Subject: Re: How to get the preceding template on the same level From: "Sebastian Rahtz" <sebastian.rahtz@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> Date: Mon, 5 Jun 2000 21:37:12 +0100 (BST) |
Frederic Schwebel writes: > <mfrac> > <mn>2</mn><mn>3</mn> > </mfrac> > <mo>×</mo> > <mn>2</mn> > > the preceding axis would give me twice mn and one mfrac. I have no way to > know if mfrac is on the same level as ×. Maybe I am being dense, but don't you just want the preceding-sibling axis? that delivers your siblings up in nearest first order. so presumably if "name(preceding-sibling::*[1])" produces "mfrac", you fire. > And also : (2/3)+2*3 > for the preceding-axis of "*", I'd still get mfrac, but it's not the FIRST > predecessor on the same tree level so I would have to ignore it... How can do a <xsl:for-each select="preceding-sibling::*"> <!-- look at each sibling in turn --> </xsl:for-each> sebastian XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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