is it possible to use a variable in xsl:include like e.g.
<xsl:variable name="home_directory"
select="document('../../XML/config.xml')//iese.config:home_dir
ectory"/>
<xsl:include
href="concat('file://',$home_directory,'class/Person/XSL/perso
n_short.xs
l')"/>
This example does not work! Does anyone know why? The parser
says something like wrong syntax for pathnames,... Any suggestions?
xsl:include must be able to be resolved at compile time, so it cannot
come from a variable.
The usual technique here is to put the variable code into stylesheets
that import the common code, rather that have a stylesheet that imports
stylesheets based on the variable.
Currently you are trying to do:
A
/ | \
/ | \
B C D
Where A imports B, C, or D based on some variable. Whereas you want to
do:
B C D
\ | /
\ | /
A
Where B, C, or D imports A. You can then decide which stylesheet to
apply to your source (b c or d) based on the variable.
cheers
andrew
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list