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RE: Escaping of URI References

From: Kay, Michael <Michael.Kay@softwareag.com>
Date: Fri, 12 Jul 2002 00:54:51 +0200
Message-ID: <DFF2AC9E3583D511A21F0008C7E6210602679F64@daemsg02.software-ag.de>
To: Bjoern Hoehrmann <derhoermi@gmx.net>, public-qt-comments@w3.org

XSLT 1.0 does URI escaping ("of non-ASCII characters") as part of the HTML
serialization method, when attributes known to be URIs are output. This has
some recognized limitations, for example browsers can be unhappy if the
fragment part of the URI reference is escaped. In XSLT 2.0 we are planning
to provide an option to switch off this automatic escaping and to give the
user control on a more selective basis using a function.

We have drafted and discussed such an escape-URI() function and unless there
are any last-minute hitches I hope it will appear in the next draft.

Michael Kay

> -----Original Message-----
> From: Bjoern Hoehrmann [mailto:derhoermi@gmx.net] 
> Sent: 11 July 2002 20:02
> To: public-qt-comments@w3.org
> Subject: Escaping of URI References
> 
> 
> 
> 
> 
> 
> Hi,
> 
> The current XQuery 1.0 and XPath 2.0 Functions and Operators 
> Working Draft states, that applying the string() function to 
> an instance of a anyURI data type results in a string without 
> any URI encoding applied to the anyURI for compatibility 
> reasons. It further notes URI escaping should be under user 
> control. However, there is no function defined for URI 
> escaping. While I do not care too much about escaping space 
> characters, it is a major shortcoming if one is not able to 
> convert a anyURI to a URI. Consider you have an XML Schema 
> based XML document like
> 
>   ...
>   <link href='http://www.hoehrmann.de/~björn/' />
>   ...
> 
> The href attribute in this example is an anyURI. anyURIs 
> allow IRI References, hence the anyURI is valid. If I now 
> want to transform this document to XHTML, which uses URIs 
> instead of anyURIs, I can do something like
> 
>   <xsl:template match='link'>
>     <xhtml:a href='{@href}'><xsl:value-of select='@href'/></xhtml:a>
>   </xsl:template>
> 
> but I will get
> 
>   <a href='http://www.hoehrmann.de/~björn/'
>     >http://www.hoehrmann.de/~björn/</a>
> 
> This is invalid XHTML since the 'ö' is disallowed in URI 
> References. The desired output would be
> 
>   <a href='http://www.hoehrmann.de/~bj%C3%B6rn/'
>     >http://www.hoehrmann.de/~björn/</a>
> 
> But is is not possible to generate this fragment using the 
> function set provided by the draft. A possible solution is to 
> add a new function 'xf:anyURI-escape' or 'xf:anyURI-toURI' 
> that converts an anyURI to an URI as defined in section 
> 3.2.17 of XML Schema Part 2. However, I do not like this 
> solution, always using this function makes style sheets, 
> XPath querys, XQuerys, etc. rather hard to read, consider
> 
>   <xsl:template match='link'>
>     <xhtml:a href='{xf:anyURI-to-URI(@href)}'
>      ><xsl:value-of select='xf:anyURI-to-URI(@href)'/></xhtml:a>
>   </xsl:template>
> 
> real ugly compared to the above. XSLT 2.0 could add some 
> convenience method to perform this conversion implicitly. 
> This could be as easy as adding a new attribute to the 
> xsl:output element if it is considered that documents will 
> allow either URIs or anyURIs. However, such function is necessary.
> 
> regards.
> 
> 
Received on Thursday, 11 July 2002 18:54:57 UTC

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