After clicking submit on my form,my php save form just displays the whole backend code instead of a simple data was saved successfully and it does not reflect

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1 Like
3

I created a data entry form with html

<!DOCTYPE html>
<html>
<head>
    <title>Contact Us Data Entry</title>
</head>
<body>

<h2>Contact Us Data Entry</h2>

<form method="post" action="contactsave.php">
    <label for="name">Name:</label>
    <input type="text" id="name" name="name" required><br><br>

    <label for="email">Email:</label>
    <input type="email" id="email" name="email" required><br><br>

    <label for="message">Message:</label><br>
    <textarea id="message" name="message" rows="4" cols="50" required></textarea><br><br>

    <input type="submit" value="Submit">
</form>

</body>
</html>
**and a data save form with php**

<?php
// Define MySQL database connection credentials
$servername = "localhost";
$username = "root";
$password = ""; // Assuming empty password
$database = "otrs";

// Create connection
$conn = new mysqli("localhost","root","","otrs");

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

// Check if the form is submitted
if ($_SERVER["REQUEST_METHOD"] == "POST") {
    // Get the form data
    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['message'];

    // Prepare SQL statement to insert data into the database
    $sql = "INSERT INTO contactustable VALUES ($name,$email,$message)";
                                                                      
    // Prepare and bind parameters to avoid SQL injection             
    $stmt = $conn->prepare($sql);                                     
    $stmt->bind_param("sss", $name, $email, $message);

    // Execute the statement
    if ($stmt->execute() === TRUE) {
        // Data inserted successfully
        echo "Data saved successfully!";
    } else {
        // Error occurred
        echo "Error: " . $sql . "<br>" . $conn->error;
    }

    // Close statement and connection
    $stmt->close();
    $conn->close();
}
?>

but the save form does not save the data to the table in SQL .

4

Hello!

You have to put the variables into $conn, not the values.

Use placeholders under VALUES( ?, ?, ?)
By using the user input directly, you are inviting the SQL injections that you are trying to avoid with the statements.

Change this and run tests. Come back to us anytime if your data still doesn’t get saved.

1 Like
5 
// Connection parameters
$servername = "localhost";
$username = "root";
$password = "";
$database = "otrs";

// Create connection
$conn = new mysqli("localhost","root","","otrs");

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

// Retrieve data from form
$bookingno = $_POST['bookingno'];
$status = $_POST['status'];
$updationdate_time = date("Y-m-d H:i:s"); // Get current date and time

// Prepare SQL statement to insert data into updation table
$sql = "INSERT INTO updation (bookingno, status, updationdate_time) VALUES ('$bookingno', '$status', '$updationdate_time')";

// Execute SQL statement
if ($conn->query($sql) === TRUE) {
    echo "Data saved successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

// Close connection
$conn->close();
?>
the data is not getting saved