Subject: Re: [xsl] problem with proper numbering From: Joerg Pietschmann <joerg.pietschmann@xxxxxx> Date: Tue, 02 Oct 2001 18:51:14 +0200 |
"Andreas Putscher" <andreasputscher@xxxxxx> wrote > Hi everybody, [...] > So I would like to transform "slidesets" and "slides" into "cards". The > problem I have is concerned with the numbering of the "cards". [...] > So the position() function does not work in here as it starts numbering the > card-id with "a" again and again That's how position() works, by design. Look it up in the spec. >, but what I want is strictly ascending numbering of the id. The default answers are: a) Use <xsl:number count="slideset|slide" level="any" format="a"/> b) Use "count(preceding::slideset|preceding::slide)" instead of "position()" c) Use <card id="{generate-id()}"> ... instead of <card> <xsl:attribute name="id"><xsl:number ...></xsl:attribute> ... The last possibility uses a shortcut called attribute value template instead of the more verbose <xsl:attribute> you used. As a rule of thumb, c) should be faster than a) which in turn should be noticable faster than b), however, generate-id() generates usually strings of several characters, which might matter in the case of generating WML. See http://www.w3c.org/TR/xslt#number, http://www.w3.org/TR/xpath#axes, http://www.w3.org/TR/xpath#section-Node-Set-Functions and http://www.w3c.org/TR/xslt#misc-func Untested, try it out by yourself. HTH anyway J.Pietschmann XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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