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Wrting an image link with an XSLT

I'm calling the following XML feed;

http://www.google.co.uk/ig/api?weather=sydney

And thanks to some great help from Gertone transforming it with the following XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output indent="yes"/>
    <xsl:strip-space elements="*"/>
    
    <xsl:template match="/*">
        <xsl:apply-templates select="node()"/>
    </xsl:template>

    <xsl:template match="*">
        <div>
            <xsl:attribute name="class">
                <xsl:value-of select="concat(name(),'_data')"/>
            </xsl:attribute>
            <xsl:apply-templates select="node()|@data"/>
        </div>
    </xsl:template>
    
    <xsl:template match="forecast_information|current_conditions|forecast_conditions|icon|forecast_date">
        <div>
            <xsl:attribute name="class">
                <xsl:value-of select="name()"/>
            </xsl:attribute>
            <xsl:apply-templates select="node()|@data"/>
        </div>
    </xsl:template>
    
    <xsl:template match="postal_code|latitude_e6|longitude_e6|temp_f"/>

    <xsl:template match="low | high">
        <div>
            <xsl:attribute name="class">
                <xsl:value-of select="concat(name(), '_data')"/>
            </xsl:attribute>
            <xsl:call-template name="farenheit-to-celsius">
                <xsl:with-param name="temp" select="@data"/>
            </xsl:call-template>
        </div>
    </xsl:template>
    
    <xsl:template match="*" mode="id">
        <xsl:value-of select="concat(name(),'_data')"/>
    </xsl:template>
    
    <xsl:template name="farenheit-to-celsius">
        <xsl:param name="temp"/>
        <xsl:choose>
            <xsl:when test="not(string(number($temp)) = 'NAN')">
                <xsl:value-of select="format-number(($temp - 32) * (5 div 9) , '##.0')"/>
            </xsl:when>
            <xsl:otherwise>
                <xsl:value-of select="$temp"/>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>
</xsl:stylesheet>

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For the node where it renders

<div class="icon">/ig/images/weather/mostly_sunny.gif</div>

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I'd like it to render

<div class="icon"><img src="/myfolderpath/ig/images/weather/mostly_sunny.gif"></div>

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ASKER CERTIFIED SOLUTION

Gertone (Geert Bormans)

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Gertone (Geert Bormans)

maybe you need to explicitely set the output method to html

XSLT needs to be wellformed (hence the element closing for <img/>)

XML : <img ... />

html doesn't want that

HTML : <img ... >

set the method to html to get around that

<xsl:output indent="yes"/>
becomes
<xsl:output indent="yes" method="html"/>

sonic1234

ASKER

Top class help Gertone, your clear code is helping me learn how this works. Much appreciated.

Gertone (Geert Bormans)

Welcome