Question

摘要

在Spring中，我可以通过指定相对于src/main/resources/目录的路径来访问资源。例如，如果我要求/public/index.html，我会得到代表FileSystemResource的{{1}}。但是，我没有办法走向相反的方向。

给定/Users/.../src/main/resources/public/index.html，有没有办法找到相对于FileSystemResource的路径？

实施例

我使用src/main/resouces/获取应用中的文件资源列表。我需要的资源位于我的应用的PathMatchingResourcePatternResolver文件夹中。

src/main/resources/public/

我可以很容易地获得绝对路径，但是我想要一种从ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver(); for (Resource r : resolver.getResources("/public/**")) { // r.getURI().getString() gives me the absolute path. }开始获取该部分的方法，因为Spring就是这样开始的。

Answer 1

我正要问的是如何做到这一点，以一种比我的解决方案更少的hacky方式（以及处理ClassPathResource s，这可能发生在我的情况下）。所以这是我的回答和问题：

ResourcePatternResolver.getResources返回的资源可以是FileSystemResource或ClassPathResource，具体取决于glob模式是否包含通配符。（在我的例子中，glob模式作为参数给出，应该支持任何模式。）所以，没有getFile;唯一可用的路径信息似乎是getDescription。

为了确定根的目录深度，我放了一个参考文件＆＃34; input-files.txt＆＃34;那里。这是因为我认为根目录资源＆＃34; classpath：/ input-files /＆＃34;在基于jar的类路径的情况下不需要存在。根据根深度，我可以从资源描述中删除那么多目录。

有没有人知道更简单或更少的hacky解决方案？

和我一样，它应该同时使用基于jar和目录的类路径。

import java.io.IOException;
import java.nio.file.Path;
import java.util.stream.Collectors;
import java.util.stream.Stream;
import javax.inject.Inject;
import org.springframework.boot.CommandLineRunner;
import org.springframework.boot.builder.SpringApplicationBuilder;
import org.springframework.context.ApplicationContext;
import org.springframework.core.io.Resource;

public class TestClasspathRelativePath implements CommandLineRunner {

    private static final String RESOURCE_ROOT = "classpath:/input-files/";
    private static final String RESOURCE_ROOT_MARKER_FILE = "input-files.txt";
    private static final String FILE_SEPARATOR_PATTERN = "[/\\\\]+";

    @Inject
    private ApplicationContext appContext;

    public static void main(String[] args) {
        new SpringApplicationBuilder()
                .sources(TestClasspathRelativePath.class)
                .web(false)
                .build()
                .run("**/*");
    }

    @Override
    public void run(String... args) throws Exception {
        for (String templateFilenamePattern : args) {
            for (Resource template : appContext.getResources(RESOURCE_ROOT + templateFilenamePattern)) {
                createInputFile(template);
            }
        }
    }

    private void createInputFile(Resource template) throws IOException {

        // A resource whose immediate parent is the root input
        // files directory, without wildcard lookup. (As of Spring 4.3.14, with
        // a directory classpath entry, this is a ClassPathResource.)
        Resource topLevelFile = appContext.getResource(RESOURCE_ROOT + RESOURCE_ROOT_MARKER_FILE);

        // A resource whose immediate parent is the root input
        // files directory, with wildcard lookup. (As of Spring 4.3.14, with a
        // directory classpath entry, this is a FileSystemResource.)
        Resource wildcardTopLevelFile =
                // match with min description length is
                // RESOURCE_ROOT/input-files.txt (probably sole match anyway)
                Stream.of(appContext.getResources(RESOURCE_ROOT + "**/" + RESOURCE_ROOT_MARKER_FILE)).min(
                        (r1, r2) -> Integer.valueOf(r1.getDescription().length()).compareTo(Integer.valueOf(r2.getDescription().length())))
                        .orElseThrow(() -> new IllegalStateException("resource " + RESOURCE_ROOT + "/" + RESOURCE_ROOT_MARKER_FILE + " not found"));

        // In the real program, both top level resources are computed only once and on-demand.

        String targetFilename = "<target-dir>/input/" + relativize(template, topLevelFile, wildcardTopLevelFile);
        System.out.println(targetFilename);
        // ... read template, process, write results to targetFilename
    }

    /**
     * Replacement for {@link Path#relativize(Path)} which should also work with
     * non-file resources.
     */
    private static String relativize(Resource child, Resource... topLevelFiles) {
        // find the top-level file for child's type
        Resource referenceFile = Stream.of(topLevelFiles)
                .filter(f -> f.getClass().isInstance(child))
                .findFirst()
                .orElseThrow(() -> new IllegalStateException("don't know how to relativize " + child));
        int rootLevel = descriptionUpToName(referenceFile).split(FILE_SEPARATOR_PATTERN, -1).length - 1;
        return Stream.of(descriptionUpToName(child).split(FILE_SEPARATOR_PATTERN, -1)).skip(rootLevel)
                .collect(Collectors.joining("/"));
    }

    /**
     * Hack to strip the suffix after the name from a resource description. The
     * prefix need not be stripped, because
     * {@link #relativize(Resource, Resource...)} uses a safer way to get rid of
     * that.
     *
     * @return e.g. "file [C:\foo\classpath\input-files\a\b.txt" if the
     *         resource's description is "file
     *         [C:\foo\classpath\input-files\a\b.txt]"
     */
    private static String descriptionUpToName(Resource resource) {
        String path = resource.getDescription();
        int i = path.lastIndexOf(resource.getFilename());
        return path.substring(0, i + resource.getFilename().length());
    }
}

找到FileSystemResource的相对路径（Spring Boot）

摘要

实施例

1 个答案: