Subject: [xsl] forwarding only existing parameter From: Guillaume Rousse <rousse@xxxxxxxxxxxxxx> Date: Mon, 17 Sep 2001 12:26:13 +0200 |
Hello list. I'm trying to forward parameters from a template to another, only if they are defined, in the most simple way. <template match="foo"> <param name="bar"/> <apply-templates select="."> <with-param name="bar" select="$bar"/> </apply-templates> </template> This one actually forward an empty bar argument when not called with a bar argument, so it's wrong. <template match="foo"> <param name="bar"/> <apply-templates select="."> <if test="$bar"> <with-param name="bar" select="$bar"/> </if> </apply-templates> </template> This one is rejected as not xsl-compliant <template match="foo"> <param name="bar"/> <choose> <when test="$bar"> <apply-templates select="."> <with-param name="bar" select="$bar"/> </apply-templates> </when> <otherwise> <apply-templates select="."/> </otherwise> </choose> </template> This one is OK, but really ugly. Isn't there any other way ? -- Guillaume Rousse <rousse@xxxxxxxxxxxxxx> GPG key http://lis.snv.jussieu.fr/~rousse/gpgkey.html XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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