Question

能否改进,只有单角功能?

var show = $("#shows ul li");   
show.addClass("active");

$(show).each(function(c){
    var cvalue = $.cookie( show  + c);
   if ( cvalue ==  closed  + c ) { 
        $(this).css({display:"none"});
        $(this).removeClass( active ).addClass( inactive );
   };
}); 

$("#shows li.active").toggle(function(){
    var num = show.index(this);
    var cookieName =  show  + num;
    var cookieValue =  closed  + num;
    $(this).slideUp(500);
    $(this).removeClass( active );
    $.cookie(cookieName, cookieValue, { path:  / , expires: 10 }); 
},function(){
    var num = $(this).index(this);
    var cookieName =  show  + num;
    $(this).slideDown(500);
    $(this).addClass("active");        
    $.cookie(cookieName, null, { path:  / , expires: 10 });
});

$("#shows li.inactive").toggle(function(){
    var num = show.index(this);
    var cookieName =  show  + num;
    $(this).slideDown(500);
    $(this).addClass("active");
    $(this).removeClass( inactive );       
    $.cookie(cookieName, null, { path:  / , expires: 10 });
},function(){
    var num = show.index(this);
    var cookieName =  show  + num;
    var cookieValue =  closed  + num;
    $(this).slideUp(500);
    $(this).removeClass( active );
    $.cookie(cookieName, cookieValue, { path:  / , expires: 10 }); 
});

Answer 1

你可以做这样的事情。

$("#shows li").toggle(function(){

    var isactive = $(this).hasClass("active") ? true : false;

    var num = show.index(this);
    var cookieName =  show  + num;
    var cookieValue = null;

    if(isactive){
        cookieValue =  closed  + num;
        $(this).slideUp(500);
        $(this).removeClass( active );
    }else{
        $(this).slideDown(500);
        $(this).addClass("active");
        $(this).removeClass( inactive );       
    }

    $.cookie(cookieName, cookieValue, { path:  / , expires: 10 }); 

},function(){
    var isactive = $(this).hasClass("active") ? true : false;

    var num = $(this).index(this);
    var cookieName =  show  + num;
    var cookieValue = null;

    if(isactive){
        $(this).slideDown(500);
        $(this).addClass("active");        
    }else{
        cookieValue =  closed  + num;
        $(this).slideUp(500);
        $(this).removeClass( active ); 
    }
    $.cookie(cookieName, cookieValue, { path:  / , expires: 10 }); 
});

Answer 2

否则,情况会好:

function func1() {
    var num = show.index(this);
    var cookieName =  show  + num;
    var cookieValue =  closed  + num;
    $(this).slideUp(500);
    $(this).removeClass( active );
    $.cookie(cookieName, cookieValue, { path:  / , expires: 10 }); 
}

function func2() {
    var num = $(this).index(this);
    var cookieName =  show  + num;
    $(this).slideDown(500);
    $(this).addClass("active");        
    $.cookie(cookieName, null, { path:  / , expires: 10 });
}

$("#shows li.active").toggle(func1,func2);
$("#shows li.inactive").toggle(func2,func1);

Answer 3

如果不看到你的标记,你就能够肯定地将这些标记减少到一个短的<条码>。例如,使用<条码>至ggleClass和<条码>滑坡Toggle<>。方法:

$("#shows li").toggle(function(){
    var num = show.index(this);
    $(this).slideToggle(500);
    $(this).toggleClass( active );
    var cookieName =  show  + num;
    var cookieValue = ($(this).hasClass("active") ?  show  :  closed ) + num;
    $.cookie(cookieName, cookieValue, { path:  / , expires: 10 }); 
}

您是否真正需要<条码>、活性和<条码>。我建议,这些要素只是为了在<代码>上(活性)/代码”类别之间(而任何内容都意味着不活动)。

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