have UX need process model called that return url to be opened in new tab.

Question

So I would like the user to click a link or a button in a UX and that launches a process model. The process model will determine a url that needs to be opened in a new tab. 
 I only want the process model to run when the user clicks the button or link. I have tried a!startprocess with success: but im not sure how to get the url opened in a new tab. 
 It successfully runs the process but nothing else seems to happen.Trying various things on the OnSuccess. 
 Httpreponse etc

Harshit Bumb (Appyzie) · Accepted Answer

For that to happen, you will need the user to click twice. 
 1. When the process needs to be initiated. 2. To open the URL in a new tab. 
 Now lets understand the sequence. First you start the process like you already doing and get the URL using onSuccess parameter. Make sure to add the chaining till the node that is generating this URL. Now when you have saved the URL in a local variable on your interface in link's onSuccess param, then you show a new link to the user which is present in the variable and add a safelink on it to open the same in a new tab. 
 NOTE - This cannot be achieved with just one click. User will have to click twice.