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
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());
}
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);
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();
}
}
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)
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);
}
