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Re: [xsl] most efficient way to get XML source's parent dir path

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At 2009-02-10 11:53 -0500, Robert Koberg wrote:

I want to get the path to the parent directory of the XML used as the
source in a transformation.

Is this the best way?

<xsl:variable
  name="path-tokens"
  select="tokenize(document-uri(/), '/')" as="xs:string*"/>
<xsl:variable
  name="source-dir-path"
  select="string-join(remove($path-tokens, count($path-tokens)),
'/')" as="xs:string"/>

I also unsuccessfully tried to use only one variable:

<xsl:variable
  name="source-dir-path-test"
  select="string-join(tokenize(document-uri(/), '/')[not(last())],
'/')" as="xs:string"/>

which results in an empty string.

Is there a better way?

How about replacing the last component of the URI with nothing:

select="replace(document-uri(/),'/[^/]*$','')"

I hope this helps.

. . . . . . . . . . Ken

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