- From: Julian Reschke <julian.reschke@gmx.de>
- Date: Mon, 23 Mar 2009 16:16:39 +0100
- To: Dan Connolly <connolly@w3.org>
- CC: public-html@w3.org
Hi Dan, I'm a bit confused by: > The parsing process described here should be more closely aligned with > the rules given in RFC 3987. > > 1. > > Strip leading and trailing space characters <#space-character> from w. > > 2. > > Percent-encode all non-URI characters in w. > > This probably needs to be laid out in more detail. > > Note: this step will replace all of the following characters with > a percent-encoded equivalent: > > * all characters with codepoints less than or equal to U+0020 > (i.e. the C0 control characters) > * all characters with codepoints greater than or equal to > U+007% (i.e. U+007?F and all non-ASCII characters in the w) > * U+0022 double quotation mark > * U+0025 percent sign > * U+003C less-than sign > * U+003E greater-than sign mark > * U+005C reverse solidus (backslash) > * U+005E circumflex accent > * U+0060 grave accent > * U+007B left curly bracket > * U+007C vertical line > * U+007D right curly bracket > > As a result of percent-encoding the percent sign, any occurrences > of percent-encoding in the Web address will be double-encoded at > this step. Why would you want that? It seems to mean that if w includes "%20" (a properly escaped space character), it will be encoded into "%2520". BR, Julian
Received on Monday, 23 March 2009 15:17:26 UTC