How to extract domain name from url?

Question

How do I extract the domain name from a url using bash? like: http://example.com/ to example.com must work for any tld, not just .com

Answer 1

You can use simple AWK way to extract the domain name as follows:

echo http://example.com/index.php | awk -F[/:] '{print $4}'

OUTPUT: example.com

:-)

Answer 2

$ URI="http://user:pw@example.com:80/"
$ echo $URI | sed -e 's/[^/]*\/\/\([^@]*@\)\?\([^:/]*\).*/\2/'
example.com

see http://en.wikipedia.org/wiki/URI_scheme

Answer 3

basename "http://example.com"

Now of course, this won't work with a URI like this: http://www.example.com/index.html but you could do the following:

basename $(dirname "http://www.example.com/index.html")

Or for more complex URIs:

echo "http://www.example.com/somedir/someotherdir/index.html" | cut -d'/' -f3

-d means "delimiter" and -f means "field"; in the above example, the third field delimited by the forward slash '/' is www.example.com.

Answer 4

echo $URL | cut -d'/' -f3 | cut -d':' -f1

Works for URLs:

http://host.example.com
http://host.example.com/hi/there
http://host.example.com:2345/hi/there
http://host.example.com:2345

Answer 5

sed -E -e 's_.*://([^/@]*@)?([^/:]+).*_\2_'

e.g.

$ sed -E -e 's_.*://([^/@]*@)?([^/:]+).*_\2_' <<< 'http://example.com'
example.com

$ sed -E -e 's_.*://([^/@]*@)?([^/:]+).*_\2_' <<< 'https://example.com'
example.com

$ sed -E -e 's_.*://([^/@]*@)?([^/:]+).*_\2_' <<< 'http://example.com:1234/some/path'
example.com

$ sed -E -e 's_.*://([^/@]*@)?([^/:]+).*_\2_' <<< 'http://user:pass@example.com:1234/some/path'
example.com

$ sed -E -e 's_.*://([^/@]*@)?([^/:]+).*_\2_' <<< 'http://user:pass@example.com:1234/some/path#fragment'
example.com

$ sed -E -e 's_.*://([^/@]*@)?([^/:]+).*_\2_' <<< 'http://user:pass@example.com:1234/some/path#fragment?params=true'
example.com

Answer 6

#!/usr/bin/perl -w
use strict;

my $url = $ARGV[0];

if($url =~ /([^:]*:\/\/)?([^\/]+\.[^\/]+)/g) {
  print $2;
}

Usage:

./test.pl 'https://example.com'
example.com

./test.pl 'https://www.example.com/'
www.example.com

./test.pl 'example.org/'
example.org

 ./test.pl 'example.org'
example.org

./test.pl 'example'  -> no output

And if you just want the domain and not the full host + domain use this instead:

#!/usr/bin/perl -w
use strict;

my $url = $ARGV[0];
if($url =~ /([^:]*:\/\/)?([^\/]*\.)*([^\/\.]+\.[^\/]+)/g) {
  print $3;
}

Answer 7

Instead of using regex to do this you can use python's urlparse:

 URL=http://www.example.com

 python -c "from urlparse import urlparse
 url = urlparse('$URL')
 print url.netloc"

You could either use it like this or put it in a small script. However this still expects a valid scheme identifier, looking at your comment your input doesn't necessarily provide one. You can specify a default scheme, but urlparse expects the netloc to start with '//' :

url = urlparse('//www.example.com/index.html','http')

So you will have to prepend those manually, i.e:

 python -c "from urlparse import urlparse
 if '$URL'.find('://') == -1 then:
   url = urlparse('//$URL','http')
 else:
   url = urlparse('$URL')
 print url.netloc"

Answer 8

there is so little info on how you get those urls...please show more info next time. are there parameters in the url etc etc... Meanwhile, just simple string manipulation for your sample url

eg

$ s="http://example.com/index.php"
$ echo ${s/%/*}  #get rid of last "/" onwards
http://example.com
$ s=${s/%\//}  
$ echo ${s/#http:\/\//} # get rid of http://
example.com

other ways, using sed(GNU)

$ echo $s | sed 's/http:\/\///;s|\/.*||'
example.com

use awk

$ echo $s| awk '{gsub("http://|/.*","")}1'
example.com

Answer 9

The following will output "example.com":

URI="http://user@example.com/foo/bar/baz/?lala=foo" 
ruby -ruri -e "p URI.parse('$URI').host"

For more info on what you can do with Ruby's URI class you'd have to consult the docs.

Answer 10

Here's the node.js way, it works with or without ports and deep paths:

//get-hostname.js
'use strict';

const url = require('url');
const parts = url.parse(process.argv[2]);

console.log(parts.hostname);

Can be called like:

node get-hostname.js http://foo.example.com:8080/test/1/2/3.html
//foo.example.com

Docs: https://nodejs.org/api/url.html

Answer 11

One solution that would cover for more cases would be based on sed regexps:

echo http://example.com/index.php | sed -e 's#^https://\|^http://##' -e 's#:.*##' -e 's#/.*##'

That would work for URLs like: http://example.com/index.php, http://example.com:4040/index.php, https://example.com/index.php

Answer 12

With Ruby you can use the Domainatrix library / gem

http://www.pauldix.net/2009/12/parse-domains-from-urls-easily-with-domainatrix.html

require 'rubygems'
require 'domainatrix'
s = 'http://www.champa.kku.ac.th/dir1/dir2/file?option1&option2'
url = Domainatrix.parse(s)
url.domain
=> "kku"

great tool! :-)

Answer 13

Pure Bash implementation without any sub-shell or sub-process:

# Extract host from an URL
#   $1: URL
function extractHost {
    local s="$1"
    s="${s/#*:\/\/}" # Parameter Expansion & Pattern Matching
    echo -n "${s/%+(:*|\/*)}"
}

E.g. extractHost "docker://1.2.3.4:1234/a/v/c" will output 1.2.3.4