如何召回一个功能并拥有 在这个例子中点击带有Jquery的“取消”按钮后,对话框会不断回来? 我相信这很容易,但仍然在这里学习一些基础知识。 感谢
function definitelyClose() {
window.location = 'http://www.google.com'
};
var autoCloseTimer;
var timeoutObject;
var timePeriod = 5000;
var warnPeriod = 10000;
$(document).ready(function() {
$('#proba').dialog({
autoOpen: false
});
setTimeout(function() {
$('#proba').attr('title', 'Warning').text('Sesion will expire').dialog('open');
$('#proba').dialog({
buttons: {
'Cancel': function() {
$(this).dialog('close');
clearTimeout(autoCloseTimer);
}
}
});
autoCloseTimer = setTimeout('definitelyClose()', warnPeriod);
}, timePeriod);
});
答案 0 :(得分:1)
您需要创建名称为
的函数所以你会得到这样的东西:
$(document).ready(function() {
var autoCloseTimer;
var timePeriod = 5000;
var warnPeriod = 10000;
function definitelyClose() {
window.location = 'http://www.google.com'
};
// You need a function with a name
function showWarning() {
$('#proba').attr('title', 'Warning')
.text('Sesion will expire')
.dialog('open');
$('#proba').dialog({
buttons: {
'Cancel': function() {
$(this).dialog('close');
clearTimeout(autoCloseTimer);
// Now you can recall the function
setTimeout(showWarning, timePeriod);
}
}
});
autoCloseTimer = setTimeout(definitelyClose, warnPeriod);
}
$('#proba').dialog({ autoOpen: false });
setTimeout(showWarning, timePeriod);
});