<?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;"> <?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~~
如果是的话,不要加上单引号。第二,你加个记录判断
if(rownum>0){
.....
}