angularjs : disable a button and show popup instead - angularjs

Question

I have this button :
html:
<button nav-direction="back" class="button yy" ui-sref="app.result" ui-sref-active="currentNav" ng-click="navResult()">
Board
</button>
I would like it to display a popup if a certain condition is, else I would like it to go to another page.
I need to keep the benefit of the class in ui-sref-active to show that this is the current page.
controller.js
$scope.navResult = function (){
console.log(sessionService.get('computed'));
if (sessionService.get('computed')) {
$scope.go('app.result');
} else {
//popup to user to tap on a board
//$scope.go('app.compute');
var popupConfig = {
title: 'Beware! ;)',
template: 'Tap on a board below'
};
var popup = $ionicPopup.show(popupConfig);
ClosePopupService.register(popup);
}
}
$scope.go = function ( state ) {
// console.log("go has been launched with : "+ state)
$state.go( state );
};

Answer 1

Simple. You just use an ng-click method instead of a ui-sref, and go to the state from there.
<button nav-direction="back" ng-class="{'your-class':classCondition}" class="button yy" ng-click="navResult()">
Board
</button>
Then in your controller....
$scope.navResult = function(){
if(something){
$scope.classCondition = false;
//code to display popup here
} else {
$state.go('app.result')
}
}
You can pass any valid state into $state.go, so if you ever want to check for a condition and perform some logic BEFORE you redirect to another page, use it inside a $scope method instead of just using the straight ui-sref.