1

如何在 Phonegap 在线系统中使用 jQuery Mobile 处理 Android 中的物理后退按钮?我想显示确认退出应用程序(是 - 否)。我尝试了很多方法,但没有任何效果。

4

7 回答 7

8

尝试这个:

document.addEventListener("deviceready", onDeviceReady, false);

function onDeviceReady() {
    document.addEventListener("backbutton", onBackKeyDown, false); //Listen to the User clicking on the back button
}

function onBackKeyDown(e) {
    e.preventDefault();
    navigator.notification.confirm("Are you sure you want to exit ?", onConfirm, "Confirmation", "Yes,No"); 
    // Prompt the user with the choice
}

function onConfirm(button) {
    if(button==2){//If User selected No, then we just do nothing
        return;
    }else{
        navigator.app.exitApp();// Otherwise we quit the app.
    }
}
于 2013-09-05T07:49:45.420 回答
5

第一次点击时设置exitApp=true,所以第二次点击后退按钮会退出应用。但是设置一个间隔来将exitApp的状态更改为false。因此,当在 1 秒内单击两次后退按钮时,将退出应用程序。

document.addEventListener('deviceready', function() {
    var exitApp = false, intval = setInterval(function (){exitApp = false;}, 1000);
    document.addEventListener("backbutton", function (e){
        e.preventDefault();
        if (exitApp) {
            clearInterval(intval) 
            (navigator.app && navigator.app.exitApp()) || (device && device.exitApp())
        }
        else {
            exitApp = true
            history.back(1);
        } 
    }, false);
}, false);
于 2014-04-30T09:41:08.463 回答
2

//设备就绪函数

document.addEventListener('deviceready', function() {

    document.addEventListener("backbutton", ShowExitDialog, false);

}, false);

//按下后退按钮时的对话框

 function ShowExitDialog() {
        navigator.notification.confirm(
                ("Do you want to Exit?"), // message
                alertexit, // callback
                'My APp', // title
                'YES,NO' // buttonName
        );

    }

// 调用退出函数

 function alertexit(button){

        if(button=="1" || button==1)
        {

            device.exitApp();
        }

}
于 2013-08-12T10:18:39.190 回答
1

不确定 phonegap 是否发生了变化,但 Suhas 和 geet 的解决方案几乎只对我有用。(但绝对给了我让它工作所需的东西,谢谢!)后退按钮基本上坏了。

以下是使它对我有用的调整:

加载您的应用程序后,请执行以下操作:

        document.addEventListener("backbutton", onBackKeyDown, false);//hijack the backbutton

        function onBackKeyDown(e){
            var page = $.mobile.activePage.attr('id');
            xStat.rec("back button clicked from page:" + page);
            if (page == 'menuPage'){//are you on the 'root page' from which phonegap will exit?
                e.preventDefault();
                $.mobile.changePage('#aboutToExitAppPage');
            } else {
                window.history.back();//restore normal back button functionality
            }
        }


//somewhere else in your code for the "aboutToExit app" page
        $('#aboutToExitAppPage').on('pageinit', function(){
            $(this).find('#exitApp').on('click', function(){
                navigator.app.exitApp();//quit the app.
            });
        });

和 HTML

<div data-role="page" id="aboutToExitAppPage">
    <div data-role="header" id="" data-position="inline" data-backbtn="true" >
        <h1 class=>About to exit app</h1>
    </div>
    <div data-role="content" style="width:100%; height:100%; padding:0;">
        <ul id="" data-role="listview" data-inset="true" data-theme="c" data-dividertheme="b" data-role="fieldContain">
            <input id="exitApp" class="" type="button" value="Exit" data-theme="">
            <a href="#menuPage" data-role='button'>Main Menu</a>
        </ul>
    </div>
</div>
于 2014-04-24T23:16:39.770 回答
0

这对我有用。

<script type="text/javascript" src="cordova.js"></script>
<script type="text/javascript" src="js/index.js"></script>

用这个

<script type="text/javascript">
    document.addEventListener("deviceready", onDeviceReady, false);

        function onDeviceReady(){
            document.addEventListener("backbutton", function(e) {
                e.preventDefault();

                        if(confirm("Exit App?")) {
                            navigator.app.exitApp();
                        }
            }, false);
        }
    </script>

或者

<script type="text/javascript">
document.addEventListener("deviceready", onDeviceReady, false);

    function onDeviceReady(){
        document.addEventListener("backbutton", function(e) {
            e.preventDefault();

            press = press + 1;
            if(press == 1) {
                alert('Press again to exit');
            }

            if(press == 2) {
                navigator.app.exitApp();
            }
        }, false);
    }
</script>
于 2020-02-28T14:51:10.070 回答
0

尝试这个:

// When Device ready
document.addEventListener('deviceready', function() {
    document.addEventListener("clickBackbutton", ExitDialogPrompt, false);
}, false); 

function ExitDialogPrompt() {
    navigator.notification.confirm(
        ("Do you want to Exit?"), // message
        prompt, // callback
        'Your title', // title
        'YES,NO' // button Name
    );
}

function alertexit(button){
    if (button=="0" || button==1) {
       navigator.app.exitApp();
    }
}
于 2016-02-24T11:32:47.013 回答
0

如果用户单击是,则退出应用程序。首先通过运行安装对话框插件cordova plugin add cordova-plugin-dialogs

document.addEventListener("deviceready", onDeviceReady, false);

function onDeviceReady() {
    document.addEventListener("backbutton", onBackKeyDown, false); //Listen to the User clicking on the back button
}

function onBackKeyDown(e) {
    e.preventDefault();
    navigator.notification.confirm("Are you sure you want to exit ?", onConfirm, "Confirmation", "Yes,No"); 
    // Prompt the user with the choice
}

function onConfirm(button) {
    if(button==1){//If User selected Yes, then exit app
        navigator.app.exitApp();
    } else {
        return;
    }
}
于 2019-08-08T05:36:18.047 回答