How do I fix notice " column 'firstname' cannot be null " with error on line 17

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Hello, I'm a beginner in PHP, HTML, and Apache. When I run the file, the XAMPP has a notice:

Fatal error: Uncaught mysqli_sql_exception: Column 'firstname' cannot be null in C:\xampp\htdocs\connect\connection.php:17 Stack trace: #0 C:\xampp\htdocs\connect\connection.php(17): mysqli_stmt->execute() #1 {main} thrown in C:\xampp\htdocs\connect\connection.php on line 17

I have two files, one in HTML and one in PHP.

What I have tried:

I double-checked the database on XAMPP but I don't know what's wrong with it.

HTML code:

<!DOCTYPE html>
<!--
Click nbfs://nbhost/SystemFileSystem/Templates/Licenses/license-default.txt to change this license
Click nbfs://nbhost/SystemFileSystem/Templates/Other/html.html to edit this template
-->
<!DOCTYPE html>
<html>
<head>
  <title>Sign-In Form</title>
  <style>
    body {
      font-family: Arial, sans-serif;
    }
    
    .container {
      max-width: 400px;
      margin: 0 auto;
      padding: 20px;
      border: 1px solid #ccc;
      border-radius: 5px;
      background-color: #f2f2f2;
    }
    
    .form-group {
      margin-bottom: 15px;
    }
    
    label {
      display: block;
      margin-bottom: 5px;
    }
    
    input[type="text"],
    input[type="email"],
    input[type="tel"],
    select {
      width: 100%;
      padding: 8px;
      border-radius: 3px;
      border: 1px solid #ccc;
    }
    
    input[type="submit"] {
      background-color: #4CAF50;
      color: white;
      border: none;
      padding: 10px 20px;
      border-radius: 5px;
      cursor: pointer;
    }
    
    input[type="submit"]:hover {
      background-color: #45a049;
    }
  </style>
</head>
<body>
   
  <div class="container">
    <h2>Sign-In Form</h2>
    <form action="connect.php" method="post" >
      <div class="form-group">
        <label for="lastname">Last Name:</label>
        <input type="text" id="lastname" name="lastname" required>
      </div>
      <div class="form-group">
        <label for="firstname">First Name:</label>
        <input type="text" id="firstname" name="firstname" required>
      </div>
      <div class="form-group">
        <label for="email">Email:</label>
        <input type="email" id="email" name="email" required>
      </div>
        <div class="form-group">
            <label for="password">Password:</label>
            <input type="text" id="password" name="password" required>
        </div>
      <div class="form-group">
        <label for="contact">Contact Number:</label>
        <input type="number" id="contact" name="contact" required>
      </div>
      <div class="form-group">
        <label for="address">Address:</label>
        <input type="text" id="address" name="address" required>
      </div>
      <div class="form-group">
        <label for="country">Country:</label>
        <input type="text" id="country" name="country" required>
      </div>
        
    </form>
 
   <input type="submit" onClick="myFunction()"/>
         <script>
      function myFunction() {
        window.location.href="https://shopee.ph/?gclid=CjwKCAjwyqWkBhBMEiwAp2yUFhWMuFTyZ0Ev3ju0oygGkwXmdptKzZimCT5xr1Wq-lTekgL3xLtbIRoCzH0QAvD_BwE";
      }
   </script>
    </div>
</body>
</html>

PHP code:

<?php
$lastname = filter_input(INPUT_POST, 'lastname');
$firstname = filter_input(INPUT_POST, 'firstname');
$email = filter_input(INPUT_POST, 'email');
$password = filter_input(INPUT_POST, 'password');
$contact = filter_input(INPUT_POST, 'contact');
$address = filter_input(INPUT_POST, 'address');
$country = filter_input(INPUT_POST, 'country');

$conn = new mysqli('localhost', 'root', '', 'entries');
if($conn->connect_error){
    die('Connection Failed : ' . $conn->connect_error);
} else{
    $stmt = $conn->prepare("insert into entries_(lastname, firstname, email, password, contact, address, country)
            values(?, ?, ?, ?, ?, ?, ?)");
    $stmt->bind_param("ssssiss", $lastname, $firstname, $email, $password, $contact, $address, $country, );
    $stmt->execute();
    echo "sign up successful";
    $stmt->close();
    $conn->close();
}   
?>

Posted 14-Jun-23 13:34pm

Member 16029663

Updated 15-Jun-23 2:02am

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2 solutions

Solution 2

OriginalGriff is spot on, your data table with a field named 'firstname' is configured to accept a value otherwise it will return as invalid. To check if each of your variables has a value, you can run -

$lastname = filter_input(INPUT_POST, 'lastname');
$firstname = filter_input(INPUT_POST, 'firstname');
$email = filter_input(INPUT_POST, 'email');
$password = filter_input(INPUT_POST, 'password');
$contact = filter_input(INPUT_POST, 'contact');
$address = filter_input(INPUT_POST, 'address');
$country = filter_input(INPUT_POST, 'country');

if (empty($lastname) || empty($firstname) || empty($email) || empty($password) || empty($contact) || empty($address) || empty($country)) {
    // At least one variable is empty
    echo "Please fill in all the required fields.";
    return;
}

// All variables have values, continue with the rest of your code

The above will return at the line of code, in your case '$firstname = filter_input(INPUT_POST, 'firstname');' as an invalid entry will stop execution of further code.

Posted 15-Jun-23 2:02am

Andre Oosthuizen

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OriginalGriff · Accepted Answer · 2023-06-14T18:41:00

Validate your data before you try to submit it: the error is pretty clear:

Error

Column 'firstname' cannot be null

You fetch the value from the user page, but you don't try to check if the user entered anything.
If he didn't, $firstname will be null, and you get an error because the DB is not setup to permit nulls in that column.