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  • Anthony Candaele 1197 posts 2049 karma points
    May 09, 2012 @ 15:19
    Anthony Candaele
    0

    how to get filename for document stored with DAMP

    Hi,

    I'm using DAMP for storing documents (doc, docx, pdf) in the Media Library.

    To render a link to the document on the page, I use this Razor script:

    @if (@Model.HasValue("activityDownload"))
    {
        var document = @Model.activityDownload.mediaItem.ActivityDocument.umbracoFile;
        <p>Download : <a href="@document">@document</a></p>
    }

    This creates a link to the document, but the problem is that the link label is also rendered as the relative path to the Media Library, while I only need the name of the file and it's file extension

    so in stead of rendering  '/media/1395/mydocument.docx' as label for the link, how can I make it render 'mydocument.docx'

    Thanks for your help,

    Anthony

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  • Anthony Candaele 1197 posts 2049 karma points
    May 09, 2012 @ 15:57
    Anthony Candaele
    0

    I solved it like this:

    @if (@Model.HasValue("activityDownload"))

    {

        var document = @Model.activityDownload.mediaItem.ActivityDocument.umbracoFile;

        string[] filename = document.Split(new char[]{'/'});

        <p>Download : <a href="@document">@filename[3]</a></p>

    }

    I was hoping that there just might be a handy Umbraco.Library helper method.

    Greetings,
    Anthony

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  • Jeroen Breuer 4909 posts 12266 karma points MVP 5x admin c-trib
    May 09, 2012 @ 16:00
    Jeroen Breuer
    1

    The path is /media/1395/mydocument.docx so that is what DAMP returns. If you want the filename you can do this:

    @using System.IO;
    var document = Path.GetFileName(Model.activityDownload.mediaItem.ActivityDocument.umbracoFile);

    This is not related to DAMP, but just a standard piece of C# to get the filename.

    Jeroen

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  • Anthony Candaele 1197 posts 2049 karma points
    May 09, 2012 @ 16:18
    Anthony Candaele
    0

    Hi Jeroen,

    thanks, your solution is more concise then my string.split() solution

    Thanks a lot,

    Anthony

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