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
我知道这有点晚了,但是,我创建了一个解决方案,适用于任何java版本。
public static String getRealtivePath(File root, File file)
{
String path = file.getPath();
String rootPath = root.getPath();
boolean plus1 = path.contains(File.separator);
return path.substring(path.indexOf(rootPath) + rootPath.length() + (plus1 ? 1 : 0));
}
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));
}
}
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();
}
我知道这有点晚了,但是,我创建了一个解决方案,适用于任何java版本。
public static String getRealtivePath(File root, File file)
{
String path = file.getPath();
String rootPath = root.getPath();
boolean plus1 = path.contains(File.separator);
return path.substring(path.indexOf(rootPath) + rootPath.length() + (plus1 ? 1 : 0));
}
在另一个答案中提到的错误是由Apache HttpComponents中的URIUtils解决的
public static URI resolve(URI baseURI,
String reference)
对象的URI引用 基本URI。解决bug的方法 java.net.URI ()
我的版本大致基于马特和史蒂夫的版本:
/**
* 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());
}