Question

I need to copy from input document to output document all attributes but one.

My input is like this:

<mylink id='nextButton' type='next' href='javascript:;' /> 

And I need output like this:

<a id='nextButton' href='javascript:;' /> 

If I use the following XSL:

<xsl:template match='mylink'>     <a><xsl:copy-of select='attribute::*'/></a> </xsl:template> 

I get all attributes to output like this:

<a id='nextButton' type='next' href='javascript:;' /> 

But I want to ignore the ‘type’ attribute. I’ve tried the following but none of them seems to work the way I need:

<xsl:copy-of select='attribute::!type'/> <xsl:copy-of select='attribute::!'type''/> <xsl:copy-of select='attribute::*[!type]'/> <xsl:copy-of select='attribute::not(type)'/> 

How should I write my stylesheet to get needed output?

Answer 1

Shortest form:

<xsl:template match='mylink'>     <a><xsl:copy-of select='@*[name()!='type']'/></a> </xsl:template> 

Longer one (that’s the first thing I came up with, I leave it for reference):

<xsl:template match='mylink'>     <a>      <xsl:for-each select='@*'>       <xsl:if test='name() != 'type''>        <xsl:attribute name='{name()}'><xsl:value-of select='.'/></xsl:attribute>       </xsl:if>       </xsl:for-each>     </a> </xsl:template>