zoukankan      html  css  js  c++  java
  • jQuery检查元素是否在视口内(屏幕可见区域内)

    $.fn.isOnScreen = function(){
         
        var win = $(window);
         
        var viewport = {
            top : win.scrollTop(),
            left : win.scrollLeft()
        };
        viewport.right = viewport.left + win.width();
        viewport.bottom = viewport.top + win.height();
         
        var bounds = this.offset();
        bounds.right = bounds.left + this.outerWidth();
        bounds.bottom = bounds.top + this.outerHeight();
         
        return (!(viewport.right < bounds.left || viewport.left > bounds.right || viewport.bottom < bounds.top || viewport.top > bounds.bottom));
         
    };
  • 相关阅读:
    变量1
    PHP 函数
    发送post请求
    XXE
    CSRF
    Html基础
    暴力破解
    Brup sute
    XSS
    URL 传参转义 (特殊符号转义)
  • 原文地址:https://www.cnblogs.com/jiji262/p/3039582.html
Copyright © 2011-2022 走看看