php – 解析错误：语法错误,128行文件中的意外$end

嗨所以这是代码：
它是一个页面,用于在下拉列表中显示数据库中的可用表,然后在表中显示结果.这样做的实际代码(在中间)可以完全独立工作,但是当我尝试在其周围添加模板时,我会收到错误…

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<Meta name="keywords" content="" >
<Meta name="description" content="" >
<Meta http-equiv="content-type" content="text/html; charset=utf-8" >
<title>SNYSB Archive</title>
<link href="style.css" rel="stylesheet" type="text/css" media="screen" >
<!-- Location of javascript. -->
<script language="javascript" type="text/javascript" src="swfobject.js" ></script>
</head>
    <div id="wrapper">
    <div id="header">
    <!-- KEEP THIS BIT [ITS FORMATTING] --> 
    </div>
    <!-- end #header -->
    <div id="menu">
            <ul>
            <li><a href="Hpage.PHP">Home</a></li>
            <li><a href="Register.PHP">Register</a></li>
        </ul>
    </div>
    <!-- end #menu -->
    <div id="page">
    <div id="page-bgtop">
    <div id="page-bgbtm">
    <div id="content">
            <div class="post">
            <div class="post-bgtop">
            <div class="post-bgbtm">
<h1 class="title">PUT heading HERE!</h1>
                <div class="entry">
                    <p class="Body">
<?PHP
$dbname = 'snysbarchive';
$conn= MysqL_connect('localhost', 'root', 'usbw');
if (!$conn) {
    echo 'Could not connect to MysqL';
    exit;
}

$sql = "SHOW TABLES FROM $dbname";
$result = MysqL_query($sql);

if (!$result) {
    echo "DB Error, Could not list tables\n";
    echo 'MysqL Error: ' . MysqL_error();
    exit;
}
if (MysqL_select_db($dbname, $conn))
{
?>
    <form method="post" action="new 2.PHP">
    <select name="tables">
    <?PHP
    while ($row = MysqL_fetch_row($result)) {
    ?>    
    <?PHP
        echo '<option value="'.$row[0].'">'.$row[0].'</option>';
    }
    ?>
    </select>
    <input type="submit" value="Show">
</form>
<?PHP
//MysqL_free_result($result);
if (isset($_POST) && isset($_POST['tables']))
{
    $tbl=$_POST['tables'];
    //echo $_POST['tables']."<br />";
    $query="SELECT * from $tbl";
    $res=MysqL_query($query);
    echo $query;
    if ($res)
    {
    ?>
    <table border="1">
    <?PHP
        while ( $row = MysqL_fetch_array($res))
        {
            echo "<tr>";
            echo "<td>".$row[0]."</td>";
            echo "<td>".$row[1]."</td>";
            echo "<td>".$row[2]."</td>";
            echo "<td>".$row[3]."</td>";        
            echo "</tr>";
        } ?>
    </table>
    <?PHP
    }
}
?>
</div>
</div>
            </div>
            </div>

        <div style="clear: both;">&nbsp;</div>
        </div>
        <!-- end #content -->
        <div id="sidebar">
            <ul>
                <li>
                    <h2>Welcome!</h2>
                    <p>Welcome to SNYSBs archive!
                       </p>
                </li>
                <li>
                    <h2>SNYSB</h2>
                    <p>
            <a href="Contact.PHP">Contact Us!</a>
                    </p>
                </li>
            </ul>
        </div>
        <!-- end #sidebar -->
        <div style="clear: both;">&nbsp;</div>
    </div>
    </div>
    </div>
    <!-- end #page -->
    <div id="footer">
        <p>copyright (c) 2008 Sitename.com. All rights reserved. Design by <a href="http://www.freecsstemplates.org/">Free CSS Templates</a>.</p>
    </div>
    <!-- end #footer -->
</div>
</body>
</html>

它一直说意想不到的结局,但我不知道如何解决它？

Error Message:Parse error: Syntax error, unexpected $end in file on line 128

谢谢

解决方法:

您没有在函数周围使用正确的大括号选择数据库.

php – 解析错误：语法错误,128行文件中的意外$end

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