键入标题时,我注意到有关此问题的一些类似问题,但它们似乎不在PHP中。那么用PHP函数解决该问题的方法是什么?
待指定。
$a="/home/apache/a/a.php"; $b="/home/root/b/b.php"; $relpath = getRelativePath($a,$b); //needed function,should return '../../root/b/b.php'
有什么好主意吗?谢谢。
试试这个:
function getRelativePath($from, $to) { // some compatibility fixes for Windows paths $from = is_dir($from) ? rtrim($from, '\/') . '/' : $from; $to = is_dir($to) ? rtrim($to, '\/') . '/' : $to; $from = str_replace('\\', '/', $from); $to = str_replace('\\', '/', $to); $from = explode('/', $from); $to = explode('/', $to); $relPath = $to; foreach($from as $depth => $dir) { // find first non-matching dir if($dir === $to[$depth]) { // ignore this directory array_shift($relPath); } else { // get number of remaining dirs to $from $remaining = count($from) - $depth; if($remaining > 1) { // add traversals up to first matching dir $padLength = (count($relPath) + $remaining - 1) * -1; $relPath = array_pad($relPath, $padLength, '..'); break; } else { $relPath[0] = './' . $relPath[0]; } } } return implode('/', $relPath); }
这将给
$a="/home/a.php"; $b="/home/root/b/b.php"; echo getRelativePath($a,$b), PHP_EOL; // ./root/b/b.php
和
$a="/home/apache/a/a.php"; $b="/home/root/b/b.php"; echo getRelativePath($a,$b), PHP_EOL; // ../../root/b/b.php
$a="/home/root/a/a.php"; $b="/home/apache/htdocs/b/en/b.php"; echo getRelativePath($a,$b), PHP_EOL; // ../../apache/htdocs/b/en/b.php
$a="/home/apache/htdocs/b/en/b.php"; $b="/home/root/a/a.php"; echo getRelativePath($a,$b), PHP_EOL; // ../../../../root/a/a.php