php编写的简单页面跳转功能实现代码

复制代码 代码如下:

//链接数据库'查询
mysql_connect('localhost','username','userpwd')or die("数据库链接失败".mysql_error());
mysql_select_db('库名');
mysql_query('set names utf8');
$sql1="select * from user ";
$query1=mysql_query($sql1);
$count=array();
while($row=mysql_fetch_assoc($query1)){
    $count[]=$row;
}
$totalnews=count($count);
//判断page
if($_GET['page']){
    $page=$_GET['page'];
}else{
    $page=1;
}
$start=($page-1)*$newnum;
   $sql="select * from user limit $start,$newnum";
   $query=mysql_query($sql);
   $ret=array();
   while($row=mysql_fetch_assoc($query)){
       $ret[]=$row;
       }
?>
//表格样式
<div>
   <table;>
     <?php foreach ($ret as $key=>$value){ ?>
       <tr>
           <td><?php echo $value['id'] ?></td>
           <td><?php echo $value['username'] ?></td>
           <td><?php echo $value['pwd'] ?></td>
           <td>删除|修改</td>
       </tr>
     <?php } ?>
     <tr >
//页面跳转
           <td colspan="4"><a href="upload.php?page=1">首页</a> <a href="upload.php?page=<?php echo $lastpage ?>">上一页</a> <?php echo $page; ?>/<?php echo $pagenum; ?> <a href="upload.php?page=<?php echo $nextpage; ?>">下一页</a> <a href="upload.php?page=<?php echo $pagenum ?>">尾页</a><input type="text" /><input type="button" value="跳转"/>
           </td>
     </tr>
   </table>
</div>