jQuery:如果在其他位置检测到单击，则隐藏弹出窗口

I'm trying to hide a div if the user clicks anywhere BUT the popup OR it's children. This is the code I have so far:$("body").click(function(){ var $target = $(event.target); if(!$target.is(".popup") || !$target.is(".popup").children()){ $("body").find(".popup").fadeOut().removeClass('active'); }});It works for the .popup div, but if any of it's children are clicked, it hides it anyway. 解决方案 You really could simplify this a bit I think:// If an event gets to the body$("body").click(function(){ $(".popup").fadeOut().removeClass("active");});// Prevent events from getting pass .popup$(".popup").click(function(e){ e.stopPropagation();});Clicking on the popup, or any of its children will cause propagation to stop before it reaches the body.Demo of stopping event-propagation: http://jsbin.com/ofeso3/edit 这篇关于jQuery:如果在其他位置检测到单击，则隐藏弹出窗口的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！