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
如果你正在编写一个Maven插件,你可以使用Plexus的PathTool:
import org.codehaus.plexus.util.PathTool;
String relativeFilePath = PathTool.getRelativeFilePath(file1, file2);
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。
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()));
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)
酷! !我需要一些类似这样的代码,但用于比较Linux机器上的目录路径。我发现这在父目录为目标的情况下不起作用。
下面是该方法的目录友好版本:
public static String getRelativePath(String targetPath, String basePath,
String pathSeparator) {
boolean isDir = false;
{
File f = new File(targetPath);
isDir = f.isDirectory();
}
// 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 - (isDir?0:1); /* only subtract 1 if it is a file. */
// 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;
}
}
//if we are comparing directories then we
if (targetPath.length() > common.length()) {
//it's OK, it isn't a directory
relative += targetPath.substring(common.length());
}
return relative;
}