Php 推荐用户视频,如youtube
我正在尝试制作一个类似youtube的主页。使用下面的代码,我想制作推荐给我的用户的视频 以下代码仅显示用户的视频Php 推荐用户视频,如youtube,php,mysql,Php,Mysql,我正在尝试制作一个类似youtube的主页。使用下面的代码,我想制作推荐给我的用户的视频 以下代码仅显示用户的视频 <?php $query = "SELECT user.uid, user.user_name, user.user_avatar, user_posts.uid_dk, user_posts.post_id
<?php $query = "SELECT
user.uid,
user.user_name,
user.user_avatar,
user_posts.uid_dk,
user_posts.post_id,
user_posts.post_name,
user_posts.post_info,
user_posts.post_time,
user_posts.post_ext,
user_posts.post_num,
user_posts.post_views
FROM user
JOIN user_posts
ON user_posts.uid_dk = user.uid
WHERE user_name='$user_name' LIMIT 5";
$run_query = mysql_query($query);
while($data=mysql_fetch_assoc($run_query)){
$post_name=$data['post_name'];
$post_time = $data['post_time'];
$post_views = $data['post_views'];
$post_numid = $data['post_num'];
$post_id = $data['post_id'];
$user_name = $data['user_name'];
$user_avatar = $data['user_avatar'];
?>
<div class="onerilent"><img src="<?php echo $user_avatar;?>"><?php echo $user_name ;?> Recommended for you</div>
<div class="onmnwrp">
<div class="onmn">
<div class="onmn_img"><img src="<?php echo $base_url.'user_uploads/'.$post_num;?>.png"></div>
<div class="onmg_tit"><?php echo $post_name;?></div>
<div class="onm_snm">gönderen: <?php echo $user_name;?></div>
<div class="onm_tim"><?php echo $post_views;?> views</div>
</div>
</div>
<?php } ?>
“>推荐给您
.png“>
哥德伦:
意见
我只想展示这一部分一次
<div class="onerilent"><img src="<?php echo $user_avatar;?>"><?php echo $user_name ;?> Recommended for you</div>
“>推荐给您
任何人都可以在这方面帮助我吗?最简单的方法是使用计数器,如下所示:
<?php
$query = "SELECT
user.uid,
user.user_name,
user.user_avatar,
user_posts.uid_dk,
user_posts.post_id,
user_posts.post_name,
user_posts.post_info,
user_posts.post_time,
user_posts.post_ext,
user_posts.post_num,
user_posts.post_views
FROM user
JOIN user_posts
ON user_posts.uid_dk = user.uid
WHERE user_name='$user_name' LIMIT 5";
$run_query = mysql_query($query);
$counter = 1;
while($data=mysql_fetch_assoc($run_query)){
$post_name=$data['post_name'];
$post_time = $data['post_time'];
$post_views = $data['post_views'];
$post_numid = $data['post_num'];
$post_id = $data['post_id'];
$user_name = $data['user_name'];
$user_avatar = $data['user_avatar'];
if($counter == 1){
$counter++;
echo '<div class="onerilent"><img src="'.$user_avatar.'">'.$user_name.' Recommended for you</div>';
}
?>
<div class="onmnwrp">
<div class="onmn">
<div class="onmn_img"><img src="<?php echo $base_url.'user_uploads/'.$post_num;?>.png"></div>
<div class="onmg_tit"><?php echo $post_name;?></div>
<div class="onm_snm">gönderen: <?php echo $user_name;?></div>
<div class="onm_tim"><?php echo $post_views;?> views</div>
</div>
</div>
<?php
}
?>
.png“>
哥德伦:
意见
注意$counter变量在循环之前设置为1,在循环内部有一个条件来检查它是否设置为值1,如果设置为值1,则返回html并递增$counter,这样它就不再设置为1了。这是非常好的解释性答案。谢谢。小心
$user\u name
从哪里来-这可能是SQL注入漏洞。@halfer如何修复它?看一看。