Javascript: extract URLs from string (inc. querystring) and return array

I just use URI.js -- makes it easy.

var source = "Hello www.example.com,\n"
    + "http://google.com is a search engine, like http://www.bing.com\n"
    + "http://exämple.org/foo.html?baz=la#bumm is an IDN URL,\n"
    + "http://123.123.123.123/foo.html is IPv4 and "
    + "http://fe80:0000:0000:0000:0204:61ff:fe9d:f156/foobar.html is IPv6.\n"
    + "links can also be in parens (http://example.org) "
    + "or quotes »http://example.org«.";

var result = URI.withinString(source, function(url) {
    return "<a>" + url + "</a>";
});

/* result is:
Hello <a>www.example.com</a>,
<a>http://google.com</a> is a search engine, like <a>http://www.bing.com</a>
<a>http://exämple.org/foo.html?baz=la#bumm</a> is an IDN URL,
<a>http://123.123.123.123/foo.html</a> is IPv4 and <a>http://fe80:0000:0000:0000:0204:61ff:fe9d:f156/foobar.html</a> is IPv6.
links can also be in parens (<a>http://example.org</a>) or quotes »<a>http://example.org</a>«.
*/

• https://github.com/medialize/URI.js
• http://medialize.github.io/URI.js/

try this

var expression = /[-a-zA-Z0-9@:%_\+.~#?&//=]{2,256}\.[a-z]{2,4}\b(\/[-a-zA-Z0-9@:%_\+.~#?&//=]*)?/gi;

you could use this website to test regexp http://gskinner.com/RegExr/

Following regular expression extract URLs from string (inc. query string) and returns array

var url = "asdasdla hakjsdh aaskjdh https://www.google.com/search?q=add+a+element+to+dom+tree&oq=add+a+element+to+dom+tree&aqs=chrome..69i57.7462j1j1&sourceid=chrome&ie=UTF-8 askndajk nakjsdn aksjdnakjsdnkjsn";

var matches = strings.match(/\bhttps?::\/\/\S+/gi) || strings.match(/\bhttps?:\/\/\S+/gi);

Output:

["https://www.google.com/search?q=format+to+6+digir&…s=chrome..69i57.5983j1j1&sourceid=chrome&ie=UTF-8"]

Note: This handles both http:// with single colon and http::// with double colon in string, vice versa for https, So it's safe for you to use. :)

You could use the regex from URI.js:

// gruber revised expression - http://rodneyrehm.de/t/url-regex.html
var uri_pattern = /\b((?:[a-z][\w-]+:(?:\/{1,3}|[a-z0-9%])|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}\/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))/ig;

String#match and or String#replace may help…