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
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));
}
}
其他回答
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));
}
}
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);
}
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。
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);
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)