用php写一个通过get请求url参数 https://xxx.com/we.php?url=33445566 获得we.php?url=后面的 33445566 参数构建一个新的url https://www.wegame.com.cn/api/forum/lua/wg_video/get_video_play_info?videoid=han_33445566&upload_type=0&platform=2 再请求链接https://www.wegame.com.cn/api/forum/lua/wg_video/get_video_play_info?videoid=han_33445566&upload_type=0&platform=2 列如获取到内容[{"url_list":["https:\/\/ugc-svp.wegame.com.cn\/_0bc35ybegaacjian37nrkjsyh3wdipxaeq2a.f0.mp4?dis_k=609924090fa95c5b93b8c625c3c6508c&dis_t=1725821351"]后截取内容https:

```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; ?> ``` [2024-09-09 04:49:12 | AI写代码神器 | 291点数解答]

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