I have two lists:


    ListA Lookup

Both lists have custom display forms. In the ListB display form, I'm using a SharePoint:FormField to display a link to the currently selected ListA item. The link, however, does not point to the correct default display form. I have verified that the form works and that if you go to the ListA view and select View Item, it displays in the proper default form. Am I doing something wrong here or missing something?

The SharePoint:FormField I'm using to display is:

<SharePoint:FormField runat="server" id="ff11{$Pos}" ControlMode="Display" FieldName="Assessment" __designer:bind="{ddwrt:DataBind('u',concat('ff11',$Pos),'Value','ValueChanged','ID',ddwrt:EscapeDelims(string(@ID)),'@Assessment')}"/>

The link that is generated from this is https://mysite/Lists/Assessments/DispForm.aspx?ID=12345, rather than the _layouts/15/listform.aspx suggested below.

  • What's the version of your SharePoint server, 2007,2010 or 2013? – JoannaW_MSFT May 3 '18 at 1:56
  • 2016 on-premises. – DrewB May 7 '18 at 11:09

What have you done on the SharePoint:FormField?

In SharePoint lists/libraries, once the lookup item link is clicked your get redirect to a page with the following url:


The PageType defines whether it’s the display or editform for the item, 4 means display form.

By default, it will open the default display form for the lookup item.

Clear the cache and re-open the site, try it again.

  • Thanks for the suggestion. I've cleared all the caches I could think of, restarted all the machines involved and it's still generating an incorrect link. For now, I just added a javascript that examines all the links on the page and substitutes the correct link for the incorrect one. It'll get the site out the door until I can figure out what's going on. – DrewB May 2 '18 at 11:27

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