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Re: How to output attribute without namespace
- To: Dmitri Ilyin <dmitri dot ilyin at memIQ dot com>
- Subject: Re: [xsl] How to output attribute without namespace
- From: Jeni Tennison <mail at jenitennison dot com>
- Date: Fri, 7 Sep 2001 09:25:56 +0100
- CC: "'xsl-list at lists dot mulberrytech dot com'" <xsl-list at lists dot mulberrytech dot com>
- Organization: Jeni Tennison Consulting Ltd
- References: <07D07AD1C8CAD4119B0900D0B7E7E8E0649A90@UHURA>
- Reply-To: xsl-list at lists dot mulberrytech dot com
Hi Dmitri,
> It works.But the problem that i have is i don't know the local-name
> of attribute....
You can work out the local name of an attribute with the local-name()
function. Try:
<!-- match all attributes in the param namespace -->
<xsl:template match="@param:*">
<!-- create an attribute of that name in no namespace -->
<xsl:attribute name="{local-name()}">
<xsl:value-of select="." />
</xsl:attribute>
</xsl:template>
I hope that helps,
Jeni
---
Jeni Tennison
http://www.jenitennison.com/
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