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我有一些代码来显示/隐藏取决于条件的输入按钮:如何以编程方式打开/关闭按钮?
settings = settingsManager.readSettings();
if (settings) {
$("#settings-back-button").show();
} else {
$("#settings-back-button").hide();
}
然而,这并不妨碍用户从按下设备后退按钮。我怎样才能做到这一点?
settings = settingsManager.readSettings();
if (settings) {
$("#settings-back-button").show();
disableDeviceBackButton(); // How to implement this?
} else {
$("#settings-back-button").hide();
reEnableDeviceBackButton(); // How to implement this?
}
重复,请参阅:http://stackoverflow.com/questions/18211984/how-to-control-back-button-event-in-jquery-mobile – sina72 2015-02-24 12:35:41
我已更新我的问题。另一个问题是询问如何控制后退按钮,但这个问题是关于后退按钮的切换。 – 2015-02-24 13:04:04