<?php
$query=mysql_query("select storage_area,storage_tel,storage_stock from storage where storage_num='$storage_num'");
$result=mysql_fetch_assoc($query);
?>
<tr>
<td style="padding:5px;"> &nbsp; <?php echo $result['storage_area']; ?> </td> ---------*
.........我想输出storage表中某个storage_num的相应的storage_area,storage_tel,storage_stock,*所指那一行有错误,该怎么改? 帮忙啊~~