Hello Jamie, You can find in the attachment an XSLT stylesheet which generates an HTML table with the requested columns. It can be applied to XML files with the same structure (result/row/*) and with any number of child elements inside the "row" element. The names of the child elements are the names of the HTML table columns (see the <xsl:for-each select="row[1]/*"> instruction). Also each "row" element of the XML source fills a row in the HTML table (see <xsl:template match="row"> and <xsl:template match="row/*">). To apply the XSLT stylesheet just open it in <oXygen/> and press the "Apply Transformation Scenario" toolbar button, which will ask you to define a transformation scenario. Press the "New" button to open the Edit scenario dialog, in the "XML URL" field of the XSLT tab select your XML file and check the "Show As XHTML" checkbox on the Output tab. Press OK to end the definition of the new scenario and press the "Transform now" button to run the transformation and see the HTML table result in a special XHTML view at the bottom of the <oXygen/> window. I hope that helps, Sorin Jamie Wang wrote:
Hi, I am new to XSLT and have a question on how to dynamically generate columns in XSL style sheet that can be applied generic xml files that has the same document structure but with different number of columns and column names. The XML file is shown below. In this case, the number of columns is 3 and column names are the elements’ name (e.g., COUNTRY_ID, COUNTRY_NAME and REGION_ID. Thanks in advance for your assistance. -Jamie
<?xml version="1.0" encoding="ISO-8859-1"?>
<?xml-stylesheet type="text/xsl" href="http://rome:18080/countries.xsl"?>
<result>
<row>
<COUNTRY_ID>AR</COUNTRY_ID>
<COUNTRY_NAME>Argentina</COUNTRY_NAME>
<REGION_ID>2.0</REGION_ID>
</row>
<row>
<COUNTRY_ID>AU</COUNTRY_ID>
<COUNTRY_NAME>Australia</COUNTRY_NAME>
<REGION_ID>3.0</REGION_ID>
</row>
result>
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