content.style.display != 'none'

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这对我来说是有效的,我用表演和隐藏来使我的 div 隐藏/可见:

if( $(this).css('display') == 'none' ){
    /* your code goes here */
} else {
    /* alternate logic   */
}

我搜索了这个, 没有一个答案 是正确的我的情况, 所以我创造了一个功能, 返回假的,如果一个人的眼睛看不到元素

jQuery.fn.extend({
  isvisible: function() {
    //
    //  This function call this: $("div").isvisible()
    //  Return true if the element is visible
    //  Return false if the element is not visible for our eyes
    //
    if ( $(this).css('display') == 'none' ){
        console.log("this = " + "display:none");
        return false;
    }
    else if( $(this).css('visibility') == 'hidden' ){
        console.log("this = " + "visibility:hidden");   
        return false;
    }
    else if( $(this).css('opacity') == '0' ){
        console.log("this = " + "opacity:0");
        return false;
    }   
    else{
        console.log("this = " + "Is Visible");
        return true;
    }
  }  
});

您需要检查... 显示和可见度 :

if ($(this).css("display") == "none" || $(this).css("visibility") == "hidden") {
    // The element is not visible
} else {
    // The element is visible
}

如果我们自动检查$( thiss) 。 is (“: 可见”) , jquery check for both the things 自动检查 。

要检查是否显示 I 使用 :

if ( !$('#book').is(':visible')) {
    alert('#book is not visible')
}

或以下是 sam , 将jquery 选择器保存在一个变量中, 以便在需要它多次时有更好的性能 :

var $book = $('#book')

if(!$book.is(':visible')) {
    alert('#book is not visible')
}

但如果元素的 css 和下面一样呢?

.element{
    position: absolute;left:-9999;    
}

因此,对于堆叠溢出问题的答案是,如何检查元素是否在屏幕之外,也应当考虑。