【发布时间】:2015-12-08 01:06:47
【问题描述】:
我正在尝试在我的 WordPress 项目中使用 ajax 来根据帖子 ID 动态加载视频。当我点击视频链接时,我收到错误 500 内部服务器错误。我通过 ajax 将帖子 ID 发送到 PHP 脚本,在那里我尝试使用 ID 获取帖子的帖子元,然后显示存储在该帖子元中的视频 url。谁能看出我哪里出错了?
下面是我的简码函数
/**
* Render the HTML code for the speaker interviews
*
* @return $output string The HTML code to be rendered, caught by ob_start();
*/
function test_view_shortcode_fn( $attributes ) {
ob_start();
$args = array( 'post_type' => 'test_video',
'post_status' => 'publish',
'posts_per_page' => 1
);
$main_video = new WP_Query( $args );
if( $main_video->have_posts() ) : while ( $main_video->have_posts() ) : $main_video->the_post();
$video = get_post_meta( get_the_ID(), '_video_url' );
$poster = get_post_meta( get_the_ID(), '_video_poster' );
?>
<div class="row">
<div class="columns small-12 medium-7 large-8 video-single">
<?php echo do_shortcode("[video src='" . $video[0] . "' poster='" . $poster[0] . "' width='800' height='640' preload='none' ]") ?>
<ul class="channel-share">
<li><a class="text-center" href="https://www.facebook.com/sharer/sharer.php?u=<?php the_permalink(); ?>" title="Share on Facebook" target="_blank"><i class="fa fa-facebook"></i></a></li>
<li><a class="text-center" href="https://twitter.com/home?status=<?php the_title(); ?>-<?php the_permalink(); ?>" title="Share on Twitter" target="_blank"><i class="fa fa-twitter"></i></a></li>
<li><a class="text-center" href="https://www.linkedin.com/shareArticle?mini=true&url=<?php the_permalink(); ?>&title=<?php the_title(); ?>" title="Share on Linked In" target="_blank"><i class="fa fa-linkedin"></i></a></li>
<li><a class="text-center" href="https://plus.google.com/share?url=<?php the_permalink(); ?>" title="Share on Google+" target="_blank"><i class="fa fa-google-plus"></i></a></li>
</ul>
<?php the_content(); ?>
</div>
<div class="columns small-12 medium-5 large-4 video-single">
<aside>
<ul class="iot-channel-sidebar">
<?php
$args = array(
'post_type' => 'test_video',
'posts_per_page' => 5,
'orderby' => 'rand'
);
$channel = new WP_Query( $args );
if( $channel->have_posts() ) : while( $channel->have_posts() ) : $channel->the_post();
$poster = get_post_meta( get_the_ID(), '_video_poster' );
$id = get_the_ID();
$video_ajax_url = VIDEO_URI . '/inc/video_ajax.php';
?>
<?php if( $poster ) { ?>
<li class="video-archive-item">
<div class="video-poster">
<a title="<?php the_title(); ?>" id="video_<?php echo $id; ?>">
<img src="<?php echo $poster[0]; ?>" alt="<?php the_title(); ?>" />
</a>
</div>
<div class="video-title">
<a title="<?php the_title(); ?>">
<h4><?php the_title(); ?></h4>
</a>
</div>
</li>
<script type="text/javascript">
jQuery("#video_<?php echo $id; ?>").click(function(){
jQuery.ajax({
url: '<?php echo $video_ajax_url; ?>',
type: 'POST',
data: {id: '<?php echo $id; ?>'},
success: function(data){
$('.wp-video').html(data); // Load data into a <div> as HTML
}
});
});
</script
<?php } ?>
<?php endwhile; endif; ?>
</ul>
</aside>
</div>
</div>
<?php endwhile; endif;
$output = ob_get_clean();
return $output;
}
我的 ajax:
<?php
global $post;
$postID = $_POST['id'];
setup_postdata( $postID );
echo $postID;
$video = get_post_meta( $postID, '_video_url' );
$poster = get_post_meta( $postID, '_video_poster' );
echo do_shortcode("[video src='" . $video[0] . "' poster='" . $poster[0] . "' width='800' height='640' preload='none' ]")
?>
【问题讨论】:
-
检查您的错误日志,500 来自 PHP 中的某些内容。
标签: javascript php jquery ajax wordpress