Escaping dots in bash variables

Question

I want to escape dots from an IP address in Unix shell scripts (bash or ksh) so that I can match the exact address in a grep command.

echo $ip_addr | sed "s/\./\\\./g"

works (outputs 1\.2\.3\.4), but

ip_addr_escaped=`echo $ip_addr | sed "s/\./\\\./g"`
echo $ip_addr_escaped

Doesn't (outputs 1.2.3.4)

How can I correctly escape the address?

Edit: It looks like

ip_addr_escaped=`echo $ip_addr | sed "s/\./\\\\\\\./g"`

works, but that's clearly awful!

Answer 1

bash parameter expansion supports pattern substitution, which will look (slightly) cleaner and doesn't require a call to sed:

echo ${ip_addr//./\\.}

Answer 2

Yeah, processing of backslashes is one of the strange quoting-related behaviors of backticks `...`. Rather than fighting with it, it's better to just use $(...), which the same except that its quoting rules are smarter and more intuitive. So:

ip_addr_escaped=$(echo $ip_addr | sed "s/\./\\\./g")
echo $ip_addr_escaped

But if the above is really your exact code — you have a parameter named ip_addr, and you want to replace . with \. — then you can use Bash's built-in ${parameter/pattern/string} notation:

ip_addr_escaped=${ip_addr//./\\.}

Or rather:

grep "${ip_addr//./\\.}" [FILE...]

Answer 3

Replace the double quotes with single quotes.