Substitution in text file **without** regular expressions

Question

I need to substitute some text inside a text file with a replacement. Usually I would do something like

sed -i 's/text/replacement/g' path/to/the/file

The problem is that both text and replacement are complex strings containing dashes, slashes, blackslashes, quotes and so on. If I escape all necessary characters inside text the thing becomes quickly unreadable. On the other hand I do not need the power of regular expressions: I just need to substitute the text literally.

Is there a way to do text substitution without using regular expressions with some bash command?

It would be rather trivial to write a script that does this, but I figure there should exist something already.

Answer 1

When you don't need the power of regular expressions, don't use it. That is fine.
But, this is not really a regular expression.

sed 's|literal_pattern|replacement_string|g'

So, if / is your problem, use | and you don't need to escape the former.

ps: about the comments, also see this Stackoverflow answer on Escape a string for sed search pattern.

---

Update: If you are fine using Perl try it with \Q and \E like this,
perl -pe 's|\Qliteral_pattern\E|replacement_string|g'
_{RedGrittyBrick has also suggested a similar trick with stronger Perl syntax in a comment here}

Answer 2

You could also use perl's \Q mechanism to "quote (disable) pattern metacharacters"

perl -pe 'BEGIN {$text = q{your */text/?goes"here"}} s/\Q$text\E/replacement/g'

Answer 3

I pieced together a few other answers and came up with this:

function unregex {
   # This is a function because dealing with quotes is a pain.
   # http://stackoverflow.com/a/2705678/120999
   sed -e 's/[]\/()$*.^|[]/\\&/g' <<< "$1"
}
function fsed {
   local find=$(unregex "$1")
   local replace=$(unregex "$2")
   shift 2
   # sed -i is only supported in GNU sed.
   #sed -i "s/$find/$replace/g" "$@"
   perl -p -i -e "s/$find/$replace/g" "$@"
}