所以我试图在PHP中运行一个查询，虽然查询本身没有错误（或者看起来是这样），但编辑器在"echo"语句中看到了一个错误。代码是这样的：
<?php
include("include/session.php");
?>
<?php
$db = new PDO('mysql:host=localhost;dbname=cvtool;charset=utf8', 'user', 'pass'); // change these to your own database details
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // later, change ERRMODE_WARNING to ERRMODE_EXCEPTION so users wont see any errors
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$department = isset($_GET['department'])? $_GET['department']: null;
$sql = 'SELECT *
FROM education
WHERE school LIKE ?;
$q = $conn->prepare($sql);
$q->execute(array('%$department%');
$q->setFetchMode(PDO::FETCH_ASSOC);
while ($r = $q->fetch()) {
echo sprintf('%$department', $r['school']);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<!--The viewport tag is used in order to scale the page properly inside any screen size -->
<meta charset="utf-8" name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1">
<title>CV Tool</title>
<link rel="shortcut icon" href="images/favicon.ico" />
<link rel="stylesheet" href="css/main.css"/>
<!--Import JQuery from stored file -->
<script src="js/jquery-1.11.1.min.js"></script>
<!--Import JQuery from Google's Content Delivery Network -->
<!--<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js">-->
<link href='http://fonts.googleapis.com/css?family=PT+Sans:400,700' rel='stylesheet' type='text/css'>
<script type="text/javascript" src="js/menu.js"></script>
<script type="text/javascript" src="js/backToTop.js"></script>
</head>
<body>
<!--Big wrapper contains the whole site (header,navigation menu,....,footer-->
<div id="big_wrapper">
<header id="top_header">
<a href="main.php"><img src="images/cvlogo.png"> </a>
</header>
<br>
<nav class="clearfix">
<ul class="clearfix">
<li><a href="main.php">Home</a></li>
<?php
/**
* User has already logged in, so display relavent links, including
* a link to the admin center if the user is an administrator.
*/
if($session->logged_in){
echo "<li><a href='"search.php'">Search</a></li>"
."<li><a href='"myCVs.php'">My CV(s)</a></li>"
."<li><a href='"userinfo.php?user=$session->username'">My Account</a></li>"
;
echo "<li><a href='"process.php'">Logout</a></li>";
}
else
?>
</ul>
<a href="#" id="pull">Menu</a>
</nav>
<section id="main_section">
<table class="table table-bordered table-condensed">
<thead>
<tr>
<th>Department</th>
</tr>
</thead>
<tbody>
<?php while ($r = $q->fetch()): ?>
<tr>
<td><?php echo htmlspecialchars($r['school'])?></td>
</tr>
<?php endwhile; ?>
</section>
<footer id="the_footer">
City CV Tool 2014
</footer>
<a href="#" class="back-to-top"></a>
</div>
</body>
</html>
问题是，无论我如何更改它，它仍然会给我一个错误。错误出现在以下特定行：
while ($r = $q->fetch()) {
echo sprintf('%$department', $r['school']);
}
这个错误可能只是我在没有意识到的情况下丢失了一些东西或添加了一些额外的东西。我知道代码是关于一个非常具体的案例的，但我们仍然感谢您的帮助。
您永远不会关闭以下字符串变量：
$sql = 'SELECT *
FROM education
WHERE school LIKE ?;