PHP联系表单连接但不提交

Question

大家好，这是我的 PHP 联系表格。它成功连接，因为我在各个点运行了一些测试以查看它是否有效，但它似乎没有提交到数据库。谁能帮帮我？

<?php

define('DB_NAME', 'x');
define('DB_USER', 'x');
define('DB_PASSWORD', 'x');
define('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!link) {
    die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
    die('Can\'t use' . DB_NAME . ': ' . mysql_error());
}

$value = $_POST['UserName'];
$value1 = $_POST['UserEmail'];
$value2 = $_POST['UserMessage'];

$sql = "INSERT INTO ContactUs (UserName, UserEmail, UserMessage) VALUES ('$value', '$value1', '$value2')";


if (!mysql_query($sql)) {
    die('Error: ' . mysql_error());
}

mysql_close();

我还附上了联系表单的实际 sn-p 代码（在模态框中）：

<div class = "modal fade" id = "contact" role="dialog">
<div class = "modal-dialog">
<div class = "modal-content">
<div class = "modal-header">
<h4>Contact form</h4>
</div>
<div class = "modal-body">
<p>HELLO HELLO HELLO HELLO</p>
</div>
   <div class="form-group">
<form action="ContactUs.php" method="post">

 <div class="form-group">
 <label for="UserEmail">Email address</label>
 <input type="text" class="form-control" id="UserEmail" name="UserEmail" placeholder="What is your E-Mail">
 </div>
 <div class="form-group">
 <label for="UserName">Name</label>
 <input type="text" class="form-control" id="UserName" name="UserName" placeholder="What is your name?">
 </div>
 <div class="form-group">
 <label for="UserMessage">Message</label>
 <input type="text" class="form-control" id="UserMessage" name="UserMessage" placeholder="What would you like to say?">
 </div>
 <input type="Submit" value="Submit" class="btn btn-default">Submit</button>

</form>
</div>

<div class = "modal-footer">
<a class = "btn btn-default" data-dismiss = "modal">Close</a>
<a class = "btn btn-primary" data-dismiss = "modal">Close</a>
</div>
</div>
</div>
</div>
</div>

看到这是一个联系表，我只是希望将信息发送到我的电子邮件，如果有人能告诉我这是否合适，那就太好了[如果不是如何调整它以使其如此]。我使用了一个视频教程[链接如下]，但他只提到了如何存储到数据库中。

https://www.youtube.com/watch?v=wp6Ngpk5XiY

Answer 1

您在检查 if (!link) 时出现拼写错误，但此外，您应该使用更新、更安全的 MySQLi 功能。试试这个：

<?php
define('DB_NAME', 'x');
define('DB_USER', 'x');
define('DB_PASSWORD', 'x');
define('DB_HOST', 'localhost');

$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if ($mysqli->connect_errno)
{
    die("Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error);
}

$userName = $_POST['UserName'];
$userEmail = $_POST['UserEmail'];
$userMessage = $_POST['UserMessage'];

if (!($stmt = $mysqli->prepare("INSERT INTO ContactUs (UserName, UserEmail, UserMessage) VALUES (?, ?, ?)")))
{
    die("Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error);
}

$stmt->bind_param("sss", $userName, $userEmail, $userMessage);
if (!$stmt->execute())
{
    die("INSERT failed: (" . $stmt->errno . ") " . $stmt->error);
}

$mysqli->close();

这使用Prepared Statements，出于所有意图和目的，它将准备好的语句定义中的? 标记替换为您绑定到该语句的变量的转义值。这有助于防止 SQL 注入。