我从递归glob函数中得到了路径列表。。。类似的东西
/a/b/aaaa.txt
/c/d/bbbb.txt
/a/z/dddd.txt
/a/b/c/d/yyyy.txt
将它们放入阵列的最简单方法是什么：
$a[a][b][] = aaaa.txt
$a[c][d][] = bbbb.txt
$a[a][z][] = dddd.txt
$a[a][b][c][d][] = yyyy.txt
如果您掌握了递归的概念，这听起来像是一个家庭作业练习测试。一旦你理解了这个概念，这类问题就很容易解决。
$paths = array(
'/a/b/aaaa.txt',
'/c/d/bbbb.txt',
'/a/z/dddd.txt',
'/a/b/c/d/yyyy.txt',
);
$a = array();
foreach ($paths as $path)
{
$newArray = buildArrayFromPath($path);
$a = array_merge_recursive($a, $newArray);
}
/**
* Make an array from a path string
*/
function buildArrayFromPath($path)
{
$path = trim($path, '/');
$parts = explode('/', $path);
return recursiveBuildArray($parts);
}
/**
* Recursively build a multidimensional array from path parts
*/
function recursiveBuildArray(array $left, $new = array())
{
$key = array_shift($left);
if (count($left) > 1) {
$new[$key] = recursiveBuildArray($left, $new);
}
else {
$new[$key][] = array_pop($left);
}
return $new;
}
// check if we got back the expected result
$benchmark['a']['b'][] = 'aaaa.txt';
$benchmark['c']['d'][] = 'bbbb.txt';
$benchmark['a']['z'][] = 'dddd.txt';
$benchmark['a']['b']['c']['d'][] = 'yyyy.txt';
// TRUE if $a and $b have the same key/value pairs in the same order and of the same types.
var_dump($a === $benchmark);
完全递归的解决方案，当我需要合并2个glob数组时，我会使用它。所有重复项仅插入一次。
function add_path_to_array($path, $array) {
$path = trim($path, '/');
$exploded = explode('/',$path);
if (count($exploded) > 1) {
$key = $exploded[0];
if (!isset($array[$key]))
$array[$key] = array();
$array[$key] = array_replace_recursive($array[$key], add_path_to_array(substr($path, strpos($path, '/')), $array[$key]));
}
else
if (!in_array($path, $array))
$array[] = $path;
return $array;
}
$list = array(
'/a/b/aaaa.txt',
'/c/d/bbbb.txt',
'/a/z/dddd.txt',
'/a/b/c/d/yyyy.txt',
);
$nested_array = array();
foreach($list as $path)
$nested_array = add_path_to_array($path, $nested_array);
var_dump($nested_array);