javascript - 怎么判断当前是否有某个class,如果没有则添加class?
ringa_lee
ringa_lee 2017-04-10 16:24:02
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判断当前元素是否有某个class,有就不添加,没有就添加当前class

ringa_lee
ringa_lee

ringa_lee

membalas semua(12)
PHPzhong

请使用hasClass

左手右手慢动作

结合上面总结了一下,我想出来的只有这几种:

function addClass0(targetEle, filterClassName){
    var target = document.getElementById(targetEle);
    var isExist = target && target.className.split('').indexOf(filterClassName)!=-1;
    if(!isExist){
        target.className = target.className + ' ' + filterClassName;
    }
}

function addClass(targetEle, filterClassName){
    var target = document.getElementById(targetEle);
    if(!target.classList.contains('add')){
         target.classList.add('add');
    }
}

function addClass1(targetEle, filterClassName){
     if(!$(targetEle).is('.'+filterClassName)){
         $(targetEle).addClass(filterClassName);
     }
}

function addClass2(targetEle, filterClassName){
     if(!$(targetEle).hasClass(filterClassName)){
          $(targetEle).addClass(filterClassName);
     }
}

function addClass3(targetEle, filterClassName){
     $(targetEle).not('.'+filterClassName).addClass(filterClassName);
}

function addClass4(targetEle, filterClassName){
     $(targetEle+':not(.'+filterClassName+')').addClass(filterClassName);
}

function addClass5(targetEle, filterClassName){
    $(targetEle).toggleClass(filterClassName);
}
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