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Rank: Devotee
Joined: 7/20/2006
Posts: 89
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Hi, Does anybody know how to get a document by its @nodeName in .net? I can run a query in XSLT and get it, or do Document.IsNode() but that is with the id ( http://forum.umbraco.org/16191 ) I tried to run the following snippet from the cutting edge example: http://old.umbraco.org/frontpage/documentation/articles/cuttingedgeensuringuniquenames.aspxCode:
foreach(umbraco.BusinessLogic.console.IconI d in documentObject.Parent.Children)
{
if (d.Id != documentObject.Id && d.Text == currentName) {
//currentName = documentObject.Text + " (" + uniqueNumber.ToString() + ")";
//uniqueNumber++;
// do my own stuff here
}
}
But nothing seems to cut it. No errors from the above example though, just no hit. Does anybody have some code to do this? Len.
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Rank: Addict
Joined: 7/19/2006
Posts: 597
Location: Bad Homburg, Germany
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With XMLDOM try umbraco.library.GetXmlNodeByXPath(string xpathQuery) or with objects iterate through all items in umbraco.presentation.nodefactory.node Thomas
• 2007/2008 MVP • www.thoehler.com • Bad Homburg, Germany
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Rank: Devotee
Joined: 7/20/2006
Posts: 89
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I trying to go in that direction: Code:
XPathNavigator xp = umbraco.content.xmlContent.CreateNavigator();
XPathNodeIterator nodes = xp.Select("//*");
string result = "";
while(nodes.MoveNext())
{
XPathNavigator node = nodes.Current;
result = result + node.GetAttribute("nodeName","").ToString();
result = result + node.GetAttribute("id","").ToString();
}
xpathResult.Text = "Result:" + result;
I will try with umbraco.presentation.nodefactory.node, so if it makes any difference Thx, Len
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Rank: Addict
Joined: 7/19/2006
Posts: 597
Location: Bad Homburg, Germany
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In xslt it is: Code:<xsl:copy-of select="umbraco.library:GetXmlNodeByXPath('.')/descendant-or-self::node [@nodeName = 'NODENAME']"/> in .Net it is Code:umbraco.library.GetXmlNodeByXPath("./descendant-or-self::node [@nodeName = 'NODENAME']") where you get an XPathNodeIterator Thomas
• 2007/2008 MVP • www.thoehler.com • Bad Homburg, Germany
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