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问题描述
大家好,请在这里需要帮助。我有一个视频播放器,我想从mysql数据库填充视频,这是我的代码。
Hi everybody, pls i need help here. i ve a video player and i want to populate video on it from mysql database, here is my code.
<ul id="playlist1" style="display:none;"> <li data-thumb-source="assets/img/ddd.jpg" data-video-source="assets/video/ddd.mp4" data-poster-source="assets/img/ddd.jpg" data-downloadable="yes"> <div data-video-short-description=""> <div> <p class="minimalDarkThumbnailTitle">ddd</p> <p class="minimalDarkThumbnailDesc">dddd.</p> </div> </div> <div data-video-long-description=""> <div> <p class="minimalDarkVideoTitleDesc">dddd</p> <p class="minimalDarkVideoMainDesc">ddd</p> <p>For more information about this please follow <a href="#" target="_blank">this link</a></p> </div> </div> </li> </ul>和我这样的mysql数据库
and i ve mysql database like this
id | data-thumb-source | data-video-source | data-poster-source | minimalDarkThumbnailTitle | minimalDarkThumbnailDesc和php代码从mysql获取数据
and php code to get the data from mysql
<ul> <?php $sql = "SELECT * FROM `video`"; foreach ($db->query($sql) as $row) { $li = '<li> data-thumb-source="' .$row['data-thumb-source']. '"'; $li .= ' class="playlistItem" data-type="local"'; $li .= ' data-video-source="' .$row['data-video-source']. '"'; $li .= ' minimalDarkThumbnailTitle="' .$row['minimalDarkThumbnailTitle']. '"'; $li .= ' minimalDarkThumbnailDesc="' .$row['minimalDarkThumbnailDesc']. '"'; echo $li; } $db = null; ?> </ul>和视频列表仍为空白,无法显示视频,请帮助我解决问题
and video list is still blank, no video to display, pls help me to get it work
推荐答案sql = SELECT * FROM`video`; foreach (
db-> query( db->query(
sql) as
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