java - 在 gs-uploading-files Spring Boot 示例上将 Java 1.8 lambda 转换为 Java 1.7

Question

我正在 https://spring.io/ 上进行一些练习 Spring Boot 。

做例子https://spring.io/guides/gs/uploading-files/当我使用 Java 8 时它工作正常，但不幸的是，我在 Web 服务中包装的代码需要 Java 7。我列出了所有错误代码，有人可以帮助我将 lambda 转换为 1.7 兼容代码并替换新库(java.util.stream.Stream 和 java.util.stream.Collectors)。

应用程序.java

@SpringBootApplication
@EnableConfigurationProperties(StorageProperties.class)
public class Application {

    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }

    @Bean
    CommandLineRunner init(StorageService storageService) {
        return (args) -> {
            storageService.deleteAll();
            storageService.init();
        };
    }
}

return (args) -> errors out saying "use source -8 or higher to enable lambda expressions"

FileUploadController.java

import java.util.stream.Collectors

//..

@GetMapping("/")
public String listUploadedFiles(Model model) throws IOException {

    model.addAttribute("files", storageService
            .loadAll()
            .map(path ->
                    MvcUriComponentsBuilder
                            .fromMethodName(FileUploadController.class, "serveFile", path.getFileName().toString())
                            .build().toString())
            .collect(Collectors.toList()));

    return "uploadForm";
}

package java.util.stream does not exist

type loadAll() is erroneous

errors out saying "use source -8 or higher to enable lambda expressions"

FileSystemStorageService.java

@Override
public Stream<Path> loadAll() {
    try {
        return Files.walk(this.rootLocation, 1)
                .filter(path -> !path.equals(this.rootLocation))
                .map(path -> this.rootLocation.relativize(path));
    } catch (IOException e) {
        throw new StorageException("Failed to read stored files", e);
    }

}

cannot find symbol walk

errors out saying "use source -8 or higher to enable lambda expressions"

StorageService.java

import java.util.stream.Stream;

public interface StorageService {

    void init();

    void store(MultipartFile file);

    Stream<Path> loadAll();

    Path load(String filename);

    Resource loadAsResource(String filename);

    void deleteAll();

}

package java.util.stream does not exist

Answer 1

Bulltorious，我很抱歉你的赏金，但你真的只有很少的选择。 Stream API 仅在 Java 8 中添加，因此在 Java 7 中不存在。您可以通过手动或几乎自动编写匿名类来解决与 lambda 相关的问题(参见 Retrolambda )。然而，对于 Stream API，您只有两个选择:

1. 手动将 Stream API 反向移植到 Java 7。或者使用其他人尝试向后移植，例如 streamsupport project在 SourceForge 上(或在 GitHub 上复制)
2. 摆脱 Stream API 并使用旧的 Java 7 类。

更新(替换Files.walk)

如果您的 Files.walk 是您唯一使用 Java-8 特定 API 的地方，您可以相对轻松地将其替换为 Java 7 API:

interface PathNameMapper
{
    String mapPath(Path path);
}


List<String> loadAll(PathNameMapper mapper)
{
    try
    {
        List<String> result = new ArrayList<>();

        Files.walkFileTree(rootLocation, EnumSet.noneOf(FileVisitOption.class), 1, new SimpleFileVisitor<Path>()
        {
            @Override
            public FileVisitResult visitFile(Path file, BasicFileAttributes attrs) throws IOException
            {
                if (!rootLocation.equals(file))
                {
                    result.add(mapper.mapPath(rootLocation.relativize(file)));
                }
                return FileVisitResult.CONTINUE;
            }
        });


        return result;
    }
    catch (IOException e)
    {
        throw new StorageException("Failed to read stored files", e);
    }
}

然后你的 listUploadedFiles 会变成这样

@GetMapping ("/")
public String listUploadedFiles(Model model) throws IOException
{

    model.addAttribute("files", storageService
            .loadAll(new PathNameMapper()
            {
                @Override
                public String mapPath(Path path)
                {
                    return MvcUriComponentsBuilder
                            .fromMethodName(FileUploadController.class, "serveFile", path.getFileName().toString())
                            .build().toString();
                }
            }));

    return "uploadForm";
}

更新 2(labmda -> 匿名类转换)

为了完整起见，还有一个如何手动将 lambda 转换为匿名类的示例:

@Bean
CommandLineRunner init(StorageService storageService)
{
    return new CommandLineRunner()
    {
        @Override
        public void run(String... args) throws Exception
        {
            storageService.deleteAll();
            storageService.init();              
        }
    };
}