如何用两个绝对路径(或URL)在Java中构造一个相对路径?


275

给定两个绝对路径,例如

/var/data/stuff/xyz.dat
/var/data

如何创建以第二条路径为基础的相对路径?在上面的示例中,结果应为:./stuff/xyz.dat


3
对于Java 7及更高版本,请参见@VitaliiFedorenko的答案。
安迪·托马斯

1
tl;博士回答: Paths.get(startPath).relativize(Paths.get(endPath))。toString()(顺便说一句,在Java 8中,对于我来说,“ ../”似乎工作得很好) ,所以...)
安德鲁(Andrew

Answers:


297

这有点round回,但是为什么不使用URI?它具有相对方法,可以为您进行所有必要的检查。

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java.nio.file.Path#relativize自Java 1.7起就存在,如@Jirka Meluzin其他答案中指出的那样。


17
请参阅彼得·穆勒(Peter Mueller)的答案。除了最简单的情况外,relativize()看起来很残破。
戴夫·雷

11
是的,仅当基本路径是第一个路径的父路径时才有效。如果您需要像“ ../../relativepath”之类的某些向后分层结构,它将无法正常工作。我找到了解决方案:mrpmorris.blogspot.com/2007/05/…–
Aurelien Ribon

4
就像@VitaliiFedorenko写道:use java.nio.file.Path#relativize(Path),它仅适用于父双点和全部。
坎帕

考虑使用toPath()代替toURI()。它完全可以创建"..\.."。但java.lang.IllegalArgumentException: 'other' has different root在询问从"C:\temp"到的相对路径时要注意异常"D:\temp"
伊戈尔(Igor)

这不能按预期工作,它在我的测试用例中返回data / stuff / xyz.dat。
令人讨厌的

238

从Java 7开始,您可以使用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

3
不错,简短,没有额外的lib +1。亚当·克鲁姆(Adam Crume)的解决方案(命中1)未通过我的测试,下一个答案(命中2)“唯一的“有效”解决方案”添加了一个新的jar,并且代码多于我的实现,我之后在这里发现……总比没有好。 )
hokr 2013年

1
但是要当心这个问题
ben3000 '16

1
检查是否可以..在必要时进行添加(可以)。
欧文

不幸的是,Android不包括java.nio.file:(
Nathan Osman

1
我发现如果在“相对化”之前未对“ pathBase”进行“标准化”,则会得到奇怪的结果。尽管在此示例中很好,但我pathBase.normalize().relativize(pathAbsolute);通常会这样做。
pstanton '17

77

在撰写本文时(2010年6月),这是唯一通过我的测试案例的解决方案。我不能保证此解决方案没有错误,但是它确实通过了包含的测试用例。我编写的方法和测试取决于Apache commons IO中FilenameUtils类。

该解决方案已使用Java 1.4进行了测试。如果您使用的是Java 1.5(或更高版本),你应该考虑更换StringBufferStringBuilder(如果你还在使用Java 1.4,你应该考虑雇主的变化,而不是)。

import java.io.File;
import java.util.regex.Pattern;

import org.apache.commons.io.FilenameUtils;

public class ResourceUtils {

    /**
     * Get the relative path from one file to another, specifying the directory separator. 
     * If one of the provided resources does not exist, it is assumed to be a file unless it ends with '/' or
     * '\'.
     * 
     * @param targetPath targetPath is calculated to this file
     * @param basePath basePath is calculated from this file
     * @param pathSeparator directory separator. The platform default is not assumed so that we can test Unix behaviour when running on Windows (for example)
     * @return
     */
    public static String getRelativePath(String targetPath, String basePath, String pathSeparator) {

        // Normalize the paths
        String normalizedTargetPath = FilenameUtils.normalizeNoEndSeparator(targetPath);
        String normalizedBasePath = FilenameUtils.normalizeNoEndSeparator(basePath);

        // Undo the changes to the separators made by normalization
        if (pathSeparator.equals("/")) {
            normalizedTargetPath = FilenameUtils.separatorsToUnix(normalizedTargetPath);
            normalizedBasePath = FilenameUtils.separatorsToUnix(normalizedBasePath);

        } else if (pathSeparator.equals("\\")) {
            normalizedTargetPath = FilenameUtils.separatorsToWindows(normalizedTargetPath);
            normalizedBasePath = FilenameUtils.separatorsToWindows(normalizedBasePath);

        } else {
            throw new IllegalArgumentException("Unrecognised dir separator '" + pathSeparator + "'");
        }

        String[] base = normalizedBasePath.split(Pattern.quote(pathSeparator));
        String[] target = normalizedTargetPath.split(Pattern.quote(pathSeparator));

        // First get all the common elements. Store them as a string,
        // and also count how many of them there are.
        StringBuffer common = new StringBuffer();

        int commonIndex = 0;
        while (commonIndex < target.length && commonIndex < base.length
                && target[commonIndex].equals(base[commonIndex])) {
            common.append(target[commonIndex] + pathSeparator);
            commonIndex++;
        }

        if (commonIndex == 0) {
            // No single common path element. This most
            // likely indicates differing drive letters, like C: and D:.
            // These paths cannot be relativized.
            throw new PathResolutionException("No common path element found for '" + normalizedTargetPath + "' and '" + normalizedBasePath
                    + "'");
        }   

        // The number of directories we have to backtrack depends on whether the base is a file or a dir
        // For example, the relative path from
        //
        // /foo/bar/baz/gg/ff to /foo/bar/baz
        // 
        // ".." if ff is a file
        // "../.." if ff is a directory
        //
        // The following is a heuristic to figure out if the base refers to a file or dir. It's not perfect, because
        // the resource referred to by this path may not actually exist, but it's the best I can do
        boolean baseIsFile = true;

        File baseResource = new File(normalizedBasePath);

        if (baseResource.exists()) {
            baseIsFile = baseResource.isFile();

        } else if (basePath.endsWith(pathSeparator)) {
            baseIsFile = false;
        }

        StringBuffer relative = new StringBuffer();

        if (base.length != commonIndex) {
            int numDirsUp = baseIsFile ? base.length - commonIndex - 1 : base.length - commonIndex;

            for (int i = 0; i < numDirsUp; i++) {
                relative.append(".." + pathSeparator);
            }
        }
        relative.append(normalizedTargetPath.substring(common.length()));
        return relative.toString();
    }


    static class PathResolutionException extends RuntimeException {
        PathResolutionException(String msg) {
            super(msg);
        }
    }    
}

通过的测试用例是

public void testGetRelativePathsUnix() {
    assertEquals("stuff/xyz.dat", ResourceUtils.getRelativePath("/var/data/stuff/xyz.dat", "/var/data/", "/"));
    assertEquals("../../b/c", ResourceUtils.getRelativePath("/a/b/c", "/a/x/y/", "/"));
    assertEquals("../../b/c", ResourceUtils.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 = ResourceUtils.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 = ResourceUtils.getRelativePath(target, base, "\\");
    assertEquals("..\\..\\Boot\\Fonts\\chs_boot.ttf", relPath);
}

public void testGetRelativePathFileToDirectory() {
    String target = "C:\\Windows\\Boot\\Fonts";
    String base = "C:\\Windows\\Speech\\Common\\foo.txt";

    String relPath = ResourceUtils.getRelativePath(target, base, "\\");
    assertEquals("..\\..\\Boot\\Fonts", relPath);
}

public void testGetRelativePathDirectoryToDirectory() {
    String target = "C:\\Windows\\Boot\\";
    String base = "C:\\Windows\\Speech\\Common\\";
    String expected = "..\\..\\Boot";

    String relPath = ResourceUtils.getRelativePath(target, base, "\\");
    assertEquals(expected, relPath);
}

public void testGetRelativePathDifferentDriveLetters() {
    String target = "D:\\sources\\recovery\\RecEnv.exe";
    String base = "C:\\Java\\workspace\\AcceptanceTests\\Standard test data\\geo\\";

    try {
        ResourceUtils.getRelativePath(target, base, "\\");
        fail();

    } catch (PathResolutionException ex) {
        // expected exception
    }
}

5
真好!但是,有一件事,如果基址和目标相同,则会中断-公用字符串以分隔符结尾,而该分隔符是规范化的目标路径所没有的,因此子字符串调用要求输入太多位数。想想我通过在函数的最后两行之前添加以下内容来修复它:if(common.length()> = normalizedTargetPath.length()){return“。}
Erhannis 2012年

4
说这是唯一可行的解​​决方案具有误导性。其他答案效果更好(当基本和目标相同时,此答案将崩溃),更简单并且不依赖commons-io。
NateS

26

使用java.net.URI.relativize时,您应该意识到Java错误: JDK-6226081(URI应该能够相对于具有部分根的路径)

目前,仅当一个方法是另一个的前缀时,的relativize()方法URI才会相对化URI。

这实质上意味着java.net.URI.relativize不会为您创建“ ..”。


6
讨厌。显然,有一种解决方法:stackoverflow.com/questions/204784/…–
skaffman

在Java 8中,Paths.get(startPath).relativize(Paths.get(endPath))。toString‌()似乎对例如“ ../”的工作正常。–
Andrew

您确定@skaffman吗?该答案引用了错误JDK-6226081,但URIUtils.resolve()提到了JDK-4708535。从源代码中,我看不到任何与回溯有关的内容(即..片段)。您是否混淆了这两个错误?
Garret Wilson

JDK-6920138被标记为JDK-4708535的副本。
Christian K.19年


17

Java 7和更高版本中,您可以简单地使用(与相比URI,它是没有错误的):

Path#relativize(Path)

10

如果您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,如您的示例所示,如果您真的想要开始时使用句号:

String s3 = ".".concat(s1.substring(s2.length()));

3
字符串s3 =“。” + s1.substring(s2.length()); 稍微更具可读性IMO
多纳尔

10

递归产生一个较小的解决方案。如果结果是不可能的(例如,不同的Windows磁盘)或不切实际的(root仅是公共目录),则将引发异常。

/**
 * 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));
    }
}

getCanonicalPath可能很重,因此当您需要处理数十万条记录时,不建议使用此解决方案。例如,我有一些清单文件具有多达一百万条记录,现在我想将它们移动为使用相对路径以实现可移植性。
user2305886 2013年

8

这是其他免费库的解决方案:

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);

产出

..\..\..\..\d\e\f2.txt

[EDIT]实际上它在更多.. \上输出,因为源是文件而不是目录。我的情况的正确解决方案是:

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);

6

我的版本大致基于MattSteve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());
 }

2
+1对我有用。只是微小的更正:而不是"/".length()您应该使用
spacer.length

5

Matt B的解决方案使要追溯的目录数量错误-它应该是基本路径的长度减去公用路径元素的数量再减去一个(对于最后一个路径元素,它可以是文件名或由""生成的结尾split) 。它发生在与工作/a/b/c//a/x/y/,但替换的参数/m/n/o/a/b/c//m/n/o/a/x/y/,你会看到这个问题。

此外,它需要else break在第一个for循环内部,否则会误处理恰好具有匹配目录名称的路径,例如/a/b/c/d//x/y/c/z- c在两个数组中位于同一插槽中,但不是实际匹配项。

所有这些解决方案都缺乏处理根不兼容的根(例如和)C:\foo\bar而无法相互相对的能力D:\baz\quux。可能只是Windows上的一个问题,但值得注意。

我在此上花费的时间比我预期的要长得多,但这没关系。我实际上需要这项工作,因此感谢所有参与的人,而且我敢肯定也会对此版本进行更正!

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;
}

以下是涵盖几种情况的测试:

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);
}

4

实际上,如果目标路径不是基本路径的子代,我的其他答案将不起作用。

这应该工作。

public class RelativePathFinder {

    public static String getRelativePath(String targetPath, String basePath, 
       String pathSeparator) {

        // find common path
        String[] target = targetPath.split(pathSeparator);
        String[] base = basePath.split(pathSeparator);

        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++;
            }
        }


        String relative = "";
        // is the target a child directory of the base directory?
        // i.e., target = /a/b/c/d, base = /a/b/
        if (commonIndex == base.length) {
            relative = "." + pathSeparator + targetPath.substring(common.length());
        }
        else {
            // determine how many directories we have to backtrack
            for (int i = 1; i <= commonIndex; i++) {
                relative += ".." + pathSeparator;
            }
            relative += targetPath.substring(common.length());
        }

        return relative;
    }

    public static String getRelativePath(String targetPath, String basePath) {
        return getRelativePath(targetPath, basePath, File.pathSeparator);
    }
}

public class RelativePathFinderTest extends TestCase {

    public void testGetRelativePath() {
        assertEquals("./stuff/xyz.dat", RelativePathFinder.getRelativePath(
                "/var/data/stuff/xyz.dat", "/var/data/", "/"));
        assertEquals("../../b/c", RelativePathFinder.getRelativePath("/a/b/c",
                "/a/x/y/", "/"));
    }

}

2
代替File.pathSeparator应该是File.separator。pathSeparator应该仅用于分割(正则表达式),至于“ ////”正则表达式(win路径正则表达式),结果路径将不正确。
亚历克斯·伊瓦瑟uv

3

凉!!我需要像这样的一些代码,但是需要比较Linux机器上的目录路径。我发现这在以父目录为目标的情况下不起作用。

这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;
}

2

我假设您有fromPath(文件夹的绝对路径)和toPath(文件夹/文件的绝对路径),并且您正在寻找一个路径,该路径将toPath中的文件/文件夹表示为相对路径from fromPath(您当前的工作目录是fromPath),则应如下所示:

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();
}

1

这里已经有很多答案,但是我发现它们并不能解决所有情况,例如基数和目标相同。该函数获取基本目录和目标路径,并返回相对路径。如果不存在相对路径,则返回目标路径。File.separator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();
}

0

这是一种从基本路径解析相对路径的方法,无论它们位于相同或不同的根中:

public static String GetRelativePath(String path, String base){

    final String SEP = "/";

    // if base is not a directory -> return empty
    if (!base.endsWith(SEP)){
        return "";
    }

    // check if path is a file -> remove last "/" at the end of the method
    boolean isfile = !path.endsWith(SEP);

    // get URIs and split them by using the separator
    String a = "";
    String b = "";
    try {
        a = new File(base).getCanonicalFile().toURI().getPath();
        b = new File(path).getCanonicalFile().toURI().getPath();
    } catch (IOException e) {
        e.printStackTrace();
    }
    String[] basePaths = a.split(SEP);
    String[] otherPaths = b.split(SEP);

    // check common part
    int n = 0;
    for(; n < basePaths.length && n < otherPaths.length; n ++)
    {
        if( basePaths[n].equals(otherPaths[n]) == false )
            break;
    }

    // compose the new path
    StringBuffer tmp = new StringBuffer("");
    for(int m = n; m < basePaths.length; m ++)
        tmp.append(".."+SEP);
    for(int m = n; m < otherPaths.length; m ++)
    {
        tmp.append(otherPaths[m]);
        tmp.append(SEP);
    }

    // get path string
    String result = tmp.toString();

    // remove last "/" if path is a file
    if (isfile && result.endsWith(SEP)){
        result = result.substring(0,result.length()-1);
    }

    return result;
}

0

通过Dónal的测试,唯一的更改-如果没有公共根,它将返回目标路径(可能已经是相对路径)

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);
    }
}

0

如果您正在编写Maven插件,则可以使用Plexus'PathTool

import org.codehaus.plexus.util.PathTool;

String relativeFilePath = PathTool.getRelativeFilePath(file1, file2);

0

如果“路径”不适用于JRE 1.5运行时或Maven插件

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();
    }

}


-1
private String relative(String left, String right){
    String[] lefts = left.split("/");
    String[] rights = right.split("/");
    int min = Math.min(lefts.length, rights.length);
    int commonIdx = -1;
    for(int i = 0; i < min; i++){
        if(commonIdx < 0 && !lefts[i].equals(rights[i])){
            commonIdx = i - 1;
            break;
        }
    }
    if(commonIdx < 0){
        return null;
    }
    StringBuilder sb = new StringBuilder(Math.max(left.length(), right.length()));
    sb.append(left).append("/");
    for(int i = commonIdx + 1; i < lefts.length;i++){
        sb.append("../");
    }
    for(int i = commonIdx + 1; i < rights.length;i++){
        sb.append(rights[i]).append("/");
    }

    return sb.deleteCharAt(sb.length() -1).toString();
}

-2

伪代码:

  1. 用路径分隔符(“ /”)分割字符串
  2. 通过遍历拆分字符串的结果来找到最大的通用路径(因此,在两个示例中,您最终将以“ / var / data”或“ / a”结尾)
  3. return "." + whicheverPathIsLonger.substring(commonPath.length);

2
这个答案充其量是hack。窗户呢?
Qix-蒙尼卡(Monica)
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