[SOLVED] mysql error occured when displaing the page


working with php and mysql as well. I have following create.php page and need save data to mysql table.

include "config.php";

if(isset($_POST['submit'])) {
    $first_name = $_POST['firstname'];
    $last_name = $_POST['lastname'];
    $email = $_POST['email'];
    $password = $_POST['password'];
    $gender = $_POST['gender'];

$sql = "INSERT INTO 'users' ('firstname','lastname','email','password','gender') VALUES ('$first_name','$last_name','$email','$password','$gender')"; // this is line 12

$result = $conn->query($sql);

if($result == TRUE) {
    echo "New record has created successfully";
else {
    echo "error:" . $sql . "<br>". $conn->error;



but got following error message

Undefined variable: first_name in C:\wamp64\www\simple\create.php on line 12 <br> error:INSERT INTO 'users' ('firstname','lastname','email','password','gender') VALUES ('','','','','') You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''users' ('firstname','lastname','email','password','gender') VALUES ('','','',''' at line 1

how to fix this?


You need to put the whole code logic inside the if(isset($_POST['submit'])) condition

What’s happening right now is: if there is no $_POST['submit'], your if won’t run, thus no variables are declared, but your SQL and rest of the code will still run and that’s why it says var not defined

if(isset($_POST['submit'])) { ... }

Coming to the next issue is of using backticks. You really shouldn’t have single quotes around the field name. You can use backticks (`) for table and column names, single quotes (‘) for strings. There is already an answer for it: When to use single quotes, double quotes, and backticks in MySQL

Answered By – Tushar

Answer Checked By – Cary Denson (BugsFixing Admin)

Leave a Reply

Your email address will not be published.