Removing an attribute from an element in xsl

[MedIt]

Hi,
I am new in using xsl.I am stuck in a simple task,which is trying to remove an atribute from a node,and keep the rest of the attributes of this element in the xslt.
I have tried a couple of options but to no use so far!
Can any one shed some light on me?
Here's a sample of my original xml:

<?xml version="1.0" encoding="UTF-8" ?>
<dom xmlns="http://www.cdisc.org/ns/odm/v1.2" xmlns:xsi="http://www.w3.org/2001/XMLSchemainstance"
xsi:schemaLocation="http://www.cdisc.org/ns/odm/v1.2 http://www.cdisc.org/schema/odm/v1.2.1/ODM121.xsd"
CreationDateTime="20060203T14:03:06" Granularity="All" AsOfDateTime="20060203T14:03:10">
/* children node here */

</dom>

And I want my transformed xml to be:
<?xml version="1.0" encoding="UTF-8" ?>
<dom xmlns="http://www.cdisc.org/ns/odm/v1.2" xmlns:xsi="http://www.w3.org/2001/XMLSchemainstance"
xsi:schemaLocation="http://www.cdisc.org/ns/odm/v1.2 http://www.cdisc.org/schema/odm/v1.2.1/ODM121.xsd"
CreationDateTime="20060203T14:03:06" Granularity="All" >
/* children node here */

</dom>

removing the 'AsOfDateTime' attribute of the dom element.
Thanks your help in advance!

MedIt

[jkmyoung]

AsOfDateTime

A typical copy template looks like so:

<xsl:template match="*">
  <xsl:copy>
    <xsl:copy-of select="@*">
    <xsl:apply-templates/>
  </xsl:copy>
</xsl:template>

So just change the copy-of to
<xsl:copy-of select="@*[name() != 'AsOfDateTime']">

[MedIt]

Thanks for your help!
I managed to alter a bit from what you specefied.
Thanks so much!

AsOfDateTime

A typical copy template looks like so:

<xsl:template match="*">
  <xsl:copy>
    <xsl:copy-of select="@*">
    <xsl:apply-templates/>
  </xsl:copy>
</xsl:template>

So just change the copy-of to
<xsl:copy-of select="@*[name() != 'AsOfDateTime']">