[bash] Escape a string for sed search pattern

Question

In my bash script I have an external (received from user) string, which I should use in sed pattern.

REPLACE="<funny characters here>"
sed "s/KEYWORD/$REPLACE/g"

How can I escape the $REPLACE string so it would be safely accepted by sed as a literal replacement?

NOTE: The KEYWORD is a dumb substring with no matches etc. It is not supplied by user.

Answer 1

Here is an example of an AWK I used a while ago. It is an AWK that prints new AWKS. AWK and SED being similar it may be a good template.

ls | awk '{ print "awk " "'"'"'"  " {print $1,$2,$3} " "'"'"'"  " " $1 ".old_ext > " $1 ".new_ext"  }' > for_the_birds

It looks excessive, but somehow that combination of quotes works to keep the ' printed as literals. Then if I remember correctly the vaiables are just surrounded with quotes like this: "$1". Try it, let me know how it works with SED.

Answer 2

The only three literal characters which are treated specially in the replace clause are / (to close the clause), \ (to escape characters, backreference, &c.), and & (to include the match in the replacement). Therefore, all you need to do is escape those three characters:

sed "s/KEYWORD/$(echo $REPLACE | sed -e 's/\\/\\\\/g' -e 's/\//\\\//g' -e 's/&/\\\&/g')/g"

Example:

$ export REPLACE="'\"|\\/><&!"
$ echo fooKEYWORDbar | sed "s/KEYWORD/$(echo $REPLACE | sed -e 's/\\/\\\\/g' -e 's/\//\\\//g' -e 's/&/\\\&/g')/g"
foo'"|\/><&!bar

Answer 3

Just escape everything in the REPLACE varible:

echo $REPLACE | awk '{gsub(".", "\\\\&");print}'