Question

I have the following pattern of a string:

"METHOD URL VERSION"

example "GET /someurl/resource.html HTTP/1.1"

How can I get the url from this string with grep.

I did the following(assuming that string contains in f.txt)

cat f.txt | grep -P '[^(( )|(\")|(HTTP\/\d\.\d)|(GET)|(POST))]+' -o

but it gives me such an output

someurl
resource
html

How can I retrieve /someurl/resource.html ?

Answer 1

This should do it:

grep -o "\/.* "

ok, if you have the pattern:

“somestring1withoutspaces somestring2withoutspaces somestring3withoutspaces” then both of these works:

grep -o " .* " 
awk '{print $2}'