使用 Spring 3
Content-Disposition=attachment
设置 filename=xyz.zip
和 FileSystemResource
最合适、最标准的方法是什么?
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 zip 文件,因此浏览器总是下载该文件,但我想明确提及该文件作为附件,并提供与文件实际名称无关的文件名。
这个问题可能有解决方法,但我想知道正确的 Spring 和
FileSystemResource
方法来实现这个目标。
附注这里使用的文件是一个临时文件,当 JVM 存在时标记为删除。
除了已接受的答案之外,Spring 还有专门用于此目的的类 ContentDisposition。我相信它涉及文件名清理。
ContentDisposition contentDisposition = ContentDisposition.builder("inline")
.filename("Filename")
.build();
HttpHeaders headers = new HttpHeaders();
headers.setContentDisposition(contentDisposition);
@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);
}
@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"));
}
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实践,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) // For Spring 5: APPLICATION_OCTET_STREAM
.body(resource);
}
对给定的答案做了一些更改,最终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 到文件的转换引用自此链接: 在 Java 中根据 blob 的内容创建文件的片段