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php - 解析错误 : syntax error, 文件第 128 行意外的 $end

转载 作者:可可西里 更新时间:2023-10-31 23:51:57 28 4
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您好,这是代码:它是一个页面,用于在下拉列表中显示我在数据库中的可用表,然后在表中显示结果。这样做的实际代码(在中间)可以完美地独立运行,但是当我尝试在它周围添加模板时,我得到了错误...

<!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>

它一直说意外结束,但我不确定如何解决?

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

谢谢

最佳答案

当服务器不支持 short-open-tags 时,如果混合使用短标签和普通开放标签,也可能会发生这种情况。 ( <? 而不是 <?php ),即使您的代码中并非如此。

<?php

$showHeader = true;

if ($showHeader) {
?>
<h1>Hello, World!</h1>
<?
}
?>

请注意,如果服务器不支持 <?,则不会注册右括号。打开标签。

关于php - 解析错误 : syntax error, 文件第 128 行意外的 $end,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9658172/

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