On Fri, Mar 22, 2013 at 2:14 PM, Martin Holmes <mholmes(_at_)uvic(_dot_)ca>
wrote:
If I have a template matching an attribute, and producing one in the output
tree, like this:
<xsl:template match="@style">
<xsl:attribute name="style" select="."/>
</xsl:template>
Is there any way to know the name of the element in the result tree which is
the parent of the attribute being created?
Some context: I'm turning TEI @style attributes into HTML @style attributes
in the output, and I'd like to handle situations in which this kind of
input:
<hi rend="text-align: center;">Centred text</hi>
results in output that doesn't work:
<span style="text-align: center;">NOT centred because it's a span</span>
If I knew the output element was a <span> or element which is inline by
default, I could add "display: block" automatically to any @style attribute
that contains a block-level CSS property such as text-align. I don't want to
add "display: block" in all cases, because e.g. a <div> element might
already have a class which floats it.
Cheers,
Martin
A two-stage stylesheet pipeline would work, provided that the context
you need is either *all* from the original document or *all* from the
result document (but not pieces/parts from each). (Also provided no
particular resource constraint.)
Might also be stylistically better than munging it all into one
stylesheet, if doing so separates different components of the
presentation.
-Brian
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail:
<mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--