从可变 HTML 表单插入到 sql 表

Question

我正在尝试根据 HTML 文件中的单选选项创建两个不同的表。这是我目前创建的 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();
}
});

所以在这个 HTML 代码中，我试图创建一个数据库。一份给校友，一份给学生。根据选中的单选框，显示不同的选项。以下是我编写（并借用）插入到我创建的两个不同表的 php 代码。我正在使用 c9.io，因为这只是一个模型，用于让我的网站启动和运行。

<?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);
?>

任何帮助将不胜感激。我对 SQL 和 PHP 还很陌生，所以这有点令人困惑。

抱歉，没有很具体。每当我填写表单并单击提交时，我都会被重定向并显示：

Cannot POST /username/project/connect.php

我的猜测是 php 文件中的 if/elseif 语句是导致问题的原因，但我无法准确确定。

Answer 1

你应该为你的表单命名，在 connect.php 中从 HTML 中提取数据，然后使用 php 插入。

这是一个快速演示，向您展示它是如何完成的！

PHP MYSQL Database Manipulation

希望对您有所帮助！