希望有人能提供帮助。我有一个客户数据库表，我试图使用 jquery 数据表在页面上显示它。该页面还将包含一个表单，我可以在其中输入数据，这将更新数据库并显示在下表中。我有 3 个正在使用的文件。这是代码:
索引.php
<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css">
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6/jquery.js"></script>
<script type="text/javascript" language="javascript" src="jquery.dataTables.js"></script>
<script type="text/javascript" charset="utf-8">
/* Table initialisation */
$(document).ready(function() {
$('#datatable').dataTable( {
"sDom": "<'row'<'span8'l><'span8'f>r>t<'row'<'span8'i><'span8'p>>"
});
});
</script>
<script type="text/javascript" language="javascript" src="jquery.form.js"></script>
<script type="text/javascript">
// prepare the form when the DOM is ready
$(document).ready(function() {
// bind form using ajaxForm
$('#htmlForm').ajaxForm({
// target identifies the element(s) to update with the server response
target: '#datatable',
resetForm: 'true',
// success identifies the function to invoke when the server response
// has been received; here we apply a fade-in effect to the new content
success: function() {
$('#datatable').fadeIn('slow');
}
});
});
</script>
</head>
<body>
<form id="htmlForm" action="customers.php" method="post">
First: <input type="text" name="first" /><br/>
Last: <input type="text" name="last" /><br/>
Phone: <input type="text" name="phone" /><br/>
Mobile: <input type="text" name="mobile" /><br/>
Fax: <input type="text" name="fax" /><br/>
Email: <input type="text" name="email" /><br/>
Web: <input type="text" name="web" /><br/><br/>
<input id="btnReload" value="Add Customer" type="submit">
</form>
<br />
<!---------------------
<-- CUSTOMER TABLE ----
----------------------->
<?php
//MySQL Database Connect
include 'customers.php';
?>
</body>
</html>
客户.php
<?php
//MySQL Database Connect
include 'config.php';
$_POST['first'] = "undefine";
$_POST['last'] = "undefine";
$_POST['phone'] = "undefine";
$_POST['mobile'] = "undefine";
$_POST['fax'] = "undefine";
$_POST['email'] = "undefine";
$_POST['web'] = "undefine";
$sql="INSERT INTO contacts (first, last, phone, mobile, fax, email, web)
VALUES
('$_POST[first]','$_POST[last]','$_POST[phone]','$_POST[mobile]','$_POST[fax]','$_POST[email]','$_POST[web]')";
if (!mysql_query($sql,$db))
{
die('Error: ' . mysql_error());
}
$query="SELECT * FROM contacts";
$result=mysql_query($query);
?>
<h2>Customers Table</h2>
<?php
$query="SELECT * FROM contacts";
$result=mysql_query($query);
?>
<table cellpadding="0" cellspacing="0" border="0" class="bordered-table zebra-striped" id="datatable">
<thead>
<tr>
<th>First</th>
<th>Last</th>
<th>Phone</th>
<th>Mobile</th>
<th>Fax</th>
<th>Email</th>
<th>Web</th>
</tr>
</thead>
<?php
$num=mysql_numrows($result);
$i=0;
while ($i < $num) {
$first=mysql_result($result,$i,"first");
$last=mysql_result($result,$i,"last");
$phone=mysql_result($result,$i,"phone");
$mobile=mysql_result($result,$i,"mobile");
$fax=mysql_result($result,$i,"fax");
$email=mysql_result($result,$i,"email");
$web=mysql_result($result,$i,"web");
?>
<tbody>
<tr>
<td><?php echo $first ?></td>
<td><?php echo $last ?></td>
<td><?php echo $phone ?></td>
<td><?php echo $mobile ?></td>
<td><?php echo $fax ?></td>
<td><?php echo $email ?></td>
<td><?php echo $web ?></td>
</tr>
<?php
$i++;
}
?>
配置.php
<?php
$mysql_hostname = "localhost";
$mysql_user = "root";
$mysql_password = "";
$mysql_database = "address";
$prefix = "";
$db = mysql_connect($mysql_hostname, $mysql_user, $mysql_password) or die("Could not connect database");
mysql_select_db($mysql_database, $db) or die("Could not select database");
?>
当我输入数据时，表单正在更新表格，但每次我重新加载页面时，它都工作得相当好，一条空白记录被插入到数据库中。有人可以就此问题的最佳解决方案提出建议。
最佳答案
正在插入空行，因为即使未提交表单，您的插入代码也会运行。您可以通过在插入之前检查以确保它是一个 POST 操作来解决此问题:
include 'config.php';
if($_SERVER['REQUEST_METHOD'] == 'POST') {
// your insertion code here
}
//or even better:
if (!empty($_POST)) {
//insertion code
}
$query="SELECT * FROM contacts";
$result=mysql_query($query);
...
但是，您有一个更严重的问题，因为您没有对输入进行清理。如果有人输入 '); DELETE FROM contacts; 在您的 web 字段中？这称为 SQL 注入(inject)攻击。
为避免这种情况，您应该使用准备好的语句或使用 mysql_real_escape_string 来保护您的应用程序免受此类攻击。
关于Php/Jquery 表单和数据表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/9055692/