Question

我有一些代码可以显示/隐藏输入按钮，具体取决于条件：

settings = settingsManager.readSettings();
if (settings) {
   $("#settings-back-button").show();
} else {
   $("#settings-back-button").hide();
}

但是，这并不妨碍用户按下设备后退按钮。我怎么能这样做？

settings = settingsManager.readSettings();
if (settings) {
   $("#settings-back-button").show();
   disableDeviceBackButton(); // How to implement this?
} else {
   $("#settings-back-button").hide();
   reEnableDeviceBackButton();  // How to implement this?
}

Answer 1

我认为此代码应该有效：

<!DOCTYPE html>
    <html>
      <head>
        <title>Back Button Example</title>

        <script type="text/javascript" charset="utf-8" src="cordova.js"></script>
    <script type="text/javascript" charset="utf-8">

    var backButtonDisabled = false;
    // Wait for device API libraries to load
    //
    function onLoad() {
        document.addEventListener("deviceready", onDeviceReady, false);
    }

    // device APIs are available
    //
    function onDeviceReady() {
        // Register the event listener
        document.addEventListener("backbutton", onBackKeyDown, false);
    }

    // Handle the back button
    //
    function onBackKeyDown(e) {

        if (backButtonDisabled) {
            e.preventDefault();
        } 
    }

    function disableDeviceBackButton() {
        backButtonDisabled = true;
    }

    function reEnableDeviceBackButton(){
        backButtonDisabled = false;
    }

</script>
  </head>
  <body onload="onLoad()">
  </body>
</html>

它可能有一些改进，比如不使用全局变量（backButtonDisabled），你可以使用sessionStorage，但重要的部分是e.preventDefault();避免后退按钮默认行为当你不想要回去。

如何以编程方式切换后退按钮开/关？

1 个答案: