Hi Muller,

I think if you have created a view(Inside the Shared or View folder) of these blocks then you can directly render them using the below syntax

@Html.propertyFor(x=>x.CurrentPage.ContentAreaPropertyName)

It will automatically render all the blocks that are selected in that contentArea property.

And if don't have the view then you can enumerate through the item in that contentArea-

@foreach (var item in Model.CurrentPage.ContentAreaPropertyName)
{              
 <img src="@Url.ContentUrl(item.MobileImageBelow)" />  
 <p>@Html.Raw(item.Title) </p>          
}

But in that case, properties are not editable from On-Page-Editing view. So I suggest you go with the first approach and create a view for the block

Hi Muller,

Do you want to perform this?

@foreach (var item in Model.MainContent.FilteredItems.OfType<BlockData>())
{
    if (item is CalloutBlock calloutBlock)
    {
        @Html.PropertyFor(x => calloutBlock.Title)
    }
    else if (item is ParagraphBlock paragraphBlock)
    {
        @Html.PropertyFor(x => paragraphBlock.CopyTop)
    }
}

Model.MainContent is the ContentArea property and you must be aware about FilteredItems (returns the publish items).

Here "is" will check if the type is a match then it will initialize that variable for example calloutBlock or paragraphBlock. You will have a strongly type object.

Thanks

I think the last example will not work because FilteredItems contains ContentAreaItems. You will have to content load them. You may use the extension method GetContent from EPiServer.Core. ContentAreaItemExtensions.

@foreach (var item in Model.MainContent.FilteredItems.Select(x => x.GetContent()))
{
    if (item is CalloutBlock calloutBlock)
    {
        @Html.PropertyFor(x => calloutBlock.Title)
    }
    else if (item is ParagraphBlock paragraphBlock)
    {
        @Html.PropertyFor(x => paragraphBlock.CopyTop)
    }
}

So, here FilteredItems will give you ContentAreaItem type in return. But to get the referenced item in backend we use ContentLoader by passing content link. Otherwise you can use the TemplateDescriptor and Tags to resolve the different view. For example, while rendering property use this

@Html.PropertyFor(x => x.MainContent, new { CssClass = "row", Tag = "MyTagName" })

For each CalloutBlockController use

[TemplateDescriptor(
AvailableWithoutTag = false,
Default = false,
ModelType = typeof(CalloutBlock),
Tags = new[] { "MyTagName"})]
public class CalloutBlockHomeController : BlockController<CalloutBlock>
{
}

For each ParagraphBlockController use

[TemplateDescriptor(
AvailableWithoutTag = false,
Default = false,
ModelType = typeof(ParagraphBlock),
Tags = new[] { "MyTagName"})]
public class ParagraphBlockHomeController : BlockController<ParagraphBlock>
{
}

Now, when you add even the CalloutBlock or ParagraphBlock into the ContentArea, it will resolve that view particular view and you will have your strongly typed object as model on that view.