使用 FileSystemResource 强制下载文件时如何设置“Content-Disposition”和“Filename”？

Question

使用 Spring 3 FileSystemResource 设置 Content-Disposition=attachment 和 filename=xyz.zip 的最合适和标准的方法是什么？

动作看起来像：

@ResponseBody
@RequestMapping(value = "/action/{abcd}/{efgh}", method = RequestMethod.GET, produces = "application/zip")
@PreAuthorize("@authorizationService.authorizeMethod()")
public FileSystemResource doAction(@PathVariable String abcd, @PathVariable String efgh) {

    File zipFile = service.getFile(abcd, efgh);

    return new FileSystemResource(zipFile);
}

虽然该文件是一个zip文件，所以浏览器总是下载该文件，但我想明确提及该文件作为附件，并提供一个与文件实际名称无关的文件名。

可能有解决此问题的方法，但我想知道实现此目标的正确 Spring 和 FileSystemResource 方法。

附：这里使用的文件是一个临时文件，当JVM存在时标记为删除。

Answer 1

@RequestMapping(value = "/action/{abcd}/{efgh}", method = RequestMethod.GET)
@PreAuthorize("@authorizationService.authorizeMethod(#id)")
public HttpEntity<byte[]> doAction(@PathVariable ObjectType obj, @PathVariable Date date, HttpServletResponse response) throws IOException {
    ZipFileType zipFile = service.getFile(obj1.getId(), date);

    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_OCTET_STREAM);
    response.setHeader("Content-Disposition", "attachment; filename=" + zipFile.getFileName());

    return new HttpEntity<byte[]>(zipFile.getByteArray(), headers);
}

Answer 2

 @RequestMapping(value = "/files/{file_name}", method = RequestMethod.GET)
    @ResponseBody
    public FileSystemResource getFile(@PathVariable("file_name") String fileName,HttpServletResponse response) {
        response.setContentType("application/pdf");      
        response.setHeader("Content-Disposition", "attachment; filename=somefile.pdf"); 
        return new FileSystemResource(new File("file full path")); 
    }

Answer 3

这是 Spring 4 的另一种方法。请注意，此示例显然没有使用有关文件系统访问的良好做法，这只是为了演示如何以声明方式设置某些属性。

@RequestMapping(value = "/{resourceIdentifier}", method = RequestMethod.GET, produces = MediaType.APPLICATION_OCTET_STREAM_VALUE)
// @ResponseBody // Needed for @Controller but not for @RestController.
public ResponseEntity<InputStreamResource> download(@PathVariable(name = "resourceIdentifier") final String filename) throws Exception
{
    final String resourceName = filename + ".dat";
    final File iFile = new File("/some/folder", resourceName);
    final long resourceLength = iFile.length();
    final long lastModified = iFile.lastModified();
    final InputStream resource = new FileInputStream(iFile);

    return ResponseEntity.ok()
            .header("Content-Disposition", "attachment; filename=" + resourceName)
            .contentLength(resourceLength)
            .lastModified(lastModified)
            .contentType(MediaType.APPLICATION_OCTET_STREAM_VALUE)
            .body(resource);
}

Answer 4

对两个给定的答案都做了一些更改，最终我在我的项目中得到了最好的结果，我需要从数据库中提取图像作为 blob，然后将其提供给客户：

@GetMapping("/images/{imageId:.+}")
@ResponseBody
public ResponseEntity<FileSystemResource>  serveFile(@PathVariable @Valid String imageId,HttpServletResponse response)
{       
    ImageEntity singleImageInfo=db.storage.StorageService.getImage(imageId);
    if(singleImageInfo==null)
    {
        return ResponseEntity.status(HttpStatus.NOT_FOUND).body(null);
    }
    Blob image=singleImageInfo.getImage();
    try {           
        String filename= UsersExtra.GenerateSession()+"xxyy"+singleImageInfo.getImage1Ext().trim();

    byte [] array = image.getBytes( 1, ( int ) image.length() );
    File file = File.createTempFile(UsersExtra.GenerateSession()+"xxyy", singleImageInfo.getImage1Ext().trim(), new File("."));
    FileOutputStream out = new FileOutputStream( file );
    out.write( array );
    out.close();
    FileSystemResource testing=new FileSystemResource(file);

    String mimeType = "image/"+singleImageInfo.getImage1Ext().trim().toLowerCase().replace(".", "");
      response.setContentType(mimeType);    

        String headerKey = "Content-Disposition";
       String headerValue = String.format("attachment; filename=\"%s\"", filename);
       response.setHeader(headerKey, headerValue);
      // return new FileSystemResource(file); 
       return ResponseEntity.status(HttpStatus.OK).body( new FileSystemResource(file));
    }catch(Exception e)
    {
        System.out.println(e.getMessage());
    }
    return null;
}

在 Kumar 的代码中使用 ResponseEntity 将帮助您使用正确的响应代码进行响应。 注意：从 blob 转换为文件引用自此链接： Snippet to create a file from the contents of a blob in Java

Answer 5

除了已接受的答案之外，Spring 还具有专门用于此目的的 ContentDisposition 类。我相信它处理文件名清理。

      ContentDisposition contentDisposition = ContentDisposition.builder("inline")
          .filename("Filename")
          .build();

      HttpHeaders headers = new HttpHeaders();
      headers.setContentDisposition(contentDisposition);