
s this would work:
sed -e '/BBB/ s/^#*/#/' -i file
Note: you don't need /g since you are doing at most one substitution per line.
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