从数组中获取值并在mysql php中选择它

从数组中获取值并在mysql php中选择它,php,mysql,video,Php,Mysql,Video,我的问题是我存储在DB列(video_id)中,它的值与该存储类似(1,3,4,7)。数字表示视频表中视频的id,该表存储视频标题和文本以及视频的所有信息,但当我编写代码选择所有视频时,结果给我一个视频 <div class="panel-body"> <? $qu="SELECT video_id FROM `training_questions` WHERE id=$questions_id AND category_id=$training_id";

我的问题是我存储在DB列(video_id)中,它的值与该存储类似(1,3,4,7)。数字表示视频表中视频的id,该表存储视频标题和文本以及视频的所有信息,但当我编写代码选择所有视频时,结果给我一个视频

<div class="panel-body">
    <?
    $qu="SELECT video_id FROM `training_questions` WHERE id=$questions_id AND category_id=$training_id";
    $query_c=  mysqli_query($con, $qu);
    while ($row7 = mysqli_fetch_assoc($query_c)) {
        $vedoes=$row['video_id'];
        $pieces = explode(",", $vedoes);
    }
    for($i=0;$i<=count($pieces);$i++){
        $query = "SELECT * FROM `video` WHERE id IN $pieces[$i] ";
        $query_che=  mysqli_query($con, $query);
        while ($row2 = mysqli_fetch_assoc($query_che)) {  ?>
            <div class="col-lg-3 col-sm-6">
                <div class="thumbnail">
                    <div class="video-container">
                        <div class="thumb"> <a href="video_details.php?id=<?= $row2['id'] ?>"> <img src="../upload/<?= $row2['img'] ?>" class="img-responsive img-rounded media-preview" alt=""> <span class="zoom-image"><i class="icon-play3"></i></span> </a> </div>
                        </div>
                    <div class="caption">
                        <h6 class="no-margin">
                            <a href ="video_details.php?id=<?= $row2['id'] ?>" class="text-default"><?= $row2['title'] ?>
                            </a>
                        </h6>
                    </div>
                </div>
            </div>
        <? 
        }
    } ?>
</div>

如何修复它?

卸下

$pieces = explode(",", $vedoes);
没用

使for循环条件如下所示

for($i=0;$i<=count($vedoes);$i++)
$i=0;$i的