如果你知道第二个字符串是第一个字符串的一部分:

String s1 = "/var/data/stuff/xyz.dat";
String s2 = "/var/data";
String s3 = s1.substring(s2.length());

或者，如果你真的想把句号放在开头，就像你的例子中那样:

String s3 = ".".concat(s1.substring(s2.length()));

Psuedo-code:

使用路径分隔符("/")拆分字符串 通过迭代拆分字符串的结果来找到最大的公共路径(因此在两个示例中，您最终会得到"/var/data"或"/a") 返回“。”+ whicheverPathIsLonger.substring (commonPath.length);

我假设你有fromPath(一个文件夹的绝对路径)，和toPath(一个文件夹/文件的绝对路径)，你正在寻找一个路径，代表文件/文件夹在toPath作为一个相对路径从fromPath(你当前的工作目录是fromPath)，然后像这样的工作:

public static String getRelativePath(String fromPath, String toPath) {

  // This weirdness is because a separator of '/' messes with String.split()
  String regexCharacter = File.separator;
  if (File.separatorChar == '\\') {
    regexCharacter = "\\\\";
  }

  String[] fromSplit = fromPath.split(regexCharacter);
  String[] toSplit = toPath.split(regexCharacter);

  // Find the common path
  int common = 0;
  while (fromSplit[common].equals(toSplit[common])) {
    common++;
  }

  StringBuffer result = new StringBuffer(".");

  // Work your way up the FROM path to common ground
  for (int i = common; i < fromSplit.length; i++) {
    result.append(File.separatorChar).append("..");
  }

  // Work your way down the TO path
  for (int i = common; i < toSplit.length; i++) {
    result.append(File.separatorChar).append(toSplit[i]);
  }

  return result.toString();
}

实际上，如果目标路径不是基路径的子路径，我的另一个答案是不成立的。

这应该有用。

public class RelativePathFinder {

    public static String getRelativePath(String targetPath, String basePath, 
       String pathSeparator) {

        // find common path
        String[] target = targetPath.split(pathSeparator);
        String[] base = basePath.split(pathSeparator);

        String common = "";
        int commonIndex = 0;
        for (int i = 0; i < target.length && i < base.length; i++) {

            if (target[i].equals(base[i])) {
                common += target[i] + pathSeparator;
                commonIndex++;
            }
        }


        String relative = "";
        // is the target a child directory of the base directory?
        // i.e., target = /a/b/c/d, base = /a/b/
        if (commonIndex == base.length) {
            relative = "." + pathSeparator + targetPath.substring(common.length());
        }
        else {
            // determine how many directories we have to backtrack
            for (int i = 1; i <= commonIndex; i++) {
                relative += ".." + pathSeparator;
            }
            relative += targetPath.substring(common.length());
        }

        return relative;
    }

    public static String getRelativePath(String targetPath, String basePath) {
        return getRelativePath(targetPath, basePath, File.pathSeparator);
    }
}

public class RelativePathFinderTest extends TestCase {

    public void testGetRelativePath() {
        assertEquals("./stuff/xyz.dat", RelativePathFinder.getRelativePath(
                "/var/data/stuff/xyz.dat", "/var/data/", "/"));
        assertEquals("../../b/c", RelativePathFinder.getRelativePath("/a/b/c",
                "/a/x/y/", "/"));
    }

}

这有点迂回，但为什么不使用URI呢?它有一个相对化方法可以帮你做所有必要的检查。

String path = "/var/data/stuff/xyz.dat";
String base = "/var/data";
String relative = new File(base).toURI().relativize(new File(path).toURI()).getPath();
// relative == "stuff/xyz.dat"

请注意文件路径是java.nio.file。正如@Jirka Meluzin在另一个答案中指出的那样，自Java 1.7以来，路径#相对化。

当使用java.net.URI.relativize时，你应该注意Java错误: JDK-6226081 (URI应该能够将路径与部分根相对化)

目前，URI的relativize()方法只在一个URI是另一个URI的前缀时才会相对化URI。

这本质上意味着java.net.URI.relativize将不会创建“…这是给你的。

我的版本大致基于马特和史蒂夫的版本:

/**
 * Returns the path of one File relative to another.
 *
 * @param target the target directory
 * @param base the base directory
 * @return target's path relative to the base directory
 * @throws IOException if an error occurs while resolving the files' canonical names
 */
 public static File getRelativeFile(File target, File base) throws IOException
 {
   String[] baseComponents = base.getCanonicalPath().split(Pattern.quote(File.separator));
   String[] targetComponents = target.getCanonicalPath().split(Pattern.quote(File.separator));

   // skip common components
   int index = 0;
   for (; index < targetComponents.length && index < baseComponents.length; ++index)
   {
     if (!targetComponents[index].equals(baseComponents[index]))
       break;
   }

   StringBuilder result = new StringBuilder();
   if (index != baseComponents.length)
   {
     // backtrack to base directory
     for (int i = index; i < baseComponents.length; ++i)
       result.append(".." + File.separator);
   }
   for (; index < targetComponents.length; ++index)
     result.append(targetComponents[index] + File.separator);
   if (!target.getPath().endsWith("/") && !target.getPath().endsWith("\\"))
   {
     // remove final path separator
     result.delete(result.length() - File.separator.length(), result.length());
   }
   return new File(result.toString());
 }

Matt B的解决方案错误地获得了回溯目录的数量——它应该是基本路径的长度减去公共路径元素的数量，减去1(对于最后一个路径元素，它是一个文件名或由split生成的尾随“”)。它恰好适用于/a/b/c/和/a/x/y/，但是将参数替换为/m/n/o/a/b/c/和/m/n/o/a/x/y/，你就会发现问题。

此外，它还需要在第一个for循环中进行else中断，否则它将错误地处理碰巧具有匹配目录名的路径，例如/a/b/c/d/和/x/y/c/z——c在两个数组中的同一个槽中，但不是真正的匹配。

所有这些解决方案都缺乏处理不能相互相对化的路径的能力，因为它们具有不兼容的根，例如C:\foo\bar和D:\baz\quux。可能只是Windows上的一个问题，但值得注意。

我花在这上面的时间比我预期的要长得多，但没关系。我实际上需要这个工作，所以感谢每个人的插话，我相信这个版本也会有修正!

public static String getRelativePath(String targetPath, String basePath, 
        String pathSeparator) {

    //  We need the -1 argument to split to make sure we get a trailing 
    //  "" token if the base ends in the path separator and is therefore
    //  a directory. We require directory paths to end in the path
    //  separator -- otherwise they are indistinguishable from files.
    String[] base = basePath.split(Pattern.quote(pathSeparator), -1);
    String[] target = targetPath.split(Pattern.quote(pathSeparator), 0);

    //  First get all the common elements. Store them as a string,
    //  and also count how many of them there are. 
    String common = "";
    int commonIndex = 0;
    for (int i = 0; i < target.length && i < base.length; i++) {
        if (target[i].equals(base[i])) {
            common += target[i] + pathSeparator;
            commonIndex++;
        }
        else break;
    }

    if (commonIndex == 0)
    {
        //  Whoops -- not even a single common path element. This most
        //  likely indicates differing drive letters, like C: and D:. 
        //  These paths cannot be relativized. Return the target path.
        return targetPath;
        //  This should never happen when all absolute paths
        //  begin with / as in *nix. 
    }

    String relative = "";
    if (base.length == commonIndex) {
        //  Comment this out if you prefer that a relative path not start with ./
        //relative = "." + pathSeparator;
    }
    else {
        int numDirsUp = base.length - commonIndex - 1;
        //  The number of directories we have to backtrack is the length of 
        //  the base path MINUS the number of common path elements, minus
        //  one because the last element in the path isn't a directory.
        for (int i = 1; i <= (numDirsUp); i++) {
            relative += ".." + pathSeparator;
        }
    }
    relative += targetPath.substring(common.length());

    return relative;
}

下面是几种情况下的测试:

public void testGetRelativePathsUnixy() 
{        
    assertEquals("stuff/xyz.dat", FileUtils.getRelativePath(
            "/var/data/stuff/xyz.dat", "/var/data/", "/"));
    assertEquals("../../b/c", FileUtils.getRelativePath(
            "/a/b/c", "/a/x/y/", "/"));
    assertEquals("../../b/c", FileUtils.getRelativePath(
            "/m/n/o/a/b/c", "/m/n/o/a/x/y/", "/"));
}

public void testGetRelativePathFileToFile() 
{
    String target = "C:\\Windows\\Boot\\Fonts\\chs_boot.ttf";
    String base = "C:\\Windows\\Speech\\Common\\sapisvr.exe";

    String relPath = FileUtils.getRelativePath(target, base, "\\");
    assertEquals("..\\..\\..\\Boot\\Fonts\\chs_boot.ttf", relPath);
}

public void testGetRelativePathDirectoryToFile() 
{
    String target = "C:\\Windows\\Boot\\Fonts\\chs_boot.ttf";
    String base = "C:\\Windows\\Speech\\Common";

    String relPath = FileUtils.getRelativePath(target, base, "\\");
    assertEquals("..\\..\\Boot\\Fonts\\chs_boot.ttf", relPath);
}

public void testGetRelativePathDifferentDriveLetters() 
{
    String target = "D:\\sources\\recovery\\RecEnv.exe";
    String base   = "C:\\Java\\workspace\\AcceptanceTests\\Standard test data\\geo\\";

    //  Should just return the target path because of the incompatible roots.
    String relPath = FileUtils.getRelativePath(target, base, "\\");
    assertEquals(target, relPath);
}

在另一个答案中提到的错误是由Apache HttpComponents中的URIUtils解决的

public static URI resolve(URI baseURI,
                          String reference)

对象的URI引用 基本URI。解决bug的方法 java.net.URI ()

酷! !我需要一些类似这样的代码，但用于比较Linux机器上的目录路径。我发现这在父目录为目标的情况下不起作用。

下面是该方法的目录友好版本:

 public static String getRelativePath(String targetPath, String basePath, 
     String pathSeparator) {

 boolean isDir = false;
 {
   File f = new File(targetPath);
   isDir = f.isDirectory();
 }
 //  We need the -1 argument to split to make sure we get a trailing 
 //  "" token if the base ends in the path separator and is therefore
 //  a directory. We require directory paths to end in the path
 //  separator -- otherwise they are indistinguishable from files.
 String[] base = basePath.split(Pattern.quote(pathSeparator), -1);
 String[] target = targetPath.split(Pattern.quote(pathSeparator), 0);

 //  First get all the common elements. Store them as a string,
 //  and also count how many of them there are. 
 String common = "";
 int commonIndex = 0;
 for (int i = 0; i < target.length && i < base.length; i++) {
     if (target[i].equals(base[i])) {
         common += target[i] + pathSeparator;
         commonIndex++;
     }
     else break;
 }

 if (commonIndex == 0)
 {
     //  Whoops -- not even a single common path element. This most
     //  likely indicates differing drive letters, like C: and D:. 
     //  These paths cannot be relativized. Return the target path.
     return targetPath;
     //  This should never happen when all absolute paths
     //  begin with / as in *nix. 
 }

 String relative = "";
 if (base.length == commonIndex) {
     //  Comment this out if you prefer that a relative path not start with ./
     relative = "." + pathSeparator;
 }
 else {
     int numDirsUp = base.length - commonIndex - (isDir?0:1); /* only subtract 1 if it  is a file. */
     //  The number of directories we have to backtrack is the length of 
     //  the base path MINUS the number of common path elements, minus
     //  one because the last element in the path isn't a directory.
     for (int i = 1; i <= (numDirsUp); i++) {
         relative += ".." + pathSeparator;
     }
 }
 //if we are comparing directories then we 
 if (targetPath.length() > common.length()) {
  //it's OK, it isn't a directory
  relative += targetPath.substring(common.length());
 }

 return relative;
}

在撰写本文时(2010年6月)，这是唯一通过我的测试用例的解决方案。我不能保证这个解决方案没有错误，但是它确实通过了包含的测试用例。我所编写的方法和测试依赖于Apache公共IO中的FilenameUtils类。

该解决方案使用Java 1.4进行了测试。如果你正在使用Java 1.5(或更高版本)，你应该考虑用StringBuilder代替StringBuffer(如果你还在使用Java 1.4，你应该考虑换一个雇主)。

import java.io.File;
import java.util.regex.Pattern;

import org.apache.commons.io.FilenameUtils;

public class ResourceUtils {

    /**
     * Get the relative path from one file to another, specifying the directory separator. 
     * If one of the provided resources does not exist, it is assumed to be a file unless it ends with '/' or
     * '\'.
     * 
     * @param targetPath targetPath is calculated to this file
     * @param basePath basePath is calculated from this file
     * @param pathSeparator directory separator. The platform default is not assumed so that we can test Unix behaviour when running on Windows (for example)
     * @return
     */
    public static String getRelativePath(String targetPath, String basePath, String pathSeparator) {

        // Normalize the paths
        String normalizedTargetPath = FilenameUtils.normalizeNoEndSeparator(targetPath);
        String normalizedBasePath = FilenameUtils.normalizeNoEndSeparator(basePath);

        // Undo the changes to the separators made by normalization
        if (pathSeparator.equals("/")) {
            normalizedTargetPath = FilenameUtils.separatorsToUnix(normalizedTargetPath);
            normalizedBasePath = FilenameUtils.separatorsToUnix(normalizedBasePath);

        } else if (pathSeparator.equals("\\")) {
            normalizedTargetPath = FilenameUtils.separatorsToWindows(normalizedTargetPath);
            normalizedBasePath = FilenameUtils.separatorsToWindows(normalizedBasePath);

        } else {
            throw new IllegalArgumentException("Unrecognised dir separator '" + pathSeparator + "'");
        }

        String[] base = normalizedBasePath.split(Pattern.quote(pathSeparator));
        String[] target = normalizedTargetPath.split(Pattern.quote(pathSeparator));

        // First get all the common elements. Store them as a string,
        // and also count how many of them there are.
        StringBuffer common = new StringBuffer();

        int commonIndex = 0;
        while (commonIndex < target.length && commonIndex < base.length
                && target[commonIndex].equals(base[commonIndex])) {
            common.append(target[commonIndex] + pathSeparator);
            commonIndex++;
        }

        if (commonIndex == 0) {
            // No single common path element. This most
            // likely indicates differing drive letters, like C: and D:.
            // These paths cannot be relativized.
            throw new PathResolutionException("No common path element found for '" + normalizedTargetPath + "' and '" + normalizedBasePath
                    + "'");
        }   

        // The number of directories we have to backtrack depends on whether the base is a file or a dir
        // For example, the relative path from
        //
        // /foo/bar/baz/gg/ff to /foo/bar/baz
        // 
        // ".." if ff is a file
        // "../.." if ff is a directory
        //
        // The following is a heuristic to figure out if the base refers to a file or dir. It's not perfect, because
        // the resource referred to by this path may not actually exist, but it's the best I can do
        boolean baseIsFile = true;

        File baseResource = new File(normalizedBasePath);

        if (baseResource.exists()) {
            baseIsFile = baseResource.isFile();

        } else if (basePath.endsWith(pathSeparator)) {
            baseIsFile = false;
        }

        StringBuffer relative = new StringBuffer();

        if (base.length != commonIndex) {
            int numDirsUp = baseIsFile ? base.length - commonIndex - 1 : base.length - commonIndex;

            for (int i = 0; i < numDirsUp; i++) {
                relative.append(".." + pathSeparator);
            }
        }
        relative.append(normalizedTargetPath.substring(common.length()));
        return relative.toString();
    }


    static class PathResolutionException extends RuntimeException {
        PathResolutionException(String msg) {
            super(msg);
        }
    }    
}

这个通过的测试用例是

public void testGetRelativePathsUnix() {
    assertEquals("stuff/xyz.dat", ResourceUtils.getRelativePath("/var/data/stuff/xyz.dat", "/var/data/", "/"));
    assertEquals("../../b/c", ResourceUtils.getRelativePath("/a/b/c", "/a/x/y/", "/"));
    assertEquals("../../b/c", ResourceUtils.getRelativePath("/m/n/o/a/b/c", "/m/n/o/a/x/y/", "/"));
}

public void testGetRelativePathFileToFile() {
    String target = "C:\\Windows\\Boot\\Fonts\\chs_boot.ttf";
    String base = "C:\\Windows\\Speech\\Common\\sapisvr.exe";

    String relPath = ResourceUtils.getRelativePath(target, base, "\\");
    assertEquals("..\\..\\Boot\\Fonts\\chs_boot.ttf", relPath);
}

public void testGetRelativePathDirectoryToFile() {
    String target = "C:\\Windows\\Boot\\Fonts\\chs_boot.ttf";
    String base = "C:\\Windows\\Speech\\Common\\";

    String relPath = ResourceUtils.getRelativePath(target, base, "\\");
    assertEquals("..\\..\\Boot\\Fonts\\chs_boot.ttf", relPath);
}

public void testGetRelativePathFileToDirectory() {
    String target = "C:\\Windows\\Boot\\Fonts";
    String base = "C:\\Windows\\Speech\\Common\\foo.txt";

    String relPath = ResourceUtils.getRelativePath(target, base, "\\");
    assertEquals("..\\..\\Boot\\Fonts", relPath);
}

public void testGetRelativePathDirectoryToDirectory() {
    String target = "C:\\Windows\\Boot\\";
    String base = "C:\\Windows\\Speech\\Common\\";
    String expected = "..\\..\\Boot";

    String relPath = ResourceUtils.getRelativePath(target, base, "\\");
    assertEquals(expected, relPath);
}

public void testGetRelativePathDifferentDriveLetters() {
    String target = "D:\\sources\\recovery\\RecEnv.exe";
    String base = "C:\\Java\\workspace\\AcceptanceTests\\Standard test data\\geo\\";

    try {
        ResourceUtils.getRelativePath(target, base, "\\");
        fail();

    } catch (PathResolutionException ex) {
        // expected exception
    }
}

从Java 7开始，你可以使用relativize方法:

import java.nio.file.Path;
import java.nio.file.Paths;

public class Test {

     public static void main(String[] args) {
        Path pathAbsolute = Paths.get("/var/data/stuff/xyz.dat");
        Path pathBase = Paths.get("/var/data");
        Path pathRelative = pathBase.relativize(pathAbsolute);
        System.out.println(pathRelative);
    }

}

输出:

stuff/xyz.dat

递归产生一个较小的解决方案。如果结果不可能(例如不同的Windows磁盘)或不切实际(根目录只是普通目录)，则抛出异常。

/**
 * Computes the path for a file relative to a given base, or fails if the only shared 
 * directory is the root and the absolute form is better.
 * 
 * @param base File that is the base for the result
 * @param name File to be "relativized"
 * @return the relative name
 * @throws IOException if files have no common sub-directories, i.e. at best share the
 *                     root prefix "/" or "C:\"
 */

public static String getRelativePath(File base, File name) throws IOException  {
    File parent = base.getParentFile();

    if (parent == null) {
        throw new IOException("No common directory");
    }

    String bpath = base.getCanonicalPath();
    String fpath = name.getCanonicalPath();

    if (fpath.startsWith(bpath)) {
        return fpath.substring(bpath.length() + 1);
    } else {
        return (".." + File.separator + getRelativePath(parent, name));
    }
}

这里已经有很多答案了，但我发现他们并不能处理所有的情况，比如基地和目标是相同的。这个函数接受一个基本目录和一个目标路径，并返回相对路径。如果不存在相对路径，则返回目标路径。文件。分隔符是不必要的。

public static String getRelativePath (String baseDir, String targetPath) {
    String[] base = baseDir.replace('\\', '/').split("\\/");
    targetPath = targetPath.replace('\\', '/');
    String[] target = targetPath.split("\\/");

    // Count common elements and their length.
    int commonCount = 0, commonLength = 0, maxCount = Math.min(target.length, base.length);
    while (commonCount < maxCount) {
        String targetElement = target[commonCount];
        if (!targetElement.equals(base[commonCount])) break;
        commonCount++;
        commonLength += targetElement.length() + 1; // Directory name length plus slash.
    }
    if (commonCount == 0) return targetPath; // No common path element.

    int targetLength = targetPath.length();
    int dirsUp = base.length - commonCount;
    StringBuffer relative = new StringBuffer(dirsUp * 3 + targetLength - commonLength + 1);
    for (int i = 0; i < dirsUp; i++)
        relative.append("../");
    if (commonLength < targetLength) relative.append(targetPath.substring(commonLength));
    return relative.toString();
}

ant有一个带有getRelativePath方法的FileUtils类。我自己还没有尝试过，但是值得一试。

http://javadoc.haefelinger.it/org.apache.ant/1.7.1/org/apache/tools/ant/util/FileUtils.html getRelativePath (java.io.File java.io.File)

下面是一个从基本路径解析相对路径的方法，不管它们在相同或不同的根路径中:

public static String GetRelativePath(String path, String base){

    final String SEP = "/";

    // if base is not a directory -> return empty
    if (!base.endsWith(SEP)){
        return "";
    }

    // check if path is a file -> remove last "/" at the end of the method
    boolean isfile = !path.endsWith(SEP);

    // get URIs and split them by using the separator
    String a = "";
    String b = "";
    try {
        a = new File(base).getCanonicalFile().toURI().getPath();
        b = new File(path).getCanonicalFile().toURI().getPath();
    } catch (IOException e) {
        e.printStackTrace();
    }
    String[] basePaths = a.split(SEP);
    String[] otherPaths = b.split(SEP);

    // check common part
    int n = 0;
    for(; n < basePaths.length && n < otherPaths.length; n ++)
    {
        if( basePaths[n].equals(otherPaths[n]) == false )
            break;
    }

    // compose the new path
    StringBuffer tmp = new StringBuffer("");
    for(int m = n; m < basePaths.length; m ++)
        tmp.append(".."+SEP);
    for(int m = n; m < otherPaths.length; m ++)
    {
        tmp.append(otherPaths[m]);
        tmp.append(SEP);
    }

    // get path string
    String result = tmp.toString();

    // remove last "/" if path is a file
    if (isfile && result.endsWith(SEP)){
        result = result.substring(0,result.length()-1);
    }

    return result;
}

这里有一个其他库免费的解决方案:

Path sourceFile = Paths.get("some/common/path/example/a/b/c/f1.txt");
Path targetFile = Paths.get("some/common/path/example/d/e/f2.txt"); 
Path relativePath = sourceFile.relativize(targetFile);
System.out.println(relativePath);

输出

..\..\..\..\d\e\f2.txt

[编辑]实际上它输出更多。因为源文件不是目录。对于我的情况，正确的解决方案是:

Path sourceFile = Paths.get(new File("some/common/path/example/a/b/c/f1.txt").parent());
Path targetFile = Paths.get("some/common/path/example/d/e/f2.txt"); 
Path relativePath = sourceFile.relativize(targetFile);
System.out.println(relativePath);

private String relative(String left, String right){
    String[] lefts = left.split("/");
    String[] rights = right.split("/");
    int min = Math.min(lefts.length, rights.length);
    int commonIdx = -1;
    for(int i = 0; i < min; i++){
        if(commonIdx < 0 && !lefts[i].equals(rights[i])){
            commonIdx = i - 1;
            break;
        }
    }
    if(commonIdx < 0){
        return null;
    }
    StringBuilder sb = new StringBuilder(Math.max(left.length(), right.length()));
    sb.append(left).append("/");
    for(int i = commonIdx + 1; i < lefts.length;i++){
        sb.append("../");
    }
    for(int i = commonIdx + 1; i < rights.length;i++){
        sb.append(rights[i]).append("/");
    }

    return sb.deleteCharAt(sb.length() -1).toString();
}

在Java 7及以后的版本中，你可以简单地使用(与URI相比，它没有错误):

Path#relativize(Path)

通过Dónal的测试，唯一的变化-如果没有公共根，它将返回目标路径(它可能已经是相对的)

import static java.util.Arrays.asList;
import static java.util.Collections.nCopies;
import static org.apache.commons.io.FilenameUtils.normalizeNoEndSeparator;
import static org.apache.commons.io.FilenameUtils.separatorsToUnix;
import static org.apache.commons.lang3.StringUtils.getCommonPrefix;
import static org.apache.commons.lang3.StringUtils.isBlank;
import static org.apache.commons.lang3.StringUtils.isNotEmpty;
import static org.apache.commons.lang3.StringUtils.join;

import java.io.File;
import java.util.ArrayList;
import java.util.List;

public class ResourceUtils {

    public static String getRelativePath(String targetPath, String basePath, String pathSeparator) {
        File baseFile = new File(basePath);
        if (baseFile.isFile() || !baseFile.exists() && !basePath.endsWith("/") && !basePath.endsWith("\\"))
            basePath = baseFile.getParent();

        String target = separatorsToUnix(normalizeNoEndSeparator(targetPath));
        String base = separatorsToUnix(normalizeNoEndSeparator(basePath));

        String commonPrefix = getCommonPrefix(target, base);
        if (isBlank(commonPrefix))
            return targetPath.replaceAll("/", pathSeparator);

        target = target.replaceFirst(commonPrefix, "");
        base = base.replaceFirst(commonPrefix, "");

        List<String> result = new ArrayList<>();
        if (isNotEmpty(base))
            result.addAll(nCopies(base.split("/").length, ".."));
        result.addAll(asList(target.replaceFirst("^/", "").split("/")));

        return join(result, pathSeparator);
    }
}

如果你正在编写一个Maven插件，你可以使用Plexus的PathTool:

import org.codehaus.plexus.util.PathTool;

String relativeFilePath = PathTool.getRelativeFilePath(file1, file2);

如果路径在JRE 1.5运行时或maven插件中不可用

package org.afc.util;

import java.io.File;
import java.util.LinkedList;
import java.util.List;

public class FileUtil {

    public static String getRelativePath(String basePath, String filePath)  {
        return getRelativePath(new File(basePath), new File(filePath));
    }

    public static String getRelativePath(File base, File file)  {

        List<String> bases = new LinkedList<String>();
        bases.add(0, base.getName());
        for (File parent = base.getParentFile(); parent != null; parent = parent.getParentFile()) {
            bases.add(0, parent.getName());
        }

        List<String> files = new LinkedList<String>();
        files.add(0, file.getName());
        for (File parent = file.getParentFile(); parent != null; parent = parent.getParentFile()) {
            files.add(0, parent.getName());
        }

        int overlapIndex = 0;
        while (overlapIndex < bases.size() && overlapIndex < files.size() && bases.get(overlapIndex).equals(files.get(overlapIndex))) {
            overlapIndex++;
        }

        StringBuilder relativePath = new StringBuilder();
        for (int i = overlapIndex; i < bases.size(); i++) {
            relativePath.append("..").append(File.separatorChar);
        }

        for (int i = overlapIndex; i < files.size(); i++) {
            relativePath.append(files.get(i)).append(File.separatorChar);
        }

        relativePath.deleteCharAt(relativePath.length() - 1);
        return relativePath.toString();
    }

}

我知道这有点晚了，但是，我创建了一个解决方案，适用于任何java版本。

    public static String getRealtivePath(File root, File file) 
    {
        String path = file.getPath();
        String rootPath = root.getPath();
        boolean plus1 = path.contains(File.separator);
        return path.substring(path.indexOf(rootPath) + rootPath.length() + (plus1 ? 1 : 0));
    }