Match a string that contains a newline using sed

Question

I have a string like this one:

    #
    pap

which basically translates to a \\t#\\n\\tpap and I want to replace it with:

Answer 1

A couple of pure bash solutions:

Concise, but somewhat fragile, using parameter expansion:

in=$'\t#\n\tpap\n' # input string

echo "${in/$'\t#\n\tpap\n'/$'\t#\n\tpap\n\tpython\n'}"

• Parameter expansion only supports patterns (wildcard expressions) as search strings, which limits the matching abilities:
• Here the assumption is made that pap is followed by \n, whereas no assumption is made about what precedes \t#, potentially resulting in false positives.
• If the assumption could be made that \t#\n\tpap is always enclosed in \n, echo "${in/$'\n\t#\n\tpap\n'/$'\n\t#\n\tpap\n\tpython\n'}" would work robustly; otherwise, see below.

Robust, but verbose, using the =~ operator for regex matching:

The =~ operator supports extended regular expressions on the right-hand side and thus allows more flexible and robust matching:

in=$'\t#\n\tpap' # input string 

# Search string and string to append after.
search=$'\t#\n\tpap'
append=$'\n\tpython'

out=$in # Initialize output string to input string.
if [[ $in =~ ^(.*$'\n')?("$search")($'\n'.*)?$ ]]; then # perform regex matching
    out=${out/$search/$search$append} # replace match with match + appendage
fi

echo "$out"