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xsl:attribute (was Re: URL query as table data)
- From: Antonio Bueno <atnbueno at webcsd dot com>
- To: Mike Brown <xsl-list at lists dot mulberrytech dot com>
- Date: Sun, 17 Feb 2002 18:10:48 +0100
- Subject: [xsl] xsl:attribute (was Re: URL query as table data)
- Organization: webcsd.com
- References: <200202170538.g1H5cmv70790@skew.org>
- Reply-to: xsl-list at lists dot mulberrytech dot com
Hello, everybody.
WARNING: XSLT newbie here.
Mike answered to Harry:
MB> Never mind. This is what you're trying to do:
MB> <A>
MB> <xsl:attribute name="HREF">
MB> <xsl:text>http://lookuphost.com/bin/lookup.cgi?term=</xsl:text>
MB> <xsl:apply-templates/>
MB> </xsl:attribute>
MB> <xsl:apply-templates/>
MB> </A>
I've tried to do the same but instead of A and HREF I was using OPTION
and VALUE. And it didn't work.
I wanted to transform
<canal id="CP">Canal +</canal>
into
<option value="CP">Canal +</option>
I'm using MSXML4 and this worked:
<xsl:template match="canal">
<option>
<xsl:value-of select="."/>
</option>
</xsl:template>
But with this I only get:
<option>Canal +</option>
Then I tried to apply Mike's answer:
<xsl:template match="canal">
<option>
<xsl:attribute name="value">
<xsl:value-of select="@id"/>
</xsl:attribute>
<xsl:value-of select="."/>
</option>
</xsl:template>
But it didn't work (the VBScript transformNode() method I'm using gave
me a numeric error).
This is what finally worked:
<xsl:template match="canal">
<xsl:element name="option">
<xsl:attribute name="value">
<xsl:value-of select="@id"/>
</xsl:attribute>
<xsl:value-of select="."/>
</xsl:element>
</xsl:template>
Why doesn't xsl:attribute works in the first try?
Is it something specific from MSXML4?
Thanks in advance
and greetings from Spain
Antonio
http://www.webcsd.com
mailto:atnbueno@webcsd.com
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