XF 2.1 Widget Display Conditions

[Jonathan Long]

Adding a couple widgets, and I need to have them display for certain situations/users.

The first one I want to display only to visitors. Got that one handled using this:

!$xf.visitor.user_id

Now, the second one I need to display only if someone IS a member of one specific user group AND NOT a member of another specific user group.

Banging my head against the wall and it is probably such a simple solution...any suggestions would be greatly appreciated.

 

[Chris D]

$xf.visitor.isMemberOf(123) AND !$xf.visitor.isMemberOf(456)

 

[Jonathan Long]

$xf.visitor.isMemberOf(123) AND !$xf.visitor.isMemberOf(456)

Thanks for the reply. the second one is a group the person is not a member of. Do I replace the “is” in the second one with “not”?

 

[Chris D]

Nope, the ! before the second one it is "not".

So in English the entire expression is "Visitor is member of 123 AND NOT member of 456".

Note: I made a typo in my previous message which I have just fixed.

 

[Jonathan Long]

Thanks, completely misses the

!

in there. Need my morning coffee. Thank you!

 

[Dubbed Navigator]

Do you have to use the user group name, or ID?

 

[Brad Padgett]

Do you have to use the user group name, or ID?

The user group ID, found by clicking on the user group in the admin panel. It's the number in the URL

 

[Dubbed Navigator]

That may be why I can’t get it to work then

 

[PapaTango]

Thanks to Chris and others!

Those ! operators are some tricky things...

My task was to ensure that a certain menu navigation button was not visible to any other but two select user groups. Using this expression form accomplishes that:

$xf.visitor.isMemberOf(2) OR $xf.visitor.isMemberOf(5)