Subject: [xsl] Positions From: "Tanzila Mohammad" <tmohammad@xxxxxxxxxxxxxxxxxxxx> Date: Tue, 13 Nov 2001 10:10:25 -0000 |
I have the following xsl: <xsl:for-each select="[full path]/literal"> <xsl:sort select="./@lang" data-type="text" order="ascending"/> ---------------------------------------------------------------------- which results in the following sorted XML: .... <literal name="number">15</literal> <literal name="table" lang="AR">ARABIC </literal> <literal name="table" lang="CH">CHINESE</literal> <literal name="table" lang="EN">TABLE</literal> <literal name="table" lang="ES"> CUADRO</literal> <literal name="table" lang="FR">TABLEAU</literal> ...... ------------------------------------------------------------------------ xsl cont...: <fo:block text-align="start"> <xsl:if test="./@name='table'"> <xsl:value-of select="."/> </xsl:if> ----------------------------------------------------------------------- Output: ARABIC CHINESE TABLE CUADRO TABLEAU ----------------------------------------------------------------------- Problem 1: I need the positions of "CUADRO" and "TABLEAU" to be reversed (see output below) I have played with position() but with no success - any suggestions?? ------------------------------------------------------------------------- Problem 2: I need the following output: ARABIC CHINESE TABLE 15 <!-- Display the 15 alongside TABLE--> CUADRO TABLEAU ------------------------------------------------------------------------- xsl cont... <xsl:if test="./@lang='EN'"> <xsl:value-of select="preceding-sibling::.[@name='number']"/> </xsl:if> </fo:block> </xsl:for-each> I am trying to get the value of the preceding sibling, where name=number, again with no success - any suggestions? Thanks Tanz XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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