摘要: var ajax = new XMLHttpRequest(); ajax.open("get", "http://gzmylike.wedei.com/zt/gzyanzhicepin/images/TheLaLaLa.mp3") ajax.responseType = "blob"; ajax. 阅读全文
posted @ 2018-06-23 11:31 MiniDuck 阅读(688) 评论(0) 推荐(0) 编辑