Hi Karen,

In a strongly typed partial it would be like this:

 @inherits Umbraco.Web.Mvc.UmbracoTemplatePage
@{
    var currentRoot = Model.Content.AncestorOrSelf(1);

    var newsCollection = currentRoot.Children.Where(x => x.DocumentTypeAlias == "NewsFolder").First().Children.Where(x => x.DocumentTypeAlias == "News");
    foreach (var newsItem in newsCollection)
    {
        <p>@newsItem.GetPropertyValue("headline")</p>
    }
}

 Then to render in your View (assuming your partial is callewd News)

 @Html.Partial("News",Model)

 Hope that helps get your started.

Thanks,

Jeavon

Hi Jeavon,

Thanks!

I tried this, but am getting an error on the following line:

var newsCollection = currentRoot.Children.Where(x => x.DocumentTypeAlias=="NewsFolder").First().Children.Where(x => x.DocumentTypeAlias=="News");

Error:

Exception Details: System.InvalidOperationException: Sequence contains no elements

So an additional queston, is how to avoid the page throwing such an error message?

Hi Karen,

Ok, so that is happening because it cannot locate your NewsFolder, can I check that your tree structure is as follows (with each item being a parent node of the next)

Content (root)
   Homepage
     NewsFolder
       News
       News
     OtherPage 

Or is NewsFolder a sibling of Homepage?

Thanks,

Jeavon

Hi,

It is:

Content (root)
- Homepage
-- Some Pages
- NewsFolder
-- News

The xslt is working, so trying to base it off that

Thanks

Ah cool, ok so try:

 var newsCollection = currentRoot.Sibling("NewsFolder").Children.Where(x => x.DocumentTypeAlias == "News");

OK - that works.

But is there a way to make it better match the xslt?  That is, get the actual root node of the content tree (vs the 1st level), then get something under it.

In xslt:

<xsl:paramname="contentRoot"select="$currentPage/ancestor-or-self::root"/>

vs this seems to get 1 level under

var currentRoot = Model.Content.AncestorOrSelf(1);

So I want something like this (but this doesn't work)

var currentRoot = Model.Content.AncestorOrSelf();

var newsFolder = currentRoot.Children.Where(x => x.DocumentTypeAlias == "NewsFolder").First();

var newsCollection = newsFolder.Children.Where(x => x.DocumentTypeAlias == "News");

Hi Karen,

No I do not believe that there is a way to specifically select the root node because it is not a real node (it has no properties). You can select all the siblings of your current root node (all the children under the root node). e.g.

 var allRootNodeChildren = Model.Content.AncestorOrSelf(1).Siblings();

And from there you have access to every node within the Umbraco install.

Hope that's helpful?

Thanks,

Jeavon

OK thank you very much for your help!

You're very welcome! Thanks for your "High five!" above, that was my 1000th karma point :-)

Hi Karen,

I just discovered that in Umbraco v6.1+ there is a new helper method called TypedContentAtXPath that can do exactly what you need:

var newsCollection = Umbraco.TypedContentAtXPath("//News");

The method lets you execute a XPath expression against your entire tree.

Hope that's helpful!

Jeavon

Awesome!  Thanks!

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