Longest common prefix of two strings in bash

Question

I have two strings. For the sake of the example they are set like this:

string1="test toast" string2="test test" 

What I want is to find the overlap starting at the beginning of the strings. With overlap I mean the string "test t" in my above example.

# So I look for the command  command "$string1" "$string2" # that outputs: "test t" 

If the strings were string1="atest toast"; string2="test test" they would have no overlap since the check starts from the beginning and the "a" at the start of string1 .

Answer 1

In sed, assuming the strings don't contain any newline characters:

string1="test toast" string2="test test" printf "%s\n%s\n" "$string1" "$string2" | sed -e 'N;s/^\(.*\).*\n\1.*$/\1/' 

Answer 2

An improved version of the sed example, this finds the common prefix of N strings (N>=0):

string1="test toast" string2="test test" string3="teaser" { echo "$string1"; echo "$string2"; echo "$string3"; } | sed -e 'N;s/^\(.*\).*\n\1.*$/\1\n\1/;D' 

If the strings are stored in an array, they can be piped to sed with printf:

strings=("test toast" "test test" "teaser") printf "%s\n" "${strings[@]}" | sed -e '$!{N;s/^\(.*\).*\n\1.*$/\1\n\1/;D;}' 

You can also use a here-string:

strings=("test toast" "test test" "teaser") oIFS=$IFS IFS=$'\n' <<<"${strings[*]}" sed -e '$!{N;s/^\(.*\).*\n\1.*$/\1\n\1/;D;}' IFS=$oIFS # for a local IFS: (IFS=$'\n'; sed -e '$!{N;s/^\(.*\).*\n\1.*$/\1\n\1/;D;}' <<<"${strings[*]}") 

The here-string (as with all redirections) can go anywhere within a simple command.