php - 从HTML变量表格插入SQL表

我正试图创建两个不同的表，基于HTML文件中的单选。这是我迄今为止创建的HTML表单：

<form action="connect.php" method=POST>
First Name: <input type="text" name="firstname"><br><br>
Last Name: <input type="text" name="lastname"><br><br>
Email: <input type="text" name="email"><br><br>
Desired Password: <input type="password" name="password"><br><br>
Re-type Password: <input type="password" name="password"><br><br>
I am a: <input type="radio" name="role" value="student">Student
<input type="radio" name="role" value="alumni">Alumni<br><br>
<div class="stud">
    Major: <input type="text" name="studmajor"><br><br>
    Emphasis: <input type="text" name="studemphasis"><br><br>
    Expected Graduation Year: <input type="text" name="studgradyear"><br><br>
    Hobbies: <input type="text" name="studhobbies">
</div>
<div class="alum">
    Gradutation Year: <input type="text" name="alumgradyear"><br><br>
    Major: <input type="text" name="alummajor"><br><br>
    Emphasis: <input type="text" name="alumemphasis"><br><br>
    Company: <input type="text" name="alumcompany"><br><br>
    Hobbies: <input type="text" name="alumhobbies">
</div>
<br>
<input type="submit" value="Submit">
</form>
<script type="text/javascript">
    $('input:radio[name=role]').change(function(){
    var role = $(this).val();
if(role=='student'){
    $('.stud').show();
    $('.alum').hide();
}
else if(role=='alumni'){
    $('.alum').show();
    $('.stud').hide();
}
});

所以在这段HTML代码中，我试图创建一个数据库。一个给校友，一个给学生。根据选中的单选框，将显示不同的选项。下面是我编写（和借用）的php代码，用于插入到我创建的两个不同表中。我正在使用c9.io，因为这只是一个模型，让我的想法，该网站的运行。

<?php

$servername = getenv('IP');
$username = getenv('C9_USER');
$password = "";
$database = "c9";
$dbport = 3306;

$db = mysqli_connect($servername, $username, $password, $database, $dbport);

if($db === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

$first_name = mysqli_real_escape_string($db, $_POST['firstname']);
$last_name = mysqli_real_escape_string($db, $_POST['lastname']);
$email_address = mysqli_real_escape_string($db, $_POST['email']);
$password = mysqli_real_escape_string($db, $_POST['password']);
$studmajor = mysqli_real_escape_string($db, $_POST['studmajor']);
$studemphasis = mysqli_real_escape_string($db, $_POST['studemphasis']);
$studgradyear = mysqli_real_escape_string($db, $_POST['studgradyear']);
$studhobbies = mysqli_real_escape_string($db, $_POST['studhobbies']);
$alumgradyear = mysqli_real_escape_string($db, $_POST['alumgradyear']);
$alummajor = mysqli_real_escape_string($db, $_POST['alummajor']);
$alumemphasis = mysqli_real_escape_string($db, $_POST['alumemphasis']);
$alumcompany = mysqli_real_escape_string($db, $_POST['alumcompany']);
$alumhobbies = mysqli_real_escape_string($db, $_POST['alumhobbies']);
$role = mysqli_real_escape_string($db, $_POST['role']);
// Post to Student
if($role == "student"){
    $sql = "INSERT INTO Student (FirstName, LastName, Email, Password, Major, Emphasis, GradYear, Hobbies)
    VALUES ('$first_name', '$last_name', '$email_address', '$password', '$studmajor', '$studemphasis', '$studgradyear', '$studhobbies');";
}
// Post to Alumni
else if($role == "alumni"){
    $sql = "INSERT INTO Alumni (FirstName, LastName, Email, Password, GradYear, Major, Emphasis, Company, Hobbies)
    VALUES ('$firstname', '$lastname', '$email', '$password', '$alumgradyear', '$alummajor', '$alumemphasis', '$alumcompany', '$alumhobbies');";
}

mysqli_close($db);
?>

任何帮助都将不胜感激。我对SQL和PHP还不太熟悉，所以这有点让人困惑。
很抱歉没有说得很具体。每当我填写表单并单击“提交”时，就会重定向并显示：

Cannot POST /username/project/connect.php

我的猜测是php文件中的if/elseif语句导致了这个问题，但我不能确切地说。

最佳答案

您应该命名表单，从connect.php中的HTML中提取数据，然后使用php插入。
这里有一个快速演示，向您展示它是如何做到的！
PHP MYSQL Database Manipulation
希望这有帮助！