display random node - Razor

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Nicky Christensen 76 posts 166 karma points

Oct 16, 2012 @ 23:36

display random node?

Hey guys...

I have a macro where i want to display a random node from umbraco, but it doesnt seem like I can use the Random() method? Any ideas to how I do this?

@* RAZOR *@
@inherits umbraco.MacroEngines.DynamicNodeContext

@{
  foreach (var item in @Model.NodeById(1399).Children.Where("NodeTypeAlias == \"FitnessAd\"").Random()) {
  
   <div class="ad">
     @Html.Raw(@item.fitnessAd)
   </div>
                                                                                        
  } 
}

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Tom Madden 253 posts 455 karma points MVP 4x c-trib

Oct 17, 2012 @ 00:22
0
Nicky,

AFAIK, Random() will return a single node, so you can't iterate through it using a foreach, instead, change your foreach line to the following:
```
varitem =@Model.NodeById(1399).Children.Where("NodeTypeAlias == \"FitnessAd\"").Random());
```
Remove the curly bracket from the end of that line as well as the closing curly from the foreach statement and you should be good to go

HTH

Tom
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Rusty Swayne 1655 posts 4993 karma points c-trib

Oct 17, 2012 @ 00:26

Hi Nicky ...

Maybe something like this - ?

@* RAZOR *@
@using umbraco.MacroEngines
@inherits umbraco.MacroEngines.DynamicNodeContext

@{
  dynamic ad = new DynamicNode(1399).ChildrenAsList.OrderBy(n => Guid.NewGuid()).FirstOrDefault();

  if(fitnessAd != null)
  {
     <div class="ad">
         @Html.Raw(@ad.fitnessAd)
     </div>
 }

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Rasmus Fjord 675 posts 1566 karma points c-trib

Oct 17, 2012 @ 08:36

I think ill try something like this.

Here you dont have to loop all items through just get a count and pick an item from your list based on the random class.(i know its not the best random and blah blah blah but here it gets the job done)

@* RAZOR *@

@inherits umbraco.MacroEngines.DynamicNodeContext

@{

int max = Model.NodeById(1399).Children.Where("NodeTypeAlias == \"FitnessAd\"").Count();

Random r = new Random();

int randomNumber = r.Next(0,max);

var item = Model.NodeById(1399).Children.Where("NodeTypeAlias == \"FitnessAd\"")[randomNumber];

    <div class="ad">

        @Html.Raw(@item.fitnessAd)

    </div>

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Nicky Christensen 76 posts 166 karma points

Oct 17, 2012 @ 10:12

1

Greay, thx alot guys :)

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Craig100 1136 posts 2523 karma points c-trib

Apr 15, 2014 @ 05:34

Is there an updated version of this for V6 MVC Razor.

Tried this but complains about the [random] on the end

@{
   if(Model.Content.AncestorOrSelf(1).Descendants("FeedbackItem").Any()) {
      int max = Model.Content.AncestorOrSelf(1).Descendants("FeedbackItem").Count(); 
      Random r = new Random();
      int randomNumber = r.Next(1,max);
      var item =  Model.Content.AncestorOrSelf(1).Descendants("FeedbackItem")[randomNumber];
      <h2>@item.feedbackBody</h2>
   }
}

Craig

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Daniel Williams 16 posts 38 karma points

Sep 24, 2014 @ 11:12

This is my solution:

var QuestionCollection = Universe.Descendants("Question").ToList();

 @if( QuestionCollection.Any() ) {
  Int32 questionCount = QuestionCollection.Count();
  Random random = new Random();
  Int32 randomNumber = random.Next( 0, questionCount );
  var Question = QuestionCollection[randomNumber];
}

@Question.Name

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Mark Downie 8 posts 28 karma points

Dec 08, 2016 @ 21:41

0

@foreach(var t in Umbraco.Content(1205).Children().Random(1))
{

@t.testimonailDetaik
@t.navigationName }

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