我试图通过我的搜索使用php / ajax_pagination链接到我的mysql数据库中的单行记录信息，我试图开始工作的代码是：
<div id="posts_content">
<?php
//Include pagination class file
include('Pagination.php');
//Include database configuration file
include('dbConfig.php');
$limit = 3;
//get number of rows
$queryNum = $db->query("SELECT COUNT(*) as postNum FROM posts");
$resultNum = $queryNum->fetch_assoc();
$rowCount = $resultNum['postNum'];
//initialize pagination class
$pagConfig = array(
'totalRows' => $rowCount,
'perPage' => $limit,
'link_func' => 'searchFilter'
);
$pagination = new Pagination($pagConfig);
//get rows
$query = $db->query("SELECT * FROM posts ORDER BY id DESC LIMIT $limit");
if($query->num_rows > 0){ ?>
<div class="posts_list">
<?php
while($row = $query->fetch_assoc()){
$postID = $row['id'];
?>
<div class="list_item"><a href="file.php?id="'.$row['id'].'"><h2><?php echo $row["title"]; ?></h2></a></div>
<?php } ?>
</div>
<?php echo $pagination->createLinks(); ?>
<?php } ?>
</div>
我的href链接显然有问题：
<a href="file.php?id="'.$row['id'].'"><h2><?php echo $row["title"]; ?></h2></a>
或者我的file.php页面：
<?php
// GET ID FROM THE URL
$id = $_GET['id'];
?>
<!DOCTYPE HTML>
<html lang="en">
<head>
<title>Ajax Pagination with Search and Filter in PHP</title>
<link href='style.css' rel='stylesheet' type='text/css'>
<script src="jquery.min.js"></script>
<script type="text/javascript"></script>
</head>
<body>
<?php
$sql =mysql_query("select * from posts where id='".$id."'");
while($row = mysql_fetch_array($sql)){
?>
<tr>
<th>title:</th>
<th>created:</th>
<th>modified:</th>
<th>statusr:</th>
</tr>
<tr>
<td><?=$row['title']?></td>
<td><?=$row['created']?></td>
<td><?=$row['modified']?></td>
<td><?=$row['status']?></td>
</tr>
<?php
}
?>
但是我无法理解为什么其他一切似乎工作得很好，也许我一直在看这个很长时间并错过了一些明显的东西...但是任何帮助都会非常感激。
使用以下代码：
<div class="list_item"><a href="file.php?id="<?php echo $row['id'] ?>"><h2><?php echo $row["title"]; ?></h2></a></div>