Question

所以我试图在PHP中运行查询，而查询本身没有错误（或者看起来似乎如此），编辑器在“echo”语句中看到错误。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>

问题在于无论我如何更改它，它仍然会给我一个错误。错误发生在以下特定行：

while ($r = $q->fetch()) {
    echo sprintf('%$department', $r['school']);
}

错误可能只是缺少的东西，或者我没有意识到的额外的东西。我知道代码是关于一个非常具体的案例，但仍然感谢任何帮助。

Answer 1

您永远不会关闭以下字符串变量：

 $sql = 'SELECT *
         FROM education
         WHERE school  LIKE ?;

PHP / MySQL中出现意外的T_CONSTANT_ENCAPSED_STRING错误

1 个答案: