Escape characters contained by bash variable in regex pattern

Question

In my bash script, i am trying to execute following Linux command:

sed -i "/$data_line/ d" $data_dir

$data_line is entered by user and it may conatain special characters that could brake regex. How can i escape all of the possible special characters in $data_line before i execute sed command?

Answer 1

You might be able to use this technique to protect the selector. The lines marked with "*****" below are the significant lines. The others are mostly for testing and demonstration. The key is to use a character that doesn't appear in the user input to delimit the selector address.

data_line='.*/ s/GOLD/LEAD/g;b;/.*'    # scary user input
candidates='/:.|@#%^&;,!~abcABC'       # *****   # (make it as long as you like)
char=$(echo "$candidates" | tr -d "$data_line")    # *****
char=${char:0:1}   # ***** choose the first candidate that doesn't appear in the user input

if [ -z "$char" ]    # ***** this test checks for exhaustion of the candidate character set
then
    echo "Unusable user input. Recommendation: cigarette and blindfold."
    exit 1
fi

# test without protection
excitement="GOLD, I tell you, thar's GOLD in them thar hills!" 
echo "$excitement" | sed "/$data_line/ d"
# output: "LEAD, I tell you, thar's LEAD in them thar hills!"

# test WITH protection
echo "$excitement" | sed "\\${char}${data_line}${char} d"    # *****
# output: "GOLD, I tell you, thar's GOLD in them thar hills!"

# test WITH protection and useful user input
data_line="secret"
mystery="The secret map is tucked in a hidden compartment in my saddle bag."
echo -e "$excitement\n$mystery" | sed "\\${char}${data_line}${char} d"
# output: "GOLD, I tell you, thar's GOLD in them thar hills!"

Answer 2

grep -v -F "$data_line" "$data_dir" > ...