如何获取数据到html< ul> <李> < /锂> < / UL>使用php从mysq

本文介绍了如何获取数据到html< ul> <李> < /锂> < / UL>使用php从mysql数据库列出的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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