我的linux机器上有这样的文件夹树结构:
/dir/yyyy/mm/dd/HH
例如: -
/dir/2014/03/01/08
/dir/2014/03/20/09
/dir/2014/03/01/10
/dir/2014/08/01/10
/dir/2014/12/15/10
/dir/2015/01/01/14
我想在php中找到最老的路径,如下所示:
最古老的路径是:2014-03-01 08
最新路径为:2015-01-01 14
怎么做?
答案 0 :(得分:0)
它可以写得更好但是有效
$paths = array(
'/dir/2014/03/01/08',
'/dir/2014/03/20/09',
'/dir/2014/03/01/10',
'/dir/2014/08/01/10',
'/dir/2014/12/15/10',
'/dir/2015/01/01/14',
);
$dates = array();
foreach($paths as $path)
{
$matches = array();
preg_match('#([^\/]+?)\/([^\/]+?)\/([^\/]+?)\/([^\/]+?)\/([^\/]+)#', $path, $matches);
$dates[$path] = strtotime(sprintf("%s-%s-%s %s:00:00", $matches[2], $matches[3], $matches[4], $matches[5]));
}
asort($dates);
$dates = array_keys($dates);
$oldest = array_shift($dates);
$newest = array_pop($dates);
它将regex的日期查找更改为unixtimestamp,然后对其进行排序并返回已排序数组的顶部和底部值。
答案 1 :(得分:0)
有点像Pascal风格)
<?php
$oldest = '';
$newest = '';
$iterator = new RecursiveIteratorIterator(new RecursiveDirectoryIterator('./dir/'));
foreach ($iterator as $file => $object) {
if ($iterator->getDepth() === 4) {
$name = $object->getPath();
if ($name > $newest) {
$newest = $name;
}
if (empty($oldest) or $name < $oldest) {
$oldest = $name;
}
}
}
var_export([$oldest, $newest]);
结果:
array (
0 => './dir/2014/03/01/08',
1 => './dir/2015/01/01/14',
)
答案 2 :(得分:0)
您所要做的就是遍历每个文件夹并找到编号最小的目录。如果您将文件路径存储在数据库中,则可以更容易,但从您的问题来看,您似乎想要搜索文件夹。
<?php
$base = 'dir';
$yearLowest = lowestDir($base);
$monthLowest = lowestDir($yearLowest);
$dayLowest = lowestDir($monthLowest);
echo $dayLowest;
function lowestDir($dir) {
$lowest = null;
$handle = opendir($dir);
while(($name = readdir($handle))) {
if($name == '.' || $name == '..') {
continue;
}
if(is_dir($dir.'/'.$name) && ($lowest == null || $name < $lowest)) {
$lowest = $name;
}
}
closedir($handle);
return $dir.'/'.$lowest;
}
?>