The bellow code detects and converts urls in String. Is there faster way or more elegant way to do this block of code?:
public static String detectAndConvertURLs(String text) {
String[] parts = text.split("\\s");
String rtn = text;
for (String item : parts)
try {
// adjustment based one of the answers
Pattern p = Pattern.compile("((mailto\\:|(news|(ht|f)tp(s?))\\://){1}\\S+)");
Matcher m = p.matcher(item);
if( m.matches() ) item = m.group(1);
URL url = new URL(item);
String link = url.getProtocol() + "://" + url.getHost() + "/" + (url.getPath() == null ? "" : url.getPath()) + (url.getQuery() == null ? "" : "?" + url.getQuery());
开发者_如何学Python rtn = StringUtils.replace(rtn, item, "<a rel=\"nofollow\" href=\"" + link + "\">" + link + "</a> ");
} catch (MalformedURLException ignore) {
}
return rtn;
}
I think I would use regular expressions such as:
public static String detectAndConvertURLs(String text) {
//Regex pattern (unescaped), matches any Internet URL:
//((mailto\:|(news|(ht|f)tp(s?))\://){1}\S+)
Pattern p = Pattern.compile( "((mailto\\:|(news|(ht|f)tp(s?))\\://){1}\\S+)" );
Matcher m = p.matcher( text );
if( m.matches() ){
return m.group(1);
}else return null;
}
I just took that regex from this site of useful regex expressions:
http://regexlib.com/Search.aspx?k=URL
A quick google search will produce many regex expression resources:
http://www.google.com/search?q=regex+match+url&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a
You might have to massage the regex a little to trim stuff at the beginning or end depending on your use.
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