Hi So this is the code: Its a page to display my available tables in the database in a drop-down, then display the results in the table. The actual code to do so (in the middle) works perfectly on its own, but when I try to add my template around it I get errors...
这是代码:它的一个页面在下拉菜单中显示我的可用表,然后在表中显示结果。实际的代码(在中间)是完全独立工作的,但是当我尝试在它周围添加模板时,我就会出错……
<!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>
It keeps saying unexpected end but I am not sure how to fix it?
它一直在说意想不到的结局,但我不知道该如何解决它?
Error Message:Parse error: syntax error, unexpected $end in file on line 128
Thanks
谢谢
6 个解决方案
#1
-1
You have not selected your database with correct braces around the function.
您没有在函数周围使用正确的括号来选择数据库。
#2
12
This may also occur when mixing short and normal open tags when the server does not support short-open-tags
(<?
instead of <?php
), even though this wasn't the case in your code.
当服务器不支持短开标签(
<?php
$showHeader = true;
if ($showHeader) {
?>
<h1>Hello, World!</h1>
<?
}
?>
Note that the closing bracket will not be registered if the server doesn't support the <?
open tag.
注意,如果服务器不支持
#3
11
Line 50: if (mysql_select_db($dbname, $conn))
has an opening bracket and not a closing one.
第50行:if (mysql_select_db($dbname, $conn))有一个开头括号,而不是关闭括号。
#4
3
You may need to change <?
to <?php
您可能需要更改
#5
0
you forget to close this block with a }
:
您忘记关闭这个块的一个}:
if (mysql_select_db($dbname, $conn))
{
?>
Try this code :
试试这段代码:
<?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($result)) {
echo '<option value="'.$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
}
}
?>
#6
0
You have missed a '}' so following if block is not closed.
您已经错过了一个“}”,如果块没有关闭,那么就会出现以下情况。
if (mysql_select_db($dbname, $conn))
{
By adding a } in Line #91 your code will be worked.
通过在第91行添加一个},您的代码将被工作。
But always try to write much cleaner code by following best practices.
但是,总是尝试通过遵循最佳实践来编写更清晰的代码。
#1
-1
You have not selected your database with correct braces around the function.
您没有在函数周围使用正确的括号来选择数据库。
#2
12
This may also occur when mixing short and normal open tags when the server does not support short-open-tags
(<?
instead of <?php
), even though this wasn't the case in your code.
当服务器不支持短开标签(
<?php
$showHeader = true;
if ($showHeader) {
?>
<h1>Hello, World!</h1>
<?
}
?>
Note that the closing bracket will not be registered if the server doesn't support the <?
open tag.
注意,如果服务器不支持
#3
11
Line 50: if (mysql_select_db($dbname, $conn))
has an opening bracket and not a closing one.
第50行:if (mysql_select_db($dbname, $conn))有一个开头括号,而不是关闭括号。
#4
3
You may need to change <?
to <?php
您可能需要更改
#5
0
you forget to close this block with a }
:
您忘记关闭这个块的一个}:
if (mysql_select_db($dbname, $conn))
{
?>
Try this code :
试试这段代码:
<?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($result)) {
echo '<option value="'.$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
}
}
?>
#6
0
You have missed a '}' so following if block is not closed.
您已经错过了一个“}”,如果块没有关闭,那么就会出现以下情况。
if (mysql_select_db($dbname, $conn))
{
By adding a } in Line #91 your code will be worked.
通过在第91行添加一个},您的代码将被工作。
But always try to write much cleaner code by following best practices.
但是,总是尝试通过遵循最佳实践来编写更清晰的代码。