php获取css文件中的所有类名

Question

假设我有一个 css 文件，如图所示...

span {
    //whatever
}

.block {
    //whatever
}

.block, .something {
    //whatever
}

.more,
h1,
h2 {
    //whatever
}

我想提取所有类名并将其放入数组中，但我想保留结构，所以数组看起来像...

["span", ".block", ".block, .something", ".more, h1, h2"]

所以有四项。

这是我的尝试...

$homepage = file_get_contents("style.css");

//remove everything between brackets (this works)
$pattern_one = '/(?<=\{)(.*?)(?=\})/s';

//this regex does not work properly
$pattern_two = "/\.([\w]*)\s*{/";

$stripped = preg_replace($pattern_one, '', $homepage);
$selectors = array();
$matches = preg_match_all($pattern_two, $stripped, $selectors);

用于模式 2 的正确正则表达式是什么？

Answer 1

像这样？

<?php
$css = "span {
    //whatever
}

.block {
    //whatever
}

.block, .something {
    //whatever
}

.more,
h1,
h2 {
    //whatever
}";

$rules = [];

$css = str_replace("\r", "", $css); // get rid of new lines
$css = str_replace("\n", "", $css); // get rid of new lines

// explode() on close curly braces
// We should be left with stuff like:
//   span{//whatever
//   .block{//whatever
$first = explode('}', $css);

// If a } didn't exist then we probably don't have a valid CSS file
if($first)
{
    // Loop each item
    foreach($first as $v)
    {
        // explode() on the opening curly brace and the ZERO index should be the class declaration or w/e
        $second = explode('{', $v);

        // The final item in $first is going to be empty so we should ignore it
        if(isset($second[0]) && $second[0] !== '')
        {
            $rules[] = trim($second[0]);
        }
    }
}

// Enjoy the fruit of PHP's labor :-)
print_r($rules);