Re: how to set the pattern to get the node
How I can get the node which parent_ID =1 or parent_ID is not 1 but No other sibling node List_ID equal to current node parent_ID
The value of current() changes during the evaluation of a Path in an XPath expression. It is not pointing to the List element for which the math pattern is being tested.
Besides, you might want to rule out the List element under consideration from the test on the siblings..
You could try the following:
<xsl:template match="List"> <xsl:variable name="self" select="."/> <!-- current() can be shortened to . --> <xsl:choose> <xsl:when test="Parent_ID = 1"> <!-- found a valid case --> </xsl:when> <xsl:when test="not(preceding-sibling::List[List_ID = $self/@Parent_ID] or following-sibling::List[List_ID = $self/@Parent_ID])"> <!-- found second valid case --> </xsl:when> <!-- xsl:otherwise? --> </xsl:choose> </xsl:template>
PURCHASE STYLUS STUDIO ONLINE TODAY!
Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced!
Download The World's Best XML IDE!
Accelerate XML development with our award-winning XML IDE - Download a free trial today!
Subscribe in XML format