get single random node from mntp

Press Ctrl / CMD + C to copy this to your clipboard.

Copied to clipboard

Flag this post as spam?

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

Chris Heijnen 6 posts 26 karma points

Mar 10, 2013 @ 09:46
0

Get single random node from MNTP
Hi Guys,

The title actually explains my issue but what i want is to display 1 Random node from multiple nodes i have selected with the MNTP.

With this code i'm getting all the node (id's) and returning the property's in it, so that part works fine. Just want to display 1 randomly.
```
@{

    var topNode = Model.AncestorOrSelf(1);

    foreach (var reference in topNode.referencesPicker)
    {

        var referenceNode = Model.NodeById(reference.InnerText);

        <h3>@referenceNode.referenceName</h3>
        <span>@referenceNode.referenceOneliner</span>
        @referenceNode.referenceText

    }

 }
```
Chris
Copy Link
gary 385 posts 916 karma points

Mar 10, 2013 @ 10:26
0
Hi Chris
```
foreach(var reference in topNode.referencesPicker).Random(1)
```
should work for you.

G
Copy Link
Chris Heijnen 6 posts 26 karma points

Mar 10, 2013 @ 10:33
0
Hi Gary,

I think that's not going to work. Getting syntax error on that. If i write it down this way:
```
 foreach (var reference in (topNode.referencesPicker).Random(1))
```
I get this error: 'umbraco.MacroEngines.DynamicXml' does not contain a definition for 'Random'

any suggestions?
Copy Link
gary 385 posts 916 karma points

Mar 10, 2013 @ 10:40

0

Ah

The MNTP throws that one!

Found this answer here which you should be able to adapt if it doesn't work directly - it explains how it works.

http://our.umbraco.org/forum/developers/razor/22237-Random-uComponents-Multi-Node-Tree-Picker

Hope it helps G

Copy Link

Chris Heijnen 6 posts 26 karma points

Mar 10, 2013 @ 10:52

Hi Gary,

I already looked at that solution but i couldn't figure it out. Now with a fresh look i made it work!

This is the working code btw:

@{
var topNode = Model.AncestorOrSelf(1);

    var nodes = new DynamicNodeList();

    foreach(var referenceItem in topNode.referencesPicker){
        var n = new DynamicNode(referenceItem.InnerText);
        nodes.Add(n);
    }

    foreach (dynamic reference in nodes.Random(1))
    {

@reference.referenceName

        @reference.referenceOneliner
        @reference.referenceText

    }
}

Thanks!

Copy Link

is working on a reply...

Please Sign in or register to post replies

Flag this post as spam?

@reference.referenceName