Re: Ahhh.. URL wants to get encoded. Does Java wanna?

Wayne <nospam@all4me.invalid>
Tue, 06 Nov 2007 03:06:34 -0500
Wayne wrote:

Roedy Green wrote:

On Tue, 06 Nov 2007 05:04:05 -0000, Fran?ois
<> wrote, quoted or indirectly quoted someone
who said :

. Just want to encode a string into a
readable URL (RFC2396:



I just tried using URI, it doesn't seem to escape/encode
an ampersand in any part of the URI. Also, what about the
new IRIs? A Java program should be robust enough to
handle legal URLs/URIs/IRIs, converting the the (upto)
nine parts of an IRI correctly. My understanding of
your (excellent) urlencoded page and the API docs means this:

      URI uri = new URI("http", "// & I 10%? wierd & wierder", null);
      System.out.println( uri.toURL() );

should produce:
But it produces:

(The ampersand is not encoded.) What did I do wrong?


I guess the answer is to encode the query part separately, if needed.
The following code seems to work:

public String encodeURL ( String initialURL, boolean parseQuery )
  // Parse the URL (without encoding):
  URL url = new URL( initialURL );
  String scheme = url.getProtocol(); // E.g., "http"
  String authority = url.getAuthority(); // E.g., "//user@host:port"
  String path = url.getPath(); // E.g., "/foo/bar.htm"
  String query = url.getQuery(); // E.g., "foo=bar" (starts with '?")
  if ( parseQuery )
     query = URLEncoder.encode( query, "UTF-8" );
  String fragment = url.getRef(); // I.e., the "anchor"

  // Assemble the encoded URL, using URI class to properly
  // encode each part:
  URI uri = new URI( scheme, authority, path, query, fragment );
  return uri.toString();


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