User just for report (not using XTRF)

alina.chiteala 2 years ago in Something Else updated by Jussi Rautio 2 years ago 10


I need a thing which I do not know how it can be done, but I do hope you can find a solution, 

I need to add a list o people, which will not be users of the software. 

I just need to be able to link data to them. 

Meaning I create X project, I want the project manager to be X user and not me. 

The user will not be allowed access to XTRF. 

Can this be done?


In order to assign project to a PM, that PM has to be an active user of the XTRF Home Portal.

Yes, but how can I mark an user as active if there no license for that user (because that user will not actually use the software).

I do not get it: how a PM cannot be an active user of a system like XTRF?

What is that person's role?

It is not about PMs. 

It is about sale persons. They do not login in XTRF, they do not have access to the software. 

But when I create a project, I need to assign it to a sales persons - so we can calculate the revenue per each sales person.

Now, I am the only one that has a license and that uses the software. I cannot add other users because of the license limitations.

Then, how can I assign a project to a sales person if that person is not a XTRF user?



The way the system works, is everyone who needs a role in your process requires a licence (not vendors or clients). We have a licence for a sales person. We have a unique licence just for administrative support.  I do wish we could purchase limited use licences for a lower fee (that would reduce our efforts to share licences among a few staff), but that's not how it currently works.

If XTRF allowed for more variable pricing for less workflow intense roles, that would be appreciated. :)

Create a custom field for the sales person and use that as a filter.

I use categories for this. Categories are unlimited, while there are limits to how many custom fields you can create.

Is there? Can you elaborate?

Jussi, you can add a maximum of 10 custom fields of each type in every scope.

OK, that's more than enough for us.