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

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;
}

从Java 7开始,你可以使用relativize方法:

import java.nio.file.Path;
import java.nio.file.Paths;

public class Test {

     public static void main(String[] args) {
        Path pathAbsolute = Paths.get("/var/data/stuff/xyz.dat");
        Path pathBase = Paths.get("/var/data");
        Path pathRelative = pathBase.relativize(pathAbsolute);
        System.out.println(pathRelative);
    }

}

输出:

stuff/xyz.dat

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();
    }

}

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)

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();
}