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
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)
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));
}
}
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());
}
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 ()
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()));
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);