Quoting special characters with sed

Question

I\'m trying to look in a variable passed to my program (the variable is $1) and to replace any special characters with quoted forms of said special character, in order to no

Answer 1

The purpose of the multiple invocations of sed is to place a literal backsplash before each occurrence of a set of characters. This can be done in one call to sed, but you need to be careful about how you specify the set.

First, let's see what the general command will look like:

newtarget=$( echo "$target" | sed -e 's/\([...]\)/\\\1/g'

where ... will be replaced with the set of characters to escape. This commands uses parentheses to capture a single instance of one of those characters, the replaces it with a backsplash followed by the captured character. To specify the set of characters, use

[]*^+\.$[-]

Two notes: first, the ] must come first so that it isn't mistaken for the end of the set, since [] is an invalid set. Second, - must come last, so that it isn't mistaken as the range operator (e.g., [a-z] is the set of lowercase letters, but [az-] is simply the three characters a, z, and -).

Putting it all together:

 newtarget=$( echo "$target" | sed -e 's/\([]*^+\.$[-]\)/\\\1/g' )