只是尝试从数据库中显示一条记录......但是收到错误(在获取数组时)
$con=mysql_connect("localhost","root","");
mysql_select_db("school");
$iid=$_REQUEST['id2']; //mysql_fetch_array(): supplied argument is not a valid
$d=mysql_query("select * from notification where noticeid=$iid");
$data=mysql_fetch_array( $d );
这是我的显示标准......
<tr>
<td class=""><?php echo $data['noticeid'] ?></td>
<td class=""><?php echo $data['notificationtitle'] ?></td>
<td class=""><?php echo $data['notificationbody'] ?></td>
</tr>
答案 0 :(得分:0)
试试这个:noticeid ='“。$ iid。” “
$d=mysql_query("select * from notification where noticeid='".$iid."' ");
$data=mysql_fetch_array( $d );
答案 1 :(得分:0)
您可能想引用变量:
如果noticeid是一个字符串:
$d=mysql_query("select * from notification where noticeid='".$iid."'");
如果noticeid是一个数字:
$d=mysql_query("select * from notification where noticeid=".$iid);