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请问我需要帮助。

我做了一个分页，我得到了这个错误：

Notice: Undefined index: sn in C:\xampp\htdocs\SchoolCare\pagination first try.php on line 116

这是我的代码：

<?php
require "../includes/dc_conect.php";

$sql1=("SELECT * FROM staffs WHERE staff_id='super_admin'");
$result1=mysql_query($sql1, $link) or die (mysql_error());
$dbfield1=mysql_fetch_assoc($result1);
$count1=mysql_num_rows($result1);

$sql=("SELECT COUNT(sn) FROM staffs ORDER BY sn ASC");
$query=mysql_query($sql, $link) or die (mysql_error());
$row=mysql_fetch_row($query);


//here we have the total row counts chyke okoli
$rows= $row[0];
//here is the number of result we wana display chyke okoli
$page_rows= 10;
// this tells us the page number of our last page---chyke okoli
$last = ceil($rows/$page_rows);
//make sure $last cannot be less than 1---chyke
if ($last<1){
   $last=1;
}
//establish the $pagenum variable--chike
$pagenum = 1;
//get the page num var from the URL if its present else its equal to 1;
if(isset($_GET['pn'])){
  $pagenum = preg_replace('#[^0-9]#', '', $_GET['pn']);
}
//make sure the pagenum is less than 1 and d last is nt more than d last
if($pagenum < 1){
  $pagenum = 1;
}elseif($pagenum > $last)
    {
    $pagenum = $last;
    }
//set the range of rows to query -----vitalischikeokoli
$limit = 'LIMIT '.($pagenum - 1) * $page_rows.','.$page_rows;   
//write a query to return one page of row
$sql=("SELECT sn, fname, sname, staff_id FROM staffs ORDER BY sn ASC $limit");
$query=mysql_query($sql, $link) or die (mysql_error());
//show the user what page they are into
$textline1 = "Staff Records (<b>$rows</b>)";
$textline2 = "Page <b>$pagenum</b> of <b>$last</b>";
//pagination Ctrl var
$paginationCtrls = '';
//if more than 1 page result
if($last != 1){
  /*First we check if we on page one if so we dont need the prev---vitalisokolichike---*/

     if($pagenum > 1){
       $previous = $pagenum - 1;
       $paginationCtrls .='<a href="'.$_SERVER['PHP_SELF'].'?pn='.$previous.'">PREV</a> &nbsp; &nbsp;';
       //render clickable number link that should appear at d left of the target page num

       for($i = $pagenum-4; $i < $pagenum; $i++)
       {
         if($i > 0){
             $paginationCtrls .='<a href="'.$_SERVER['PHP_SELF'].'?pn='.$i.'">'.$i.'</a> &nbsp; &nbsp;';
         }
       }
     }
     //render tha target page number without link
     $paginationCtrls .=''.$pagenum.' &nbsp; ';
     //render d clikable number
     for($i = $pagenum+1; $i <= $last; $i++){
       $paginationCtrls .='<a href="'.$_SERVER['PHP_SELF'].'?pn='.$i.'">'.$i.'</a> &nbsp; &nbsp;';
        if($i >= $pagenum+4){
        break;
        }
     }
     //check if we are on the last page then adding NEXT
      if($pagenum != $last){
        $next = $pagenum + 1;
        $paginationCtrls .='&nbsp; &nbsp;<a href="'.$_SERVER['PHP_SELF'].'?pn='.$next.'">NEXT</a>';
      }
}

?>

然后我把输出放在一张桌子上

<table height="182" id="man_staff_table">
               <tr>
                   <td height="39"><div align="right" id="dboard">
                     <div id="div2"><a href="../admin/admin_dash_board.php">Dashboard</a></div>
                   </div></td>
                   <td height="39">&nbsp;</td>
                   <td height="39">&nbsp;</td>
                   <td height="39">&nbsp;</td>
                   <td height="39">&nbsp;</td>
                   <td height="39">&nbsp;</td>
                   <td height="39"><div id="dboard"><a href="add_staff.php">Add a Staff</a></div></td>
               </tr>
               <tr>
                 <td width="119" height="17">&nbsp;</td>
                 <td width="268">&nbsp;</td>
                 <td width="13">&nbsp;</td>
                 <td width="31">&nbsp;</td>
                 <td width="15">&nbsp;</td>
                 <td width="195">&nbsp;</td>
                 <td width="112">&nbsp;</td>
           </tr>

               <tr>
                 <td height="39" colspan="7"><h3>Manage Staff Records</h3><hr></td>
           </tr>
               <tr>
                 <td height="17" colspan="7"><div align="right"><?php echo $textline2; ?></div></td>
           </tr>
               <tr>
                 <td height="21" colspan="7"><table width="960" align="center" class="imagetable">
                   <tr>
                     <th width="49">S/N</th>
                     <th width="101">Surname</th>
                     <th width="118">First Name </th>
                     <th width="80">Staff ID </th>
                     <th colspan="4">&nbsp;</th>
                   </tr>
                   <?php do{ ?>
                   <tr>
                     <td><?php echo @$row['sn']; ?></td>
                     <td><?php echo @$row['sname']; ?></td>
                     <td><?php echo @$row['fname']; ?></td>
                     <td><?php echo @$row['staff_id']; ?></td>
                     <td width="273">&nbsp;</td>
                     <td width="31"><div align="center"><img src="../image/view_icon.png" alt="View" width="15" height="12"></div></td>
                     <td width="31"><div align="center"><img src="../image/edit_icon.png" alt="Edit" width="15" height="17"></div></td>
                     <td width="31"><div align="center"><img src="../image/delete_icon.png" alt="Delete" width="15" height="15"></div></td>
                     <?php } while($row=mysql_fetch_assoc($query)) ?>
                   </tr>
                 </table></td>
           </tr>
               <tr>
                 <td height="17" colspan="7"><div align="center" id="pagination_control"><?php echo $paginationCtrls; ?></div></td>
           </tr>
         </table>

请问我需要帮助我只是使用@函数来使输出静音，但是那个特定的角色是空的，而且看起来不太好！[在此处输入图像描述][1]

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1 回答 1

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直到循环结束，do{} while()循环才会真正获取行，因此您正在使用不存在的数据（或者更确切地说，来自您的 SELECT COUNT(*) 数据，因为那是您最后一次分配 $行）在您获取 $row 之前。切换do{} while()到while：

while($row=mysql_fetch_assoc($query)){
  // now output your rows
}
于 2013-11-01T19:06:02.800 回答