Webapi之文件下载

该方法有问题！！！直接输入api没有问题，但是用axios调用api则下载的excel没有内容，如果是直接点击a标签（标签href值为api）也没问题

范例说明：

• 前端：vue.js + element-ui
• 后端：c# webapi
• 其它：只尝试本地调试，未在服务器上验证

前端部分

// html
<el-button
          size="mini" round
          icon="el-icon-download"
          @click="downloadExcel">导出</el-button>

// 引用外部js文件，在exprot default{}前
import EXCEL from '../../common/HandleExcel.js'
// 调用api
downloadTemplate: function () {
        this.axios.get(this.docApi + '/template?filename=StudentTemplate.xlsx', {
          responseType: 'blob'
        }).then((res)=>{
          EXCEL.downloadFile(res, '学生信息导入模板');
        },(res)=>{
          console.log(res);
        });
      }

// 外部js文件,HandleExcel.js
const EXCEL = {
  downloadFile: function (data, fileName) {
    if(!data){
      return;
    }
    let url = window.URL.createObjectURL(new Blob([data]));
    let link = document.createElement('a');
    link.style.display = 'none';
    link.href = url;
    link.setAttribute('download', fileName);
    document.body.appendChild(link);
    link.click();
    document.body.removeChild(link);
  }
}
export default EXCEL;

后端部分

[RoutePrefix("api/handledoc")]
    public class HandleDocumentController : ApiController
    {
		static string fileUrl = @"/document/";  //文件路径
		//  api/handledoc/template?filename={...}
		        [Route("template")]
		        [HttpGet]
		        public HttpResponseMessage getDocTemplate(string filename)
		        {
		            var FilePath = System.Web.Hosting.HostingEnvironment.MapPath(fileUrl + filename);     //获取完整绝对路径
		            var stream = new FileStream(FilePath, FileMode.Open);
		            HttpResponseMessage response = new HttpResponseMessage(HttpStatusCode.OK);
		            response.Content = new StreamContent(stream);
		            response.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
		            response.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
		            {
		                FileName = HttpUtility.UrlEncode(Path.GetFileName(FilePath))
		            };
		            return response;
		        }
	}