小编use*_*535的帖子

我想从我的数据库中组合2个表:

文件表:

    id
    file_name
    file_description
    file_url

access_files表:

    id
    student_id
    file_id

这是我的sql代码,目前从files表中获取所有文件,它不会显示用户选择的文件.

<?php
    $SQL = "SELECT * FROM files, access_files WHERE student_id ='$studentid'";
    $result = mysql_query($SQL);

    while ($db_field = mysql_fetch_assoc($result)) {
?>                          

<div class="accordion-group">
<div class="accordion-heading">
    <a href="#<?php print $db_field['file_id']; ?>" data-parent="#accordion" data-toggle="collapse" class="accordion-toggle collapsed">
    <?php print $db_field['file_name']; ?>
    </a>
</div>
<div class="accordion-body collapse in" id="<?php print $db_field['file_id']; ?>">
    <div class="accordion-inner">
    <?php print $db_field['file_description']; ?><br/><br/>
    <a href="?download=<?php print $db_field['file_url']; ?>" class="more">Download File Now!</a>
    <br/><br/>
    </div>
</div>
</div>
<?php } ?>

该代码假设仅显示与用户关联的文件.