如何在Javascript中使用Ajax使Android手机的后退按钮工作?

我有以下代码来处理Javascript框架7中设备的后退按钮。我想知道,如果所有后续页面都在使用Ajax,如何使用“后退”按钮返回到以前的页面。此时,按下后退按钮,如果用户想要退出应用程序,总是会提示用户。该应用程序有20个页面,但它们都显示为Ajax页面,而不是HTML页面。

<!-- We don't need full layout here, because this page will be parsed with Ajax-->

请帮忙。谢谢你的帮助!

        function onLoad() {
            document.addEventListener("deviceready", onDeviceReady, false);
        }
        // device APIs are available
        function onDeviceReady() {
            // Register the event listener
            document.addEventListener("backbutton", onBackKeyDown, false);
            document.addEventListener("menubutton", onMenuKeyDown, false);
        }
        // Handle the back button
        function onBackKeyDown(e) {
            e.preventDefault();
            navigator.notification.confirm("Are you sure you want to exit?", onConfirm, "Confirmation", "Yes,No");
            //navigator.app.backHistory();

            if (document.getElementById('#homepage')) {
                e.preventDefault();
                navigator.app.exitApp();
                navigator.notification.confirm("Are you sure you want to exit?", onConfirm, "Confirmation", "Yes,No");
            } else {
                navigator.app.backHistory();//This line doesn't work when tried on the if function - it seems useless - maybe not supported anymore?
                //the code never gets here coz other pages are simply loaded as Ajax, and there is only 1 index.html - how to fix?
            }
        }

        function onConfirm(button) {
            if (button == 2) { //If the user selected No, then we just do nothing
                return;
            } else {
                navigator.app.exitApp(); // Otherwise we quit the app.
            }
        }
        // Handle the menu button
        function onMenuKeyDown() {}