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Updating variable value in loop in xslt

Hi,

For first article , I was creating url variable

<xsl:variable name="url" select="concat($base, $pathIs)"/>

which depends on other 2 variables as below:

<xsl:variable name="firstNews" select="article[1]" />
<xsl:variable name="pathIs" select="$firstNews/@xml"></xsl:variable>

Now, I have article variable in loop. I understand the variable value once assigned cannot be changed in XSLT
Therefore, I need to setup value for variable firstNews as per article pasition.
Something like this:
<xsl:variable name="firstNews" select="article[position()]" />

Please can you advise that below code is correct. If not please help me correcting.
I cannot test it before tomorrow morning.

<xsl:for-each select="article[position() &gt; 1 and position() &lt; 3]">

	<xsl:variable name="firstNews" select="article[position()]" />
	<xsl:variable name="pathIs" select="$firstNews/@xml"></xsl:variable>
	<xsl:variable name="url" select="concat($base, $pathIs)"/>
	
	<img width="246" height="175" class="latest-news-item-thumb">
		<xsl:attribute name="src">
			<xsl:value-of select="concat($imgPath, '/', document($url)//image[@type='S']/src)"/>
		</xsl:attribute>
	</img>
	
</xsl:for-each>

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I am working on ASP.net2.0 using C#

Please advise.

Regards,

Gertone (Geert Bormans)

Hmm, I don't see what you are trying to do

<xsl:for-each select="article[position() > 1 and position() < 3]">
<xsl:variable name="firstNews" select="article[position()]" />

inside the for each the context is the article
so the variable should rather be like this
<xsl:variable name="firstNews" select="." />

but I don't see you using the variable inside the for-each.
Notice that a variable does not live outside its scope (in this case the variable $firstNews can only be used inside the for-each)

tia_kamakshi

ASKER

Hi Gertone,

Thanks for your reply

I have variable called $base where it has the value where image resides for all the articles

My below xslt is reading summary xml something like this

<?xml version="1.0" encoding="utf-16"?>
<newsindex>
	<article id="944714"  xml="Website\xslt\ReadingAnotherXmlWithinXSLT\944714.xml" date="20120802134700" image="Phuket-tn_tcm133-944615.jpg" publish="20120802134700">
		<title>News 1</title>
		<summary>Lorem1 ipsum dolor sit amet.</summary>
	</article>
	<article id="944130"  xml="Website\xslt\ReadingAnotherXmlWithinXSLT\9447145.xml" date="20120802134700" image="Phuket-tn_tcm133-944634.jpg" publish="20130802134700">
		<title>News 2</title>
		<summary>Lorem2 ipsum dolor sit amet.</summary>
	</article>
	
</newsindex>

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In the xml attribute above, I have the path of detail xml where all the details can be seen.
From this detail xml I wanted to fetch the image name having [@type='S']/src

So, from the below code basically I wanted to fetch the image path:
            <xsl:attribute name="src">
                  <xsl:value-of select="concat($imgPath, '/', document($url)//image[@type='S']/src)"/>
            </xsl:attribute>

<xsl:for-each select="article[position() &gt; 1 and position() &lt; 3]">

	<xsl:variable name="firstNews" select="article[position()]" />
	<xsl:variable name="pathIs" select="$firstNews/@xml"></xsl:variable>
	<xsl:variable name="url" select="concat($base, $pathIs)"/>
	
	<img width="246" height="175" class="latest-news-item-thumb">
		<xsl:attribute name="src">
			<xsl:value-of select="concat($imgPath, '/', document($url)//image[@type='S']/src)"/>
		</xsl:attribute>
	</img>
	
</xsl:for-each>

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Please advise

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Gertone (Geert Bormans)

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tia_kamakshi

ASKER

Many Many Thanks

Gertone (Geert Bormans)

welcome