Re: Parsing XML files as text with collection()

Thanks Martin, I hadn't thought of uri-collection(). I'll use the 
approach that you suggested.

On 09.10.2020 16:17, Martin Honnen wrote:
> Am 09.10.2020 um 14:31 schrieb Imsieke, Gerrit, le-tex:
> 
>> I want a recursive directory scan to read in all XML files as unparsed
>> text. Then I want to remove the DOCTYPE declaration and named entities
>> with replace(). Then I want to parse them with parse-xml(). The reason
>> is that all kinds of DTDs at different locations will be referred, but I
>> don't have access to these DTDs, and they don't matter for the purpose
>> of my analysis.
>>
>> The documentation says that Saxon will use heuristics to infer the type
>> of file. How can I force Saxon to read every *.xml file, although it may
>> appear as an XML document, as text?
> 
> With Saxon (tested with HE 10.2 Java) from the command line I have no
> problems to use uri-collection and unparsed-text in XQuery or XSLT e.g.
> 
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
>      xmlns:xs="http://www.w3.org/2001/XMLSchema"
> exclude-result-prefixes="#all" version="3.0">
> 
>      <xsl:param name="uri" as="xs:string"
>          >file:/C:/SomePath/SomeFolder/</xsl:param>
> 
>      <xsl:param name="pattern" as="xs:string"><![CDATA[<\?xml
> .*?>\s*<!DOCTYPE .*?>]]></xsl:param>
> 
>      <xsl:template name="xsl:initial-template">
>          <xsl:sequence
>              select="uri-collection($uri || '?select=*.xml') !
> (unparsed-text(.) => replace($pattern, '') => parse-xml())"
>          />
>      </xsl:template>
> 
> </xsl:stylesheet>
> 
> 


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