Hi all, I'm a bit stuck.  I'm dabbling in Umbraco and would like to have one of my navigation items point to an external link.  I have managed to do this by creating a new document type (Linkpage) using this as a template:

<%@ Master Language="C#" MasterPageFile="~/umbraco/masterpages/default.master" AutoEventWireup="true" %>
<%@ Import Namespace="umbraco.BusinessLogic" %>
<%@ Import Namespace="umbraco.cms.businesslogic.web" %>
<%@ Import Namespace="umbraco.presentation.nodeFactory" %>
<script runat="server">
  void Page_Load(object sender, System.EventArgs e)
  {
    // Get the current page
    Node currentNode = Node.GetCurrent();
    // verify that the property "URL" is not null and that it is not empty; then redirect 
    if (currentNode.GetProperty("URL") != null && currentNode.GetProperty("URL").Value.ToString().Trim() != "")
    {
        Response.Redirect(currentNode.GetProperty("URL").Value.ToString().Trim());
    }
  }
</script>
<asp:Content ContentPlaceHolderID="ContentPlaceHolderDefault" runat="server">
</asp:Content>

...and this as the navigation xslt file:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [
    <!ENTITY nbsp "&#x00A0;">
]>
<xsl:stylesheet
  version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:msxml="urn:schemas-microsoft-com:xslt"
  xmlns:umbraco.library="urn:umbraco.library"
  exclude-result-prefixes="msxml umbraco.library">


    <xsl:output method="xml" omit-xml-declaration="yes" />

    <xsl:param name="currentPage"/>

    <!-- Input the documenttype you want here -->
    <xsl:variable name="level" select="1"/>

    <xsl:template match="/">
   
 <ul id="topNavigation">
       <li class="home">
         <xsl:if test="$currentPage/@id = $currentPage/ancestor-or-self::* [@level=$level]/@id">
             <xsl:attribute name="class">home current</xsl:attribute>
         </xsl:if>
         <a href="/">Home</a>
       </li>
      <xsl:for-each select="$currentPage/ancestor-or-self::* [@level=$level]/* [@isDoc and string(umbracoNaviHide) != '1']">
  <li>
     <xsl:if test="@id = $currentPage/@id">
        <xsl:attribute name="class">current</xsl:attribute>
      </xsl:if>
<xsl:choose>
  <xsl:when test="name() = 'Linkpage'">
    <a href="{current()/uRL}">
      <xsl:value-of select="@nodeName" />
    </a>
  </xsl:when>
  <xsl:otherwise>
    <a href="{umbraco.library:NiceUrl(@id)}">
      <xsl:value-of select="@nodeName" />
    </a>
  </xsl:otherwise>
</xsl:choose>
  </li>
  
  
</xsl:for-each>
</ul>

    </xsl:template>

</xsl:stylesheet>

This works, which is awesome.

However I'd also like the option to make the link open in a new window using a checkbox in the content properties, like this:

As you can see, I've already made a new property called NewTab, but I can't work out how to use this property to influence the link and insert the vital 'target="_blank"' code.  Is there a way to do this?  My mind (and optimism) tells me it must be possible, but I just don't understand the code I'm working with well enough to figure it out.

I'm working with Umbraco 4.7.0.0, and can work with html and css easily enough, but have little to no experience with xslt or C#.

Please help!