Get description out of 'Hyperlink or Picture' column in search template

Question

I have a simple list with a column of type 'Hyperlink or Picture' named 'TestHyperlink'. I've created a Managed Property at the Search Service Application screen for this column so it can be used later inside a Search Template. I do an index reset and fully crawl to make sure the system is aware of my newly created property.

Inside the Search Template I add the property to the ManagedPropertyMapping tag and create the following line:

console.log(ctx.CurrentItem.TestHyperlink)

The log shows me the url but not the description. I've read on the Internet it should return a csv value like this: 'url, description' but not in SP2016.

So my question in short; how can i get the description out of a 'Hyperlink or Picture' column in SP2016?

Answer 1

Try:

console.log(ctx.CurrentItem["TestHyperlink.desc"]);

or also, maybe:

console.log(ctx.CurrentItem.TestHyperlink.desc);

Answer 2

While waiting for a concrete answer I've used the following work around. I've split column 'TestHyperlink' into two other columns named 'TestHyperlink_Url' and 'TestHyperlink_Desc' formatted as type 'Text'. Afterwards I've created a Managed Property for mapping the new columns. As they are simple text columns it's now possible to use them in my Search Display Template.

Answer 3

Doing my own testing the Managed Property associated with the URL field only stores the URL and throws away the Description. In fact, if I search for the description I don't find it, but if I search for the URL I find that. So that means that during content processing it appears that the URL is the only part that is kept.

Answer 4

You should map your Managed Property to Crawled property ows_q_URLH_TestHyperlink.

Then your managed property will contain URL and description in format: "Url, Desc".

Then, in display template you can add some js to extract this url\description:

var UrlFull = ctx.CurrentItem.YourManagedProperty;
var UrlPath = (UrlFull .split(','))[0];
var UrlDesc = (UrlFull .split(','))[1];

to build hyperlink for example:

<div>Link: <a href="_#= $htmlEncode(UrlPath) =#_"> _#= UrlDesc =#_ </a></div>

Answer 5

You solution means that a need für every list und every link column to make a managed property und to make a special JS File for the libary to get the decription of the linkcolumn and the URL opened in a new tab.

my conclusio : it is not possible to get a generic soltion which renders all link typ column in a libary where it is connected als JS Link. :-(

But thank so much for bringing clarity in my mind !

best greetings chris