我通过AJAX加载元素。其中一些只有当你向下滚动页面时才能看到。有什么方法可以知道元素现在是否在页面的可见部分?


当前回答

我需要检查可滚动DIV容器内元素的可见性

    //p = DIV container scrollable
    //e = element
    function visible_in_container(p, e) {
        var z = p.getBoundingClientRect();
        var r = e.getBoundingClientRect();

        // Check style visiblilty and off-limits
        return e.style.opacity > 0 && e.style.display !== 'none' &&
               e.style.visibility !== 'hidden' &&
               !(r.top > z.bottom || r.bottom < z.top ||
                 r.left > z.right || r.right < z.left);
    }

其他回答

这是我的纯JavaScript解决方案,如果它隐藏在一个可滚动的容器。

演示在这里(尝试调整窗口的大小)

var visibleY = function(el){
  var rect = el.getBoundingClientRect(), top = rect.top, height = rect.height, 
    el = el.parentNode
  // Check if bottom of the element is off the page
  if (rect.bottom < 0) return false
  // Check its within the document viewport
  if (top > document.documentElement.clientHeight) return false
  do {
    rect = el.getBoundingClientRect()
    if (top <= rect.bottom === false) return false
    // Check if the element is out of view due to a container scrolling
    if ((top + height) <= rect.top) return false
    el = el.parentNode
  } while (el != document.body)
  return true
};

编辑2016-03-26:我已经更新了解决方案,以考虑滚动过去的元素,所以它隐藏在可滚动容器的顶部。 编辑2018-10-08:更新到当滚动到屏幕上方的视图外时处理。

在打印稿

  private readonly isElementInViewPort = (el: HTMLElement): boolean => {
      const rect = el.getBoundingClientRect();
      const elementTop = rect.top;
      const elementBottom = rect.bottom;
      const scrollPosition = el?.scrollTop || document.body.scrollTop;
      return (
        elementBottom >= 0 &&
        elementTop <= document.documentElement.clientHeight &&
        elementTop + rect.height > elementTop &&
        elementTop <= elementBottom &&
        elementTop >= scrollPosition
      );

};

检查元素是否在屏幕上,而不是公认的检查div是否完全在屏幕上的方法(如果div比屏幕大,这将不起作用)。在纯Javascript中:

/**
 * Checks if element is on the screen (Y axis only), returning true
 * even if the element is only partially on screen.
 *
 * @param element
 * @returns {boolean}
 */
function isOnScreenY(element) {
    var screen_top_position = window.scrollY;
    var screen_bottom_position = screen_top_position + window.innerHeight;

    var element_top_position = element.offsetTop;
    var element_bottom_position = element_top_position + element.offsetHeight;

    return (inRange(element_top_position, screen_top_position, screen_bottom_position)
    || inRange(element_bottom_position, screen_top_position, screen_bottom_position));
}

/**
 * Checks if x is in range (in-between) the
 * value of a and b (in that order). Also returns true
 * if equal to either value.
 *
 * @param x
 * @param a
 * @param b
 * @returns {boolean}
 */
function inRange(x, a, b) {
    return (x >= a && x <= b);
}

我在我的应用程序中有这样一个方法,但它不使用jQuery:

/* Get the TOP position of a given element. */
function getPositionTop(element){
    var offset = 0;
    while(element) {
        offset += element["offsetTop"];
        element = element.offsetParent;
    }
    return offset;
}

/* Is a given element is visible or not? */
function isElementVisible(eltId) {
    var elt = document.getElementById(eltId);
    if (!elt) {
        // Element not found.
        return false;
    }
    // Get the top and bottom position of the given element.
    var posTop = getPositionTop(elt);
    var posBottom = posTop + elt.offsetHeight;
    // Get the top and bottom position of the *visible* part of the window.
    var visibleTop = document.body.scrollTop;
    var visibleBottom = visibleTop + document.documentElement.offsetHeight;
    return ((posBottom >= visibleTop) && (posTop <= visibleBottom));
}

编辑:此方法适用于I.E.(至少版本6)。请阅读评论以了解FF的兼容性。

我更喜欢使用jQuery expr

jQuery.extend(jQuery.expr[':'], {  
    inview: function (elem) {
        var t = $(elem);
        var offset = t.offset();
        var win = $(window); 
        var winST = win.scrollTop();
        var elHeight = t.outerHeight(true);

        if ( offset.top > winST - elHeight && offset.top < winST + elHeight + win.height()) {
            return true;    
        }    
        return false;  
    }
});

你可以这样用

$(".my-elem:inview"); //returns only element that is in view
$(".my-elem").is(":inview"); //check if element is in view
$(".my-elem:inview").length; //check how many elements are in view

你可以很容易地在滚动事件函数中添加这样的代码等,以检查它每次用户将滚动视图。