Javascript 存在时隐藏视频容器'；数据库中没有要显示的视频

下面是显示数据库中的练习视频的html代码：

<video id = "video" width="640" height="360" controls = "controls" src="admin/exercise_content/<?php echo $data['exercise_video']?>" alt="Exercise" type="video/mp4"> Your browser does not support the video tag. </video>

使用以下命令：

<? if ( trim($data['exercise_video']) != '' ) : ?>
<video id = "video" width="640" height="360" controls = "controls" src="admin/exercise_content/<?php echo $data['exercise_video']?>" alt="Exercise" type="video/mp4"> Your browser does not support the video tag. </video>
<? endif; ?>


使用以下命令：

<? if ( trim($data['exercise_video']) != '' ) : ?>
<video id = "video" width="640" height="360" controls = "controls" src="admin/exercise_content/<?php echo $data['exercise_video']?>" alt="Exercise" type="video/mp4"> Your browser does not support the video tag. </video>
<? endif; ?>


试试这个

<?php if ( $data['exercise_video'] != '' ) { ?>
<video id = "video" width="640" height="360" controls = "controls"src="admin/exercise_content/<?php echo $data['exercise_video']?>" alt="Exercise" type="video/mp4"> Your browser does not support the video tag.       </video>
<?php }  ?>


试试这个

<?php if ( $data['exercise_video'] != '' ) { ?>
<video id = "video" width="640" height="360" controls = "controls"src="admin/exercise_content/<?php echo $data['exercise_video']?>" alt="Exercise" type="video/mp4"> Your browser does not support the video tag.       </video>
<?php }  ?>


使用良好

<?php if (!empty($data['exercise_video']): ?>
<video id = "video" width="640" height="360" controls = "controls" src="admin/exercise_content/<?php echo $data['exercise_video']?>" alt="Exercise" type="video/mp4"> Your browser does not support the video tag. </video>
<?php endif ;?>

使用

<?php if (!empty($data['exercise_video']): ?>
<video id = "video" width="640" height="360" controls = "controls" src="admin/exercise_content/<?php echo $data['exercise_video']?>" alt="Exercise" type="video/mp4"> Your browser does not support the video tag. </video>
<?php endif ;?>

使用起来更好！空（$data['exercise_video']），因为如果代码不好用，它会抛出一个通知！空（$data['exercise_video']），因为如果未设置，代码将抛出通知