PHP一键搞定：从URL参数提取到视频链接截取全流程

```php <?php // 获取url参数 $url = $_GET['url']; // 构建新的url $new_url = "https://www.wegame.com.cn/api/forum/lua/wg_video/get_video_play_info?videoid=han_" . $url . "&upload_type=0&platform=2"; // 发起请求 $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $new_url); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); $response = curl_exec($ch); curl_close($ch); // 解析响应内容 $json_data = json_decode($response, true); $video_url = $json_data[0]; // 截取内容里面的https://ugc-svp $start = strpos($video_url, 'https://'); $end = strpos($video_url, '.f0.mp4'); $result = substr($video_url, $start, $end - $start); echo $result; ?> ``` ######[AI写代码神器 | 282点数解答 | 2024-09-09 04:55:07]