如何使用PHP检查目录是否为空?

问题描述：

我正在使用以下脚本读取目录.如果目录中没有文件，则应说为空.问题是，即使里面有ARE文件，它总是说目录是空的，反之亦然.

I am using the following script to read a directory. If there is no file in the directory it should say empty. The problem is, it just keeps saying the directory is empty even though there ARE files inside and vice versa.

<?php
$pid    =       $_GET["prodref"];
$dir    =       '/assets/'.$pid.'/v';
$q     =        (count(glob("$dir/*")) === 0) ? 'Empty' : 'Not empty';

    if ($q=="Empty")
        echo "the folder is empty"; 
    else
        echo "the folder is NOT empty";
?>

答

似乎您需要scandir而不是glob，因为glob无法看到Unix隐藏文件.

It seems that you need scandir instead of glob, as glob can't see unix hidden files.

<?php
$pid = basename($_GET["prodref"]); //let's sanitize it a bit
$dir = "/assets/$pid/v";

if (is_dir_empty($dir)) {
  echo "the folder is empty"; 
}else{
  echo "the folder is NOT empty";
}

function is_dir_empty($dir) {
  if (!is_readable($dir)) return NULL; 
  return (count(scandir($dir)) == 2);
}
?>

请注意，此代码不是效率的最高标准，因为不必读取所有文件即可知道目录是否为空.因此，更好的版本应该是

Note that this code is not the summit of efficiency, as it's unnecessary to read all the files only to tell if directory is empty. So, the better version would be

function dir_is_empty($dir) {
  $handle = opendir($dir);
  while (false !== ($entry = readdir($handle))) {
    if ($entry != "." && $entry != "..") {
      closedir($handle);
      return FALSE;
    }
  }
  closedir($handle);
  return TRUE;
}

顺便说一句，不要使用 words 来代替 boolean 值.后者的目的是告诉您是否有空.

By the way, do not use words to substitute boolean values. The very purpose of the latter is to tell you if something empty or not. An

a === b

表达式已经分别以编程语言FALSE或TRUE的形式返回Empty或Non Empty-因此，您可以将结果直接用于IF()这样的控制结构而没有任何中间值

expression already returns Empty or Non Empty in terms of programming language, FALSE or TRUE respectively - so, you can use the very result in control structures like IF() without any intermediate values

如何使用PHP检查目录是否为空?

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