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();
}

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));
    }

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();
}

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();
}

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);
}

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()));