I've got a little bit of a problem when I try to run this code. I receive this error:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in
My problem is in the while statement on line 32:
$connection = mysqli_connect($host, $username, $password, $db_name); $sql = 'SELECT * FROM provinsi'; while ($r = mysqli_fetch_array($connection, $sql)) { // line 32 $id = $r['id']; } mysqli_fetch_array()'s 1st parameter must be a result of a query. What you are doing is you are passing the connection (which doesn't make sense) and the query command itself.
Read the doc here: http://php.net/manual/en/mysqli-result.fetch-array.php
To fix this, execute the query first, then store the result to a variable then later fetch that variable.
$sql = "select * from privinsi"; $result = mysqli_query($connection,$sql); while($r = mysqli_fetch_array($result)) { // your code here } 
You wrote SQL query but you didn't execute it.
$sql = "select * from privinsi"; $results = mysqli_query($connection, $sql); while($r = mysqli_fetch_array($results){ //enter your code here } 
$sql = "select * from privinsi"; $result = mysqli_query($connection,$sql); if (!$result) { printf("Error: %s\n", mysqli_error($connection)); exit(); }else{ while($row = mysqli_fetch_array($result)){ // your code here } } This helps you to display your error so you would be able to find out the exact reason a query is not working. This is helpful when you are using conditions or trying to find specific columns in a table. Tired eyes can make you skip the fact that a particular column is not on your current database and other factors that you can casually overlook when writing a query. If your error is not detailed, you would inevitably run around in circles looking for a solution that is not beyond you.
