Sign in Register
Copied to clipboard

Flag this post as spam?

This post will be reported to the moderators as potential spam to be looked at


  • George 28 posts 71 karma points
    Jun 08, 2011 @ 10:32
    George
    0

    Getting all current nodes instead of all current documents

    Hi all;

    I have some C# code like the following:

    public enum DocumentType
        {
            CONTACT = 1221,
            EVENT = 1223,
            JOB = 1227,
            HOME_PAGE = 1256,
        };
    ...later on...
    from d in Document.GetDocumentsOfDocumentType((int)DocumentType.JOB)
    where d.Published && !d.IsTrashed
    select d
    // where document is umbraco.cms.businesslogic.web.Document

    i.e., I get all the current "job" documents for display in a list.

    Unfortunately, I get a lot of "duplicate" documents in the returned list. For example, I get the "same" document with ID 1575, 1568, 1601... etc. I think that most of these are previous revisions of the document.

    Does anyone know how I would change this code to only get "current" documents? I should probably be using something in the "Node" interface for this, but I didn't see a Node.GetNodesOfNodeType method?

    My Umbraco version is 4.6.1.

    Copy Link
  • Robert Foster 459 posts 1820 karma points MVP 3x admin c-trib
    Jun 08, 2011 @ 14:01
    Robert Foster
    0

    Bah. Now I have to write all that out again.  So here goes:

    Just using 

    from d in Document.GetDocumentsOfDocumentType((int)DocumentType.JOB)
    select d;

    should work ( at least it does in 4.7).  To get the results you are after from the Node framework, you need to write some xpath and check the results:

    var list = new List<Node>();
    var docAlias = new ContentType(DocumentType.JOB).Alias;

    XPathNodeIterator itNode = library.GetXmlNodeByXPath(string.format("\\{0}", docAlias));
    while (itNode.MoveNext()) {
        XmlNode nodeXmlNode = nodeXmlNode = ((IHasXmlNode)itNode.Current).GetNode();
        if (nodeXmlNode != null)
        {
            list.Add(new Node(nodeXmlNode));
        }
    }

    That should do it.  Not as pretty, and you could probably using LinqToXml instead to make it a little neater.  I haven't come across an umbraco library method to do this, but there may be one in uComponents?

    - Rob.

    Copy Link
  • George 28 posts 71 karma points
    Jun 17, 2011 @ 07:19
    George
    0

    OK, my bad, the original code works fine - my bug was elsewhere.

    I'm an idiot, sorry for any confusion. :)

    Copy Link

  • This forum is in read-only mode while we transition to the new forum.

    You can continue this topic on the new forum by tapping the "Continue discussion" link below.

Please Sign in or register to post replies