给出两条绝对路径，例如
/var/data/stuff/xyz.dat
/var/data
如何创建一个以第二条路径为基础的相对路径?在上面的例子中，结果应该是:./stuff/xyz.dat
当前回答
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);
}
其他回答
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/", "/"));
}
}
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()));
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);
}
}
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