在目录树中查找具有给定扩展名的文件 [英] Find files with a given extension in a directory tree

问题描述

我有一个文件夹，其中有一些文件和子文件夹.我需要返回具有给定扩展名的文件名或相对地址(如果文件位于子文件夹中)，例如 .html .

I have a folder in which I have some files and sub-folders. I need to return the names of the files or relative address (if the file is in sub-folder) that have a given extension, for example .html.

例如，这是给定文件夹的结构:

For example this is structure of given folder:

• /text/something.ncx
• /text/123.html
• content.opf
• toc.ncx
• Flow_0.html
• Flow_1.html
• Flow_2.html

因此，代码应返回此值(理想情况下是在数组中):

So the code should return this (in an array ideally):

• text/123.html
• Flow_0.html
• Flow_1.html
• Flow_2.html

它只是提取扩展名为 .html 的名称，但我真的不知道如何解决这个问题.

It's just pulling out names with .html extension, but I really don't how to solve this.

推荐答案

您可以使用 RecursiveDirectoryIterator 递归获取所有文件，然后使用 RegexIterator 过滤结果code> 仅遍历 *.html 文件.要获取相对地址，请 RecursiveDirectoryIterator :: getSubPathname() 可以使用.

You could use a RecursiveDirectoryIterator to get all files recursively, then filter the result with a RegexIterator to iterate over only the *.html files. To get the relative adress, RecursiveDirectoryIterator::getSubPathname() can be used.

$directory  = '../path/to/your/folder';
$all_files  = new RecursiveIteratorIterator(new RecursiveDirectoryIterator($directory));
$html_files = new RegexIterator($all_files, '/\.html$/');

foreach($html_files as $file) {
    echo $html_files->getSubPathname(), PHP_EOL;
}

如果您确实需要对数组进行迭代(通常不需要迭代对象)，则可以轻松地在 foreach 中构建一个对象，而不用 echo -遍历每个子路径

If you really need and array (with iterable objects you often don't), you can easily build one up in the foreach rather than echo-ing each subpath.

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