sql - PHP Warning: mysqli_stmt_bind_result() -


i new programing not php @ all. doing siple project of mine improve myself. did onec want make queries work prepared statements new me...

$create_stmt = mysqli_prepare($connection, "insert `users`(`user_name`, `password`) values (?,?)");     if(!$create_stmt){         echo 'error';         exit;     }     mysqli_stmt_bind_param($create_stmt, 'ss',$username,$password);     mysqli_stmt_execute($create_stmt);     mysqli_stmt_bind_result($create_stmt, $new_uname, $new_unamepass);     mysqli_stmt_fetch($create_stmt);

so here deal. having kind of error:

warning: mysqli_stmt_bind_result(): number of bind variables doesn't match number of fields in prepared statement.

since want add 2 values 'users' table declaring 2 new variables $new_uname, $new_unamepass, somehow not correct...

also mysqli_stmt_fetch($create_stmt); not true should if ok (obviously not ok..) please if can me or give me advice, great!

try this:

$mysqli = new mysqli('localhost', 'my_user', 'my_password', 'world');  /* check connection */ if (mysqli_connect_errno()) {     printf("connect failed: %s\n", mysqli_connect_error());     exit(); }  $stmt = $mysqli->prepare("insert `users`(`user_name`, `password`) values (?,?)"); $stmt->bind_param('ss', $user, $pass);  $user='username'; $pass = 'password1234';  /* execute prepared statement */ $stmt->execute();  printf("%d row inserted.\n", $stmt->affected_rows);  /* close statement , connection */ $stmt->close();

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