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XSLT - Loop through unknown node names

Posted on 2009-05-11
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Last Modified: 2013-11-18
My XSLT/XPATH is a bit rusty. The sample xml contains several <price> elements and I  need to figure out how sum up these values, but only the CHEAPEST of each service type.
 
So far, i've done this:

<xsl:for-each select="AvailabilitySearch/Services">
      <xsl:value-of select="sum(child::*/child::*/child::Price)"/>
</xsl:for-each>

This gets me the sum of all the price elements. However, I need to extend this to:
1/ Only Sum up the cheapest of each service i.e. the cheapest book, the cheapest cd, the chepest game
2/ Be able to skip a specified service e.g. Don't include the Books in the calculation, only sum up the cheapest cd and game.

To be able to sum up only the cheapest, i'm guessing that I should be doing a sort. As a test, I tried this:
<xsl:for-each select="child::*/child::*">
            <xsl:sort data-type="number" order="ascending" select="Price"/>
      <xsl:value-of select="Price"/>
      <br/>
</xsl:for-each>

However this sorts the WHOLE of the results by price, rather than the services individually.

Any help would be great. If anything isn't clear please let me know and i'll try and explain further.

Thanks.
<AvailabilitySearch>
	<Services>
		<Books>
			<Book id="1">
				<Name>Book 1</Name>
				<Code>1234</Code>
				<Price>10</Price>
			</Book>
			<Book id="1">
				<Name>Book 1</Name>
				<Code>5678</Code>
				<Price>20</Price>
			</Book>
		</Books>
		<CDS>
			<CD>
				<Name>CD 1</Name>
				<Code>9101112</Code>
				<Price>5</Price>
			</CD>
			<CD>
				<Name>CD 2</Name>
				<Code>13141516</Code>
				<Price>7</Price>
			</CD>
		</CDS>
                                           <Games>
			<Game>
				<Name>Game 1</Name>
				<Price>39.99</Price>
			</Game>
			<Game>
				<Name>Game 2</Name>
				<Price>42.99</Price>
			</Game>
			<Game>
				<Name>Game 3</Name>
				<Price>24.99</Price>
			</Game>
		</Games>
	</Services>
</AvailabilitySearch>

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Question by:hendrix500
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5 Comments
 
LVL 60

Assisted Solution

by:Geert Bormans
Geert Bormans earned 800 total points
ID: 24356019
Here is how to loop only the cheapest in their kind
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:template match="Services">
        <xsl:for-each select="*/*[not(Price > preceding-sibling::*/Price)][not(Price > following-sibling::*/Price)][1]">
            <xsl:copy-of select="."/>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

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LVL 60

Assisted Solution

by:Geert Bormans
Geert Bormans earned 800 total points
ID: 24356041
Here is your sum
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:template match="Services">
        <xsl:value-of select="sum(*/*[not(Price > preceding-sibling::*/Price)][not(Price > following-sibling::*/Price)][1]/Price)"/>
     </xsl:template>
</xsl:stylesheet>

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LVL 60

Accepted Solution

by:
Geert Bormans earned 800 total points
ID: 24356057
excluding books
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:template match="Services">
        <xsl:value-of select="sum(*[not(self::Books)]/*[not(Price > preceding-sibling::*/Price)][not(Price > following-sibling::*/Price)][1]/Price)"/>
     </xsl:template>
</xsl:stylesheet>

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LVL 1

Author Closing Comment

by:hendrix500
ID: 31580197
Perfect. Thanks for your quick answer.
0
 
LVL 60

Expert Comment

by:Geert Bormans
ID: 24365549
welcome
0

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