Subject: Re: [xsl] Converting XML to an Escacped URL From: Colin Paul Adams <colin@xxxxxxxxxxxxxxxxxx> Date: 30 Oct 2005 18:50:57 +0000 |
>>>>> "Gary" == Gary Stewart <the.stewarg@xxxxxxxxx> writes: Gary> On 30 Oct 2005 18:13:24 +0000, Colin Paul Adams Gary> <colin@xxxxxxxxxxxxxxxxxx> wrote: Gary> Yeah; I am aware of that as a problem. I'm using it for Gary> testing really and the documents are quite short. I'll have Gary> to get POST working in the near future though :). >> Anyway, you have 3 functions in XPath 2.0 for escaping >> characters. I take it you are using XML 1.1? Gary> XML 1.0 though I could start using 1.1. Is there a way of Gary> getting the nodeset as a string then? The version of XML is irrelevant for this. I was wondering if 1.0 element names need escaping at all. But the text certainly will. So I think you want a stylesheet that specifies xsl:output method="text". Then a template somthing like: <xsl:template match="/"> data:text/html;charset=US-ASCII;base64,<xsl:apply-templates/> </xsl:template Then in subsequent templates write out the base64 encoding (you can write an xsl:function for this) of each node. Alternatively, you can omit the base64, and %encode everything with escape-html-uri (if that's the right function for the data URI scheme. (The RFC says: "without ";base64", the data (as a sequence of octets) is represented using ASCII encoding for octets inside the range of safe URL characters and using the standard %xx hex encoding of URLs for octets outside that range." ) -- Colin Adams Preston Lancashire
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