xsl match template by attribute except one - xslt

Question

I've a global match on an attribut in my stylesheet but I want to exclude the f - element. How can I do that?
Example XML:
<a>
<b formatter="std">...</b>
<c formatter="abc">...</c>
<d formatter="xxx">
<e formatter="uuu">...</e>
<f formatter="iii">
<g formatter="ooo">...</g>
<h formatter="uuu">...</h>
</f>
</d>
</a>
Current solution:
<xsl:template match="//*[#formatter]">
...
</xsl:template>
I've tried something like this, but that didn't worked.
<xsl:template match="f//*[#formatter]">
...
</xsl:template>
<xsl:template match="//f*[#formatter]">
...
</xsl:template>

Answer 1

Either //f[#formatter] or f[#formatter] would have worked (the // is not necessary). When this XSLT is run on your example input:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="#* | node()">
<xsl:copy>
<xsl:apply-templates select="#* | node()" />
</xsl:copy>
</xsl:template>
<xsl:template match="*[#formatter]">
<xsl:element name="transformed-{local-name()}">
<xsl:apply-templates select="#* | node()" />
</xsl:element>
</xsl:template>
<xsl:template match="f[#formatter]">
<xsl:apply-templates select="node()" />
</xsl:template>
</xsl:stylesheet>
The result is:
<a>
<transformed-b formatter="std">...</transformed-b>
<transformed-c formatter="abc">...</transformed-c>
<transformed-d formatter="xxx">
<transformed-e formatter="uuu">...</transformed-e>
<transformed-g formatter="ooo">...</transformed-g>
<transformed-h formatter="uuu">...</transformed-h>
</transformed-d>
</a>
As you can see, the f is excluded. Does this answer your issue, or have I misunderstood what you want to do?