PHP code submitting to MySQL isn't working

Question

I have been tying to figure out what is wrong for a long time, but I cannot figure it out. I have been trying to connect to a local MySQL database. When I run this, no error is thrown. Also, I know that the database and the table exist, and I know that my username and password are correct.

<?php 
// Connects to your Database 
$con = mysqli_connect("localhost", "**MY_USERNAME**", "**MY_PASSWORD**", "gamesite_data") or die(mysql_error()); 
$username = $_POST["username"];
$password = $_POST["password"];
$first_name = $_POST["first_name"];
$last_name = $_POST["last_name"];
$age = $_POST["age"];

mysqli_query($con, "INSERT INTO information (`username`, `password`, `first_name`, `last_name`, `age`) VALUES ($username, $password, $firstname, $last_name, $age);"
);
?> 
<?php
mysqli_close($con)
?>

Answer 1

If the values of username, password, first_name, last_name are strings, you have to surround them with single quotes here:

mysqli_query($con, "INSERT INTO information (`username`, `password`, `first_name`, `last_name`, `age`) VALUES ('".$username."', '".$password."', '".$firstname."', '".$last_name."', ".$age.")");

And if you use mysqli_connect , you have to use also mysqli_error instead of mysql_error .

Answer 2

try this

   <?php 
 // Connects to your Database 
 $con = new mysqli("localhost", "**MY_USERNAME**", "**MY_PASSWORD**", "gamesite_data") or die(mysql_error()); 
 $username = $_POST["username"];
 $password = $_POST["password"];
 $first_name = $_POST["first_name"];
 $last_name = $_POST["last_name"];
 $age = $_POST["age"];

 $query="INSERT INTO information (`username`, `password`, `first_name`, `last_name`, `age`) VALUES ('$username', '$password', '$firstname', '$last_name', '$age')";
 $con->query($query);
 $con->close();
 ?>

Answer 3

Here:

mysqli_query($con, "INSERT INTO information (`username`, `password`, `first_name`, `last_name`, `age`) VALUES ($username, $password, $firstname, $last_name, $age);"
);

Should be

mysqli_query($con, "INSERT INTO information (`username`, `password`, `first_name`, `last_name`, `age`) VALUES ($username, $password, $first_name, $last_name, $age);"
);

You dont have $firstname variable, you have $first_name

Also keep in ming that you need to use mysqli_real_escape_string() for all $_POST values.

And i think is a good practice to pass variable into query as strings :

"INSERT INTO information (`username`, `password`, `first_name`, `last_name`, `age`) VALUES ('$username', '$password', '$first_name', '$last_name', '$age');"

Answer 4

i noticed your code here you made 2 mistake first one is you take wrong variable name into your sql query .you are declare $first_name variable but you are taking $firstname and second one you are taking extra semicolon (;) just after your value's end.

please use this code

<?php
$con = mysqli_connect("localhost", "root", "", "gamesite_data");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$username = $_POST["username"];
$password = $_POST["password"];
$first_name = $_POST["first_name"];
$last_name = $_POST["last_name"];
$age = $_POST["age"];
$query = "INSERT INTO information (`username`, `password`, `first_name`, `last_name`,  `age`) VALUES ($username, $password, $first_name, $last_name, $age)";
$result = mysqli_query($con, $query);
?>   

hope its worked