摘要: function ajaxFn(type,url,bool) { var xhr = new XMLHttpRequest(); xhr.open(type, url, bool); xhr.send(null); xhr.onreadystatechange = function() { if(xhr.readyState == 4) { console.log(... 阅读全文
posted @ 2018-07-11 09:56 adongP 阅读(141) 评论(0) 推荐(0) 编辑