Question

【php学习】PHP中判断目录是否为空的函数？

Answer 1

大体就是这样的：function is_dir_null($dir){
if (is_dir($dir)) {
if ($dh = opendir($dir)) {
//var_dump($dh);
$s='';
while (($file = readdir($dh))) {
if($file!='.' && $file!='..'){
$s.="<br/>".$file."<br/>";
}

}
if($s==''){
return "该目录为空";
}else{
return "目录中有内容".$s;

}
closedir($dh);
}
}}
echo is_dir_null("./");

Answer 2

function dir_list($path, $exts = '', $list = array()) {
$path = dir_path($path);
$files = glob($path.'*');
if ($files) {
foreach($files as $v) {
if (empty($exts) || pathinfo($v, PATHINFO_EXTENSION) == $exts) {
$list[] = $v;
if (is_dir($v)) {
$list = dir_list($v, $exts, $list);
}
}
}
}
return $list;
}
然后看返回值是什么，自己判断就行了

Answer 3

if($函数 != ''"){执行为空的操作}

Answer 4

function my_judge_empty_dir($directory)
{
$handle = opendir($directory);
while (($file = readdir($handle)) !== false)
{
if ($file != "." && $file != "..")
{
closedir($handle);
return false;
}
}
closedir($handle);
return true;
} 用这个方法吧

Answer 5

function my_judge_empty_dir($directory)
{
$handle = opendir($directory);
while (($file = readdir($handle)) !== false)
{
if ($file != "." && $file != "..")
{
closedir($handle);
return false;
}
}
closedir($handle);
return true;
} 用这个方法吧