angularjs弹出190天后回来

Question

这就是我希望这样做的方式，例如，如果你在短时间内关闭一个弹出窗口2天，例如1或2，它就不应该再这样做了直到190天。

var app = angular.module('PopUpUsers', []);
app.controller('ViewPopUp', ['$scope', '$timeout', function ($scope, $timeout) {
    $timeout(function () {
        $scope.PopUpViewDiv = true;
    }, 12500);
    $scope.ClosePopUp = function () {
        $scope.PopUpViewDiv = false;
    };
}]);

我认为我现在必须忽略这一点，这与您点击近距离并以某种形式的饼干记住它的次数相反。

问题只是：我如何完成以便他们能够记住弹出窗口关闭了多少次？

How can I get the click count

Create cookie with AngularJS

更新代码：

var app = angular.module('PopUpUsers', []);
app.controller('ViewPopUp', ['$scope', '$timeout', '$cookies', function ($scope, $timeout, $cookies) {
    var cookieWObject = $cookies.getObject('test');
    var count = 3;


    if (cookieWObject == count) {
        $scope.PopUpViewDiv = false;
    } else {
        $timeout(function() {
                $scope.PopUpViewDiv = true;
        }, 12500);

        $scope.ClosePopUp = function () {
            $scope.PopUpViewDiv = false;
            $cookies.putObject("test", count);
        };
    }
}]);