Subject: Re: [xsl] xsl:attribute (was Re: URL query as table data) From: Oleg Tkachenko <olegt@xxxxxxxxxxxxx> Date: Sun, 17 Feb 2002 20:26:46 +0200 |
<xsl:template match="canal"> <option value="{@id}"> <xsl:value-of select="."/> </option> </xsl:template>
-- Oleg Tkachenko Multiconn International, Israel
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@xxxxxxxxxx
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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