Generate XSL file from XML

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Hi All,

I have the following XMl File, I have to generate xsl with the list of all Titles and all their artist.

Can some one please help me

XML

<?xml version="1.0" encoding="ISO-8859-1"?>
<!-- Edited by XMLSpy® -->
<catalog>
    <cd>
        <title>Empire Burlesque</title>
        <artist>Bob Dylan</artist>
        <country>USA</country>
        <company>Columbia</company>
        <price>10.90</price>
        <year>1985</year>
    </cd>
    <cd>
        <title>Stop</title>
        <artist>Sam Brown</artist>
        <country>UK</country>
        <company>A and M</company>
        <price>8.90</price>
        <year>1988</year>
    </cd>
    <cd>
        <title>Stop</title>
        <artist>Sam12 Brown</artist>
        <country>UK</country>
        <company>A and M</company>
        <price>8.90</price>
        <year>1988</year>
    </cd>

</catalog>

Posted 14-Sep-10 11:06am

ArchanCM

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Comments

AspDotNetDev 14-Sep-10 17:37pm

XSL is used to transform one document into a document of another form (e.g., XML to HTML). What are you using the XSL for?

ArchanCM 14-Sep-10 17:41pm

Yes , I am using for the same . To display/transfer data from XMl to HTMl using XSL.

3 solutions

Solution 2

I have tried this

XML

<?xml version="1.0" encoding="ISO-8859-1"?>
<!-- Edited by XMLSpy® -->
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">
  <html>
  <body>
  <xsl:for-each select="catalog/cd">
               <xsl:value-of select="title"/>
               <xsl> : </xsl>
               <xsl:value-of select="artist"/>
               <xsl> , </xsl>
 </xsl:for-each>
</body>
  </html>
</xsl:template>
</xsl:stylesheet>

Output will be like : Empire Burlesque : Bob Dylan , Hide your heart : Bonnie Tyler , Hide your heart : Bonnie111 Tyler ,

I want output like
Empire Burlesque : Bob Dylan , Hide your heart : Bonnie Tyler , Bonnie111 Tyler
Can you please give some suggestion on this...

Posted 14-Sep-10 12:18pm

ArchanCM

Comments

AspDotNetDev 15-Sep-10 13:37pm

So you want to combine artists when they have the same song, and you want to remove the trailing comma?

ArchanCM 15-Sep-10 13:39pm

Yes.
Please suggest ...

Solution 4

Solution :

XML

<?xml version="1.0" encoding="ISO-8859-1"?>
<!-- Edited by XMLSpy® -->
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:key name="contacts-by-surname" match="cd" use="title" />
<xsl:template match="catalog">
	<xsl:for-each select="cd[count(. | key('contacts-by-surname', title)[1]) = 1]">
		<xsl:value-of select="title" />
		<b>  :  </b>
		<xsl:for-each select="key('contacts-by-surname', title)">
			<xsl:value-of select="artist" />
			<xsl:if test="position() != last()"> <b> , </b></xsl:if>
		</xsl:for-each>
		<xsl:if test="position() != last()">  <br /></xsl:if>
	</xsl:for-each>
</xsl:template>
</xsl:stylesheet>

Posted 16-Sep-10 11:54am

ArchanCM

Updated 16-Sep-10 13:20pm

v2

Comments

AspDotNetDev 16-Sep-10 19:22pm

Neat, I didn't know there was a "last()" function.

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AspDotNetDev · Accepted Answer · 2010-09-15T14:00:00

This page shows how to group items:

http://www.jenitennison.com/xslt/grouping/muenchian.html

Store the value in a string variable, then trim the last 3 characters off using string-length and substring:

As an alternative to string trimming, you can use an if statement, the position function, and the count function to determine whether or not to append the last comma (or the first comma, which would prevent you from having to use the count function):

Next time, be a little more specific when asking your question... there is no way I could have known that's what you wanted by reading your original question.

Generate XSL file from XML

3 solutions

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