我的加载事件的问题是,当我测试页面时它不起作用(不要隐藏预加载器图像),但是当我把函数放在.ready()中时,函数可以工作(它隐藏)
这是代码:
JAVASCRIPT:
$(document).load(function(){
$("#loaderHolder").hide("fast");
});
$(document).ready(function(){
$('#slider').cycle();
$('.sf-menu').superfish({
autoArrows: false
});
$('.scroll').slimScroll({
height: '590px',
wheelStep:5,
size:'15px',
width:'590px',
position: 'left',
railColor:'#c5c5c5',
color:'#a2a1a1',
railVisible:true,
alwaysVisible:true,
distance: '565px'
});
$('.scroll').css('width','550px');
$('.gallery').colorbox();
$('#gallery img').hover(function(){ $(this).fadeTo(500, 0.3)}, function(){$(this).stop().fadeTo(500, 1)})
$("#home-link").click(function(){
if ($(".active").length == 0)
{
return ;
}
else
{
var active = $(".active");
active.css("display","inline-block");
active.hide("slide",{},700);
active.attr("class","vanished");
}
});
$("#about-link").click(function(){
if ($(".active").length == 0)
{
var hidden = $("#about");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
else
{
if ($("#about").attr("class") == "active")
{
return ;
}
else
{
var active = $(".active");
active.css("display","inline-block");
active.hide("slide",{},700);
active.attr("class","vanished");
var hidden = $("#about");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
}
})
$("#starters-link").click(function(){
if ($(".active").length == 0)
{
var hidden = $("#starters");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
else
{
if ($("#starters").attr("class") == "active")
{
return ;
}
else
{
var active = $(".active");
active.css("display","inline-block");
active.hide("slide",{},700);
active.attr("class","vanished");
var hidden = $("#starters");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
}
})
$("#gallery-link").click(function(){
if ($(".active").length == 0)
{
var hidden = $("#gallery");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
else
{
if ($("#gallery").attr("class") == "active")
{
return ;
}
else
{
var active = $(".active");
active.css("display","inline-block");
active.hide("slide",{},700);
active.attr("class","vanished");
var hidden = $("#gallery");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
}
})
$("#contacts-link").click(function(){
if ($(".active").length == 0)
{
var hidden = $("#contacts");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
else
{
if ($("#contacts").attr("class") == "active")
{
return ;
}
else
{
var active = $(".active");
active.css("display","inline-block");
active.hide("slide",{},700);
active.attr("class","vanished");
var hidden = $("#contacts");
hidden.show("slide",{},700);
hidden.attr("class","active");
}
}
})
});
答案 0 :(得分:1)
尝试:
$(window).load
而不是
$(document).load
答案 1 :(得分:1)
尝试$(window).load()而不是$(document).load()
$(window).load(function () {
// run code
});
答案 2 :(得分:0)
我认为$ .load是实际对给定url(第一个参数)执行AJAX请求的方法。如果你想在document.load事件上做一些事情,你必须使用$(document).ready()