java - 如何让用户以ajax成功方法下载文件？

Question

我正在尝试让用户从服务器下载文件。 我在我的 Controller 中使用 ServletOutputStream(这里是代码)

@RequestMapping(value = "/get-backup-file", method = RequestMethod.GET)
@ResponseBody
public void getBackupFile(
    HttpServletRequest request, 
    HttpServletResponse response) throws MalformedURLException, IOException {

    File backupFile = new File("PATH_TO_FILE");        

    ServletOutputStream out = response.getOutputStream();

    response.setContentType("application/octet-stream");
    response.setContentLength((int)backupFile.length());
    response.setHeader("Content-Disposition", "attachment; filename=\"" + "database backup" + "\"");

    FileInputStream in = new FileInputStream(backupFile);
    byte[] buffer = new byte[4096];

    int length;
    while( (length = in.read(buffer) ) > 0) {
        out.write(buffer, 0, length);
    }
    in.close();
    out.flush();        
} 

我的客户端是这样的:

      $.ajax({
        url: 'URL_TOCONTROLLER_METHOD',
        contentType: "application/octet-stream; charset=utf-8",
        type: 'GET',
        success: function(data) {
            console.log(data);
        },
        error: function(data) {   
            console.log("error");
        }
    });

当我 console.log 数据时，它具有文件的内容，但我希望用户下载该文件，而不仅仅是打印。如何让用户将数据保存为文件？

Answer 1

你必须发送文件存储的路径并打开它的成功功能，然后用户才能下载它

如果成功就是这样

{"status":"success","path":"temp\/Vehicle_Units_2013_11_04.xls"}

脚本是

success: function(msg)
                  {
                      if(msg.status=="session-expired")
                      {
                      window.location.replace("index.jsp");
                      }
                      if(msg.status=="success")
                      {
                          window.open(msg.path);
                      }

                  }