"Gary" == Gary Stewart <the(_dot_)stewarg(_at_)gmail(_dot_)com> writes:
Gary> On 30 Oct 2005 18:13:24 +0000, Colin Paul Adams
Gary> <colin(_at_)colina(_dot_)demon(_dot_)co(_dot_)uk> 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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