我一直在试图永远解决这个问题而且它让我烦恼.代码是显示从dropbox中选择的表格工作正常..但是当我把它放入我的布局/模板时,它会抛出一个错误,我不明白为什么!
这是代码:
<!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;"> </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;"> </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>
(只需在开头和底部用我的模板粘贴原始页面的代码)
这是它不喜欢的线:
if (isset($_POST['tables']))
谢谢
编辑:新错误消息:解析错误:语法错误,128行上的FILENAME意外$结束
您看到的错误可能是因为$ _POST没有名为"tables"的键.
echo"<form method="post" action="show_tabs.php">";
echo"<select name="tables">";
我运行时都会产生错误.请注意语法突出显示SO上的颜色更改,表示并非所有颜色都被解释为字符串文字.您需要转义内部引号或将外部引号切换为单引号:
echo '<form method="post" action="show_tabs.php">';
echo '<select name="tables">';
要么
echo "<form method=\"post\" action=\"show_tabs.php\">";
echo "<select name=\"tables\">";
同样的道理
echo"<input type="submit" value="Show">";
...
echo"<table border="1">";
也,
if ($res)
}
应该读
if ($res)
{
此外,您在使用mysql_*函数时出错.我可以看到你在运行查询后才选择你的数据库,而且我没有看到调用mysql_connect.顺序是connect,select_db,query,fetch,free,close,如http://www.php.net/manual/en/mysql.examples-basic.php中所示.
echo <<< TEMPLATE
<h1 class="title">PUT HEADING HERE!</h1>
<div class="entry">
<p class="Body">
使用
?>
<h1 class="title">PUT HEADING HERE!</h1>
<div class="entry">
<p class="Body">
<?php
一个定界符的内部代码没有得到执行(虽然变量取代的),因此没有一个代码从<<< TEMPLATE到TEMPLATE;正在运行.只有变量才能进入heredoc.