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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] How to select distinct combined element values
Dear All,
I got a problem in getting distinct element group values.
XML source:
+++++++++++
<match>
<date>1-11-2000</date>
<team>Brazil</team>
<group>A</group>
<score>1</score>
</match>
<match>
<date>2-11-2000</date>
<team>Brazil</team>
<group>A</group>
<score>1</score>
</match>
<match>
<date>1-11-2000</date>
<team>Brazil</team>
<group>B</group>
<score>1</score>
</match>
<match>
<date>1-11-2000</date>
<team>Argentina</team>
<group>A</group>
<score>1</score>
</match>
<match>
<date>1-11-2000</date>
<team>Argentina</team>
<group>A</group>
<score>1</score>
</match>
for distinct team :
In XSL, I can use
<xsl:variable name="teams" select = "//team[not(.=preceding::team)]"/>
to get
Brazil
Argentina
However, if I want to get distinct team, group, that is
Brazil A
Brazil B
Argentina A
How can I do it ? would Mike give me some suggestion ?
TIA
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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