Java 如何在浏览器中显示上载的图像URL

我是新的多部分在春季开机，我复制了下面的代码从互联网上传文件和它的工作良好。我的要求是，在存储完文件后，我只想在浏览器中粘贴图像URL时显示它们，但使用下面的代码，在粘贴图像URL时下载它们

如何只显示上载的文件而不显示下载

控制器 删除该行：

.header(HttpHeaders.CONTENT_DISPOSITION, "attachment; filename=\"" + resource.getFilename() + "\"")

内容处置HTTP头通常触发下载行为。因此您不想使用它。

删除行

.header（HttpHeaders.CONTENT\u处置，“附件；文件名=\”“+资源.getFilename（）+”）

。内容处置HTTP标头通常会触发下载行为。@Codo您应该将其移动到“您的答案”amigo:）或者您也可以尝试。标头（HttpHeaders.Content_Disposition，“inline”）；

 @Service
public class FileStorageService {

    private final Path fileStorageLocation;

    @Autowired
    public FileStorageService(FileStorageProperties fileStorageProperties) {
        this.fileStorageLocation = Paths.get(fileStorageProperties.getUploadDir())
                .toAbsolutePath().normalize();

        try {
            Files.createDirectories(this.fileStorageLocation);
        } catch (Exception ex) {
            throw new FileStorageException("Could not create the directory where the uploaded files will be stored.", ex);
        }
    }

    public String storeFile(MultipartFile file) {
        // Normalize file name
        String fileName = StringUtils.cleanPath(file.getOriginalFilename());

        try {
            // Check if the file's name contains invalid characters
            if(fileName.contains("..")) {
                throw new FileStorageException("Sorry! Filename contains invalid path sequence " + fileName);
            }

            // Copy file to the target location (Replacing existing file with the same name)
            Path targetLocation = this.fileStorageLocation.resolve(fileName);
            Files.copy(file.getInputStream(), targetLocation, StandardCopyOption.REPLACE_EXISTING);

            return fileName;
        } catch (IOException ex) {
            throw new FileStorageException("Could not store file " + fileName + ". Please try again!", ex);
        }
    }


    public Resource loadFileAsResource(String fileName) {
        try {
            Path filePath = this.fileStorageLocation.resolve(fileName).normalize();
            Resource resource = new UrlResource(filePath.toUri());
            if(resource.exists()) {
                return resource;
            } else {
                throw new MyFileNotFoundException("File not found " + fileName);
            }
        } catch (MalformedURLException ex) {
            throw new MyFileNotFoundException("File not found " + fileName, ex);
        }
    }
}

.header(HttpHeaders.CONTENT_DISPOSITION, "attachment; filename=\"" + resource.getFilename() + "\"")