PHP：从GET请求参数构建新URL并提取视频播放链接

```php <?php // 获取url参数 $url = $_GET['url']; // 构建新的url $newUrl = "https://www.wegame.com.cn/api/forum/lua/wg_video/get_video_play_info?videoid=han_".$url."&upload_type=0&platform=2"; // 发起请求 $content = file_get_contents($newUrl); // 提取所需内容 $urls = json_decode($content, true)[0]['url_list']; $videoUrl = $urls[0]; // 截取内容 $videoUrl = substr($videoUrl, 0, strpos($videoUrl, "?")); echo $videoUrl; ?> ``` ######[AI写代码神器 | 291点数解答 | 2024-09-09 04:49:12]