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
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
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。
其他回答
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。
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();
}
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);
}
}
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);
}
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();
}