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我有一些代碼來顯示/隱藏取決於條件的輸入按鈕:如何以編程方式打開/關閉按鈕?
settings = settingsManager.readSettings();
if (settings) {
$("#settings-back-button").show();
} else {
$("#settings-back-button").hide();
}
然而,這並不妨礙用戶從按下設備後退按鈕。我怎樣才能做到這一點?
settings = settingsManager.readSettings();
if (settings) {
$("#settings-back-button").show();
disableDeviceBackButton(); // How to implement this?
} else {
$("#settings-back-button").hide();
reEnableDeviceBackButton(); // How to implement this?
}
重複,請參閱:http://stackoverflow.com/questions/18211984/how-to-control-back-button-event-in-jquery-mobile – sina72 2015-02-24 12:35:41
我已更新我的問題。另一個問題是詢問如何控制後退按鈕,但這個問題是關於後退按鈕的切換。 – 2015-02-24 13:04:04