IF condition with Linktoview function

[abood89]

Hello 

is it possible to use this expression to determine a user permission then show the view ?

if(IN(LOOKUP(USEREMAIL(), Employees, Email, Department), LIST("manager","reception")),LINKTOVIEW("Home"),LINKTOVIEW("Enough Permission"))

I'm getting error message 

The expression is valid but its result type 'App' is not one of the expected types: Yes/No

[Rafael_ANEIC-PY]

Yes. it is possible, but you don't use a LINKTOVIEW in an view's showif, you use it on a navigation action, a view showif expects a yes or no value to determine if it must be visible or not, which is exactly the error message you're receiving, try this instead:

IN(LOOKUP(USEREMAIL(), Employees, Email, Department), LIST("manager","reception"))

[abood89]

im using the expression in a security filter not show if view. what im trying to achive is tht if the user has the permission then he can see the view otherwise a pop up message says you dont have enough permission view 

[Rafael_ANEIC-PY]

That's not how security filters work my friend, a security filter expects a Yes/No answer for each row in order to display it on any views that it concerns, and that's all that it does, there's no pop up if the user isn't allowed by a filter to see the rows, they just don't appear.

Most people just hide the views they don't want their users to see I believe, I'm not aware of a way to show a pop up just to inform the user that they can't see a specific view.

[abood89]

the pop up message is displayed in other view that's why i made a view with a name enough permission so if the user does not have permission then the enough permission view will show up 

[Rafael_ANEIC-PY]

I see, but the thing is you can't run an action by clicking on a view directly, unless you try to go to a view directly from an action using a deep link type of action.