我创建了一个小脚本来处理低分辨率设备的响应式设计,并显示切换菜单而不是全宽菜单。以下脚本对我来说工作正常,但这对我来说看起来有点乱。我需要哪些简写或实践来使此代码更加简约和高效?
resetMenu() 函数正在处理以保留特定分辨率的 CSS,同时浏览器从正常模式调整大小到切换模式,这样做是一个好习惯吗?
$(document).ready(function($) {
function resetMenu() {
$('#top-menu li, #search-form, .social').css({"display":"block"});
$('#top-menu li').css({"display":"inline-block"});
};
$(window).resize(function () {
if($(window).width() > 640){
$(resetMenu());
}
else{
$('#top-menu li, #search-form, .social').css({"display":"none"}).fadeOut(1000);
$('#top-menu li:nth-child(2)').css({"display":"none"});
}
});
$(".togglebutton").toggle(
function () {
if($(window).height() < 360){
$('#top-menu li').css({"display":"inline-block"}).fadeIn(500);
$('#top-menu li:nth-child(2)').css({"display":"none"});
$('#search-form, .social').css({"display":"block"}).fadeIn(500);
$('#top-menu li').css({"border":"none"});
}
else{
$('#top-menu li, #search-form, .social').css({"display":"block"}).fadeIn(1000);
$('#top-menu li:nth-child(2)').css({"display":"none"});
}
},
function () {
$('#top-menu li, #search-form, .social').css({"display":"none"}).fadeOut(1000);
$('#top-menu li:nth-child(2)').css({"display":"none"});
}
)});