import io.Sourceimport scala.util.matching.Regex// gets the first ~10K of a pagedef get_page(url: String) : String = { try { Source.fromURL(url).take(10000).mkString } catch { case e => { println(" Problem with: " + url) "" } }}// staring URL for the crawlerval startURL = """http://www.inf.kcl.ac.uk/staff/urbanc/"""// regex for URLsval http_pattern = """\"https?://[^\"]*\"""".rval my_urls = """urbanc""".rval email_pattern = """([a-z0-9_\.-]+)@([\da-z\.-]+)\.([a-z\.]{2,6})""".r// http://net.tutsplus.com/tutorials/other/8-regular-expressions-you-should-know/def unquote(s: String) = s.drop(1).dropRight(1)def get_all_URLs(page: String) : Set[String] = { (http_pattern.findAllIn(page)).map { unquote(_) }.toSet}// naive version - seraches until a given depth// visits pages potentially more than oncedef crawl(url: String, n: Int) : Unit = { if (n == 0) () //else if (my_urls.findFirstIn(url) == None) () else { println("Visiting: " + n + " " + url) val page = get_page(url) println(email_pattern.findAllIn(page).mkString("\n")) for (u <- get_all_URLs(page)) crawl(u, n - 1) }}// can now deal with depth 3// start on command linecrawl(startURL, 3)