我通过从i注册的用户有一个表格,该表格的代码是如下:数据不是插入到从DB形式
<div class="contentArea">
<form action="NewUser.php" method="POST">
<div class="regd">
<ul>
<li class="regdTitle">Please enter your details:</li>
<li class="regdLable">First Name</li>
<li><input type="Text" class="regdInput" name="first_name" /></li>
<li class="regdLable">Last Name</li>
<li><input type="Text" class="regdInput" name="last_name" /></li>
<li class="regdLable">Password</li>
<li><input type="Password" class="regdInput" name="pass" /></li>
<li class="regdLable">Re-Type Password</li>
<li><input type="Password" class="regdInput" name="re_pass" /></li>
<li class="regdLable">Email</li>
<li><input type="Text" class="regdInput" name="email" /></li>
<li class="regdLable">Country</li>
<li><select name="country" style="width: 300px; height: 20px;">
<option value=" " selected="selected"></option>
<option value="ACT">ACT</option>
<option value="NSW">NSW</option>
<option value="NT">NT</option>
<option value="QLD">QLD</option>
<option value="SA">SA</option>
<option value="TAS">TAS</option>
<option value="VIC">VIC</option>
<option value="WA">WA</option>
</select></li>
<li><input type="button" value="Submit" class="regdBtn" alt="Submit"
title="Submit" name="submit"/></li>
<li><input type="button" value="Cancle" class="regdBtn" alt="Cancle"
title="Cancle" onclick="location.href='index.php'" /></li>
</ul>
</div>
</form>
</div>
,我通过具有NewUser.php文件从i插入代码的形式,以数据库的值,是如下:
<?php
// Includes
require_once ('Includes/dbconn.php');
require_once ('Includes/functions.php');
$first_name = $_POST ['first_name'];
$last_name = $_POST ['last_name'];
$pass = $_POST ['pass'];
$repass = $_POST ['re_pass'];
$email= $_POST ['email'];
$country = $_POST ['country'];
$query = "INSERT INTO user (first_name, last_name, pass, email, country)
VALUES ('{$first_name}','{$last_name}','{$pass}','{$repass}','{$email}','{$country}')";
$result = mysql_query ($query, $conn);
if (isset ($result)) {
echo "Registration sucessful";
//redirect_to (UserRegd.php);
} else {
echo "<p>User Registration Failed" . mysql_error() . "</p>";
}
?>
<?php mysql_close($conn);?>
连接码是如下:
<?php
include_once 'Constants/ConConst.php';
$conn = mysql_connect(DB_SERVER, DB_USER, DB_PASS);
if(!$conn)
{
die('Connection not set'. mysql_error());
}
$db = mysql_select_db(DB_NAME, $conn);
if(!$db)
{
die('Database not found'.mysql_error());
}
?>
我米无法插入表单值到我的数据库...我可以知道我的代码中的错误在哪里。
任何有意义的回应将高度赞赏,并提前感谢。有没有像jsfiddle.com这样的在线编辑器,我们可以把我们的代码放在演示中。
你应该做的第一件事就是在代码中放入'die($ query)'来查看你的实际SQL语句的样子。这样做,然后在这里发布结果 – JohnP 2011-04-13 15:02:22
你应该使用PDO而不是旧的方法联系mysql数据库 - http://net.tutsplus.com/tutorials/php/why-you-should-be-using-phps- pdo-for-database-access/ – Calum 2011-04-13 15:07:20
您的查询似乎很容易受到SQL注入的影响 - 请务必在代码生效之前解决此问题:) – 2011-04-13 15:09:27