<xsl:template match="Field">
<xsl:element name="{(_at_)Name}">
<xsl:value-of select="."/>
</xsl:element>
</xsl:template>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: dan(_at_)streampad(_dot_)com [mailto:dan(_at_)streampad(_dot_)com]
Sent: 16 January 2006 22:53
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] trouble getting at node
I am trying to access the child node of specific nodes based on their
attributes, but am having trouble. Here is the xml:
<Item>
<Field Name="Filename">E:\Rap\DJ Mark Farina\Connect\Mark
Farina - Connect
- 08 - Martin Venetjoki , Really Don't Stop.mp3</Field>
<Field Name="Artist">DJ Mark Farina</Field>
<Field Name="Album">Connect</Field>
<Field Name="Name">Martin Venetjoki , Really Don't Stop</Field>
<Field Name="File Type">mp3</Field>
<Field Name="Genre">Chill</Field>
<Field Name="Date">36526</Field>
<Field Name="Bitrate">128</Field>
<Field Name="Media Type">Audio</Field>
<Field Name="File Size">4196480</Field>
<Field Name="Duration">262</Field>
<Field Name="Track #">8</Field>
<Field Name="Date Created">1055524929</Field>
<Field Name="Date Modified">1054230842</Field>
<Field Name="Date Imported">1055529132</Field>
</Item>
I would like to transform that into:
<Item>
<Filename>E:\Rap\DJ Mark Farina\Connect\Mark Farina - Connect - 08 -
Martin Venetjoki , Really Don't Stop.mp3</Filename>
<Artist>DJ Mark Farina</Artist>
...
</Item>
Is it possible to do this?
thanks,
Dan
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