我不想扫描目录及其子目录中的所有文件.并在数组中获取它们的路径.就像数组中目录中文件的路径将只是
I want not scan all the files in a directory and its sub-directory. And get their path in an array. Like path to the file in the directory in array will be just
路径 -> text.txt
path -> text.txt
而子目录中文件的路径为
while the path to a file in sub-directory will be
somedirectory/text.txt
somedirectory/text.txt
我可以扫描单个目录,但它返回所有文件和子目录,无法区分.
I am able to scan single directory, but it returns all the files and sub-directories without any ways to differentiate.
if ($handle = opendir('fonts/')) {
/* This is the correct way to loop over the directory. */
while (false !== ($entry = readdir($handle))) {
echo "$entry<br/>";
}
closedir($handle);
}
获取目录和子目录中所有文件及其路径的最佳方法是什么?
What is the best way to get all the files in the directory and sub-directory with its path?
使用 SPL 中的 DirectoryIterator 可能是最好的方法:
Using the DirectoryIterator from SPL is probably the best way to do it:
$it = new RecursiveIteratorIterator(new RecursiveDirectoryIterator('.'));
foreach ($it as $file) echo $file."
";
$file
是一个 SPLFileInfo-object.它的 __toString() 方法将为您提供文件名,但还有其他几种方法也很有用!
$file
is an SPLFileInfo-object. Its __toString() method will give you the filename, but there are several other methods that are useful as well!
更多信息参见:http://www.php.net/manual/en/class.recursivedirectoryiterator.php
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