
编辑2:首先-
请参阅底部的编辑1-这不是正确的方法。但是,如果无法使序列化程序正常工作,则可以使用以下解决方案,在该解决方案中,您将XML文件读入字符串,并提示用户保存它:
@RequestMapping(value = "/files", method = RequestMethod.GET)public void saveTxtFile(HttpServletResponse response) throws IOException { String yourXmlFileInAString; response.setContentType("application/xml"); response.setHeader("Content-Disposition", "attachment;filename=thisIsTheFileName.xml"); BufferedReader br = new BufferedReader(new FileReader(new File(YourFile.xml))); String line; StringBuilder sb = new StringBuilder(); while((line=br.readLine())!= null){ sb.append(line); } yourXmlFileInAString = sb.toString(); ServletOutputStream outStream = response.getOutputStream(); outStream.println(yourXmlFileInAString); outStream.flush(); outStream.close();}那应该做的。但是请记住,浏览器会缓存URL内容-因此,最好在每个文件中使用唯一的URL。
编辑:
经过进一步检查,您还应该能够将以下代码添加到Action中,以使其起作用:
response.setContentType("text/plain");(或用于XML)
response.setContentType("application/xml");因此,完整的解决方案应该是:
@RequestMapping(value = "/files", method = RequestMethod.GET)@ResponseBody public FileSystemResource getFile(HttpServletResponse response) { response.setContentType("application/xml"); return new FileSystemResource(new File("try.xml")); //Or path to your file }欢迎分享,转载请注明来源:内存溢出
微信扫一扫
支付宝扫一扫
评论列表(0条)