我在包含每个弹出窗口中的链接的页面上有几个弹出窗口。当我打开popover并单击链接时，我会转到正确的页面。然后，当我在浏览器中单击后退按钮并单击任何酥料饼的链接我得到的错误：angular-ui弹出窗口：点击返回按钮后点击弹出窗口中的链接时发生错误
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)
如果我回去与大多数其他任何手段popovers页面，更改URL，通过一个链接，或点击后退按钮并刷新，然后当我点击一个链接时，我不会收到错误信息。