CODEWITHSUNDEEP
bashshellsed
paul590
paul590

Reputation: 1425

sed command bad option in substitution expression when replacing back slash

I am currently running the following command and getting a "bad option in substitution" error. What I would like to do is add a backslash to the string I am replacing but I am unable to do so. Any ideas?

Code:

 VAR="/host/test"
 echo ${VAR} | sed -e "s/\//\\\//g"

Upvotes: 0

Views: 2610

Answers (1)

melpomene
melpomene

Reputation: 85767

Why use sed at all?

$ echo "${VAR//\//\\/}"
\/host\/test

It looks a bit horrible, but it works fine. See ${parameter/pattern/string} in https://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion.

Your problem happens because you're using double quotes (") around your sed code, not single quotes (').

In double quotes, \\ is interpreted as the escape sequence for a single \, so the code that sed ends up seeing is:

$ echo "s/\//\\\//g"
s/\//\\//g
#^  ^  ^

sed reads this as "search for \/ and replace it by \\, with /g as options". This is an error because / is not a valid option.

Upvotes: 3

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