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How to replace # in particular line?

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# 8  
Old 01-10-2020
A decent specification will help everybody to save time and effort. Try


Code:
sed '/&@@10.10.10.10/ s/^/#/' file

This User Gave Thanks to RudiC For This Post:
yash_message
# 9  
Old 01-10-2020
Add #
Code:
sed '/^&@@10.10.10.10/ s/^/#/' /etc/rsyslog.conf

Remove #
Code:
sed '/^#&@@10.10.10.10/ s/^#//' /etc/rsyslog.conf

The & is special in sed in the replacement string. This is avoided now.
But:
The $ is special in a RE (the search string) if it's the last character.
The . is special in a RE: it means any character (not only a . character).

Better use the shell:
Code:
while IFS= read -r line
do
  case $line in
  # add a # character
  ( '&@@10.10.10.10'* ) line=#${line};;
  # remove a # character
  # ( '#&@@10.10.10.10'* ) line=${line#\#};;
  esac
  printf "%s\n" "$line"
done < /etc/rsyslog.conf

Within a 'string' the remaining problem character is '. The quick workaround is '\''.
This User Gave Thanks to MadeInGermany For This Post:
jim mcnamara
# 10  
Old 01-10-2020
Welldone sir, thank you so much :-).
you are great.
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