我在包含每個彈出窗口中的鏈接的頁面上有幾個彈出窗口。當我打開popover並單擊鏈接時，我會轉到正確的頁面。然後，當我在瀏覽器中單擊後退按鈕並單擊任何酥料餅的鏈接我得到的錯誤：angular-ui彈出窗口：點擊返回按鈕後點擊彈出窗口中的鏈接時發生錯誤
Cannot read property 'isOpen' of null:TypeError: Cannot read property 'isOpen' of null at https://dev.techdynamism.com/Assessment/Scripts/angular-ui/ui-bootstrap-tpls.js:3546:34 at Scope.$digest (https://dev.techdynamism.com/Assessment/Scripts/angularjs/angular.js:14275:36) at Scope.$apply (https://dev.techdynamism.com/Assessment/Scripts/angularjs/angular.js:14488:24) at HTMLHtmlElement. (https://dev.techdynamism.com/Assessment/Scripts/angularjs/angular.js:11351:24) at HTMLHtmlElement.jQuery.event.dispatch (https://dev.techdynamism.com/Assessment/Scripts/jquery/jquery.js:4430:9) at HTMLHtmlElement.elemData.handle (https://dev.techdynamism.com/Assessment/Scripts/jquery/jquery.js:4116:28)
如果我回去與大多數其他任何手段popovers頁面，更改URL，通過一個鏈接，或點擊後退按鈕並刷新，然後當我點擊一個鏈接時，我不會收到錯誤信息。