简单实现Ajax无刷新分页效果

以下是我自己做的一个简单分页展示

<?php

 header("Content-type:text/html; charset=utf-8");
 $link = mysqli_connect('localhost','root','123','good');
if (!$link)
{
 die("连接错误: " . mysqli_connect_error());
}

$sql = "SELECT * FROM ecs_category";
$qry = mysqli_query($link,$sql);
$total = mysqli_num_rows($qry);
$per = 10;

$page = pageDivide($total,$per);

$sql = "SELECT cat_id,cat_name FROM category limit $sqlfirst,$shownu";
 $result = mysqli_query($link,$sql);

 //$goods = mysqli_fetch_all($result,MYSQLI_ASSOC);

 //mysqli_free_result($result);

echo <<<eof
 <style type="text/css">
 table{width:500px;margin:auto;border: 1px solid black; border-collapse:collapse;text-align:center;}
 td{border: 1px solid black;}
 </style>
 <table>
  <tr><td>序号</td><td>商品编号</td><td>商品类型</td></tr>
eof;

 $num = ($page - 1) * $per;
// foreach ($goods as $k => $v) {
 while($goods = mysqli_fetch_assoc($result)){
 echo "<tr>";
 echo "<td>".++$num."</td>";
 echo "<td>$goods[cat_id]</td>";
 echo "<td>$goods[cat_name]</td>";
 echo "</tr>";
 }

 echo "<tr><td colspan='3'>$pagecon</td></tr>";
 echo "</table>";

页面是不刷新跳转的，URL不会变，可以看到网站的数据交互

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