Question

以下是形成的请求：

请求网址：https://remoteserverurl.docx 请求方法：GET 状态码：200 OK 远程地址：10.232.4.216:7317 推荐人政策：降级时不推荐人接受：text / html，application / xhtml + xml，application / xml; q = 0.9，image / webp，image / apng， / ; q = 0.8 接受编码：gzip，deflate，br 接受语言：en-US，en; q = 0.9 连接：保持活动状态 Cookie：isPageflowTouch = true; schemaId = 1; updCtx = true; typeId = 91433788276151561974313054830 主机：domain.test.com 不安全升级请求：1 用户代理：Mozilla / 5.0（Windows NT 6.1; Win64; x64）AppleWebKit / 537.36（KHTML，如Gecko）Chrome / 68.0.3440.106 Safari / 537.36

Answer 1

function getdocx(url){
var oReq = new XMLHttpRequest();
var arrayBuffer;
oReq.open('GET', url, true);
oReq.responseType = "arraybuffer";
oReq.onload = function (oEvent) {
  arrayBuffer = oReq.response; // Note: not oReq.responseText
   var binary = '';
   var bytes = new Uint8Array( arrayBuffer );
   var len = bytes.byteLength;
 for (var i = 0; i < len; i++) {
    binary += String.fromCharCode( bytes[ i ] );
 }
var contentfromDocx = window.btoa( binary );
//do what ever you want with this
oReq.send(null);

}

如何使用Ajax

1 个答案: