js判断回车,判断焦点控件

  document.onkeydown=function(event){
        e = event ? event :(window.event ? window.event : null);
        if(e.keyCode==13){ //判断是否回车
          
            if(document.activeElement.id="EColChar6D"){ //焦点所在ID
              
                return false;
            }
            return false;
        }
    }

原文地址:https://www.cnblogs.com/jasonlwings/p/3199983.html