Render page content via link field

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Joakim

Posted on 05/02/2026 09:32:15

Our client is running version 10.21.10. We're trying to render page content from a page recieved via link in backoffice.

So in Swift_ProductListInfo.cshtml we're using the ID from field CMS content page to render all content from that page.

This is easily solved with javascript but we're wondering if there is a method or another way to go about this?

The current code we've landed with is:

var page = Dynamicweb.Content.Services.Pages.GetPage(cmsContentPageIdValueParsed);
var pageViewModel = Dynamicweb.Frontend.ContentViewModelFactory.CreatePageInfoViewModel(page);

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Nicolai Pedersen

Posted on 05/02/2026 10:18:22

In your template you can call @RenderGrid(pageId) and get the grid output - but not the rest of the page.

You can get the markup of the page using PageView.GetPageviewAsTemplate()

var pageviewAsTemplate = PageView.GetPageviewAsTemplate(pageId);
<div><@pageviewAsTemplate.Output()</div>

You can execute a pageview:

var pageview = PageView.GetPageviewByPageID(pageId)
<div>pageview.Output()</div>

Joakim

Posted on 05/02/2026 10:48:16

Neat!

However, we wish to only render the content inside #content

Would this be achievable?

Nicolai Pedersen

Posted on 05/02/2026 15:37:43

This post has been marked as an answer

Depending on your implementation, that would be RenderGrid

Votes for this answer: 1

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Render page content via link field

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