Subject: [xsl] How to test if a value itn't in a nodeset From: "Jaime Stuardo" <jstuardo@xxxxxxxxxxx> Date: Thu, 7 Apr 2005 10:31:05 -0400 |
Hi all... If I have this xml <doc> <valid_combination> <ROW> <CAD>a</CAD> <COB>b</COB> </ROW> <ROW> <CAD>a</CAD> <COB>c</COB> </ROW> <ROW> <CAD>b</CAD> <COB>d</COB> </ROW> </valid_combination> <current_values> <ROW> <COB>b</COB> <DESC>d1</DESC> </ROW> <ROW> <COB>e</COB> <DESC>d2</DESC> </ROW> <ROW> <COB>f</COB> <DESC>d3</DESC> </ROW> <ROW> <COB>c</COB> <DESC>d4</DESC> </ROW> <ROW> <COB>d</COB> <DESC>d5</DESC> </ROW> </current_values> </doc> where CAD field is used to group mutually exclusive elements of current_value set, for instance, that means that I have to show 2 listings of elements of the current_values group. The first listing has to show these COB's: b, e, f, c ("d" is missing because "d" is mutually exclusive with "b" and "c" according to valid_combination configuration). The second listing has to show these COB's: e, f, d. If I use this XSLT: <xsl:apply-templates select="current_values/ROW[COB = //valid_combination[CAD = 'a']/COB]" /> <xsl:apply-templates select="current_values/ROW[COB = //valid_combination[CAD = 'b']/COB]" /> It show:s 1 list.- "b" and "c" 2 list.- "d" But I need to do the union with the other elements, not present in valid_combination group. I tried, for example for the first listing: <xsl:apply-templates select="current_values/ROW[COB = //valid_combination[CAD = 'a']/COB] | current_values/ROW[COB != //valid_combination/COB]" /> But it didn't work. I interpret that union as: "show elements of 'current_values' where their COB is present in 'valid_combination' under CAD = 'a' category (to include 'b' and 'c'), plus the elements where COB isn't in 'valid_combination' (to include only 'e' and 'f', not 'c')." That is the simplified scenario what I'm trying to achieve. Any help would be greatly appreciated. Thanks in advance Jaime
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