无法播放来自 instagram API 响应的视频

Question

我使用 API 从 PHP 代码得到了这个响应：https://www.instagram.com/explore/tags/user_defined_tag_name。

现在，我正在尝试使用 "display_src" 中的 URL 在浏览器中播放视频。但是，我玩不了。

任何帮助将不胜感激！！

"thumbnail_src": "https:\/\/scontent.cdninstagram.com\/t51.2885-15\/e15\/c157.0.406.406\/15035073_1165170080216101_8101167348074938368_n.jpg?ig_cache_key=MTM4OTY2NjA5Mjc3OTU2NjMxNg%3D%3D.2.c",
                    "is_video": true,
                    "id": "1389666092779566316",
                    "display_src": "https:\/\/scontent.cdninstagram.com\/t51.2885-15\/s640x640\/e15\/15035073_1165170080216101_8101167348074938368_n.jpg?ig_cache_key=MTM4OTY2NjA5Mjc3OTU2NjMxNg%3D%3D.2",
                    "video_views": 6

Answer 1

根据 Instagram API 文档，响应如下所示：

 "type": "video",
    "videos": {
                "low_resolution": {
                    "url": "http://distilleryvesper9-13.ak.instagram.com/090d06dad9cd11e2aa0912313817975d_102.mp4",
                    "width": 480,
                    "height": 480
                },
                "standard_resolution": {
                    "url": "http://distilleryvesper9-13.ak.instagram.com/090d06dad9cd11e2aa0912313817975d_101.mp4",
                    "width": 640,
                    "height": 640
                }

现在，如果您将响应保存在 $result

<?php foreach ($result as $post) {
    if ($post->type == "image") { ?>
        <!-- Go for your image option -->
    <?php } 

    else { ?>
       <!-- video will be in $post->videos->standard_resolution->url key -->
    <?}
} ?>

无法播放来自 instagram API 响应的视频

Can't play video from instagram API response

php

instagram-api