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

用香草语回答:

function isScrolledIntoView(el) {
    var rect = el.getBoundingClientRect();
    var elemTop = rect.top;
    var elemBottom = rect.bottom;

    // Only completely visible elements return true:
    var isVisible = (elemTop >= 0) && (elemBottom <= window.innerHeight);
    // Partially visible elements return true:
    //isVisible = elemTop < window.innerHeight && elemBottom >= 0;
    return isVisible;
}

其他回答

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。

isScrolledIntoView是一个非常必要的函数,所以我尝试了它,它适用于不高于视口的元素,但如果元素比视口大,它就不起作用了。要解决这个问题,只需改变条件

return ((elemBottom <= docViewBottom) && (elemTop >= docViewTop));

:

return (docViewBottom >= elemTop && docViewTop <= elemBottom);

请看这里的演示:http://jsfiddle.net/RRSmQ/

function isScrolledIntoView(elem) {
    var docViewTop = $(window).scrollTop(),
        docViewBottom = docViewTop + $(window).height(),
        elemTop = $(elem).offset().top,
     elemBottom = elemTop + $(elem).height();
   //Is more than half of the element visible
   return ((elemTop + ((elemBottom - elemTop)/2)) >= docViewTop && ((elemTop + ((elemBottom - elemTop)/2)) <= docViewBottom));
}

仅限Javascript:)

function isInViewport(element) {
  var rect = element.getBoundingClientRect();
  var html = document.documentElement;
  return (
    rect.top >= 0 &&
    rect.left >= 0 &&
    rect.bottom <= (window.innerHeight || html.clientHeight) &&
    rect.right <= (window.innerWidth || html.clientWidth)
  );
}

在这个伟大答案的基础上,你可以使用ES2015+进一步简化它:

function isScrolledIntoView(el) {
  const { top, bottom } = el.getBoundingClientRect()
  return top >= 0 && bottom <= window.innerHeight
}

如果你不关心顶部是否跳出窗口而只关心底部是否被看到,这可以简化为

function isSeen(el) {
  return el.getBoundingClientRect().bottom <= window.innerHeight
}

或者甚至是单行语句

const isSeen = el => el.getBoundingClientRect().bottom <= window.innerHeight