PHP mysql cannot insert data - php

Question

// Do not put any echo code other than the last line.
// didnt include $con in the post.
$userlvlid = $_POST["userlvlid"];
$username = $_POST["username"];
$password = $_POST["password"];
$lname = $_POST["lname"];
$fname = $_POST["fname"];
$mname = $_POST["mname"];
$birthdate = $_POST["birthdate"];
$streename = $_POST["streetname"];
$province = $_POST["province"];
$city = $_POST["city"];
$barangay = $_POST["barangay"];
$organization_name = $_POST["orgname"];
$email_address = $_POST["email_address"];
$license = $_POST["license"];
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}
else{
// means connection successful.
echo "sucess";
}
$response = array();
$response["success"] = false;
$sql = "SELECT * FROM users WHERE username = '$username' ";
$result = $con->query($sql);
if ($result->num_rows > 0) {
// means username already exists.
}
else {
$passwordHash = password_hash($password, PASSWORD_DEFAULT);
$sql = "INSERT INTO accounts (`userlvlid`,
`username`,
`password`,
`lname`,
`fname`,
`mname`,
`birthdate`,
`streetname`,
`region`,
`province`,
`city`,
`barangay`,
`orgname`,
`email`,
`license`)
VALUES
('$userlvlid',
'$username',
'$password',
'$lname',
'$fname',
'$mname',
'$birthdate'
'$streetname',
'$region',
'$province',
'$city',
'$barangay',
'$organization_name',
'$email_address',
'$license')";
if (mysqli_query($con, $sql)) {
$response["success"] = true;
}
else {
}
}
}
echo json_encode($response);
?>
Can someone explain why "mysqli_query($con, $sql)" is returning false? I can't find what is wrong in the code. my database contains all of the fields and here maybe a syntax error. The code doesn't give me any errors and it doesn't add the information to the database.

Answer 1

A few things, first this part is not needed:
$sql = "SELECT * FROM users WHERE username = '$username' ";
$result = $con->query($sql);
if ($result->num_rows > 0) {
// means username already exists.
}
Why? Because in between the time that you check for the user's existence and the time that you insert a new record, another client may create a user with the same username. So your second query the insert will fail if it happens (provided of course that you have a unique index on your username as you should).
Secondly, you are not escaping parameters. This leaves you open to SQL injection attacks. And it could also lead to malformed queries that do not get executed correctly - result data is not inserted. Use prepared statements instead.
$sql = mysqli_prepare($con,"INSERT INTO accounts (`userlvlid`,
`username`,
`password`,
`lname`,
`fname`,
`mname`,
`birthdate`,
`streetname`,
`region`,
`province`,
`city`,
`barangay`,
`orgname`,
`email`,
`license`)
VALUES
(?,?,?,?,?,?,?,?,?...)";
Then you need to bind the params
mysqli_bind_params($stmt,('$userlvlid',"ssssssssssssss",
$userlvlid
$username,
$password,
$lname,
$fname,
$mname,
$birthdate,
$streetname,
$region,
$province,
$city,
$barangay,
$organization_name,
$email_address,
$license));
Ugly isn't it? That's why one should use PDO instead of mysqli but using mysqli without prepared statements is just horrible so we have to slog through this.
Now while going through this copy paste, I discovered the real cause of your problem
'$mname',
'$birthdate' /*** no comma here ***/
'$streetname',

Answer 2

Try using echo for the $sql and place the exit after that statement. So that it breaks the query execution after that.
By doing this you can find whether any error is available in the query and you can rectify it.
Note: You first put echo to the Insert Statement and then break the execution by putting the exit; and you copy the statement that is echoed and place it in SQL of the DB and then check whether any error occurs in insertion. If no error occurs remove the echo and delete the exit;
If you have any Table Column mismatch or and Data Discrepancy you can rectify with the help of this echoed query.
Ensure that you provide values that are matching the values that are given into the DB.
**Note: **(E.g) if userlvlid - varchar(11) - You should provide value which is equal to 11 or less than 11 if you enter the value greater than 11 it will not be inserting and the query will fail from there on.