Get substring from string with grep command

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 ?

ennuikiller · Accepted Answer · 2011-09-11 21:13:09Z

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}'

answered Sep 11, 2011 at 21:04

ennuikiller

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1 Comment

url can start with http and with www or simply the name of the file like image.png