class database {
public $connection;
function __construct() {
$this->connection = mysqli_connect(DBHOST,
DBUSER,
DBPASS,
DBNAME) or
die('Database Connection Error: '.mysqli_connect_error());
}
public function close_database() {
return mysqli_close($this->connection);
}
public function query($query) {
$query = mysqli_query($this->connection ,$query) or die($this->show_errors('Query Execution Error: '.mysqli_error($this->connection),'E'));
return $query;
}
public function fetch_assoc($query) {
$query = mysqli_fetch_assoc($query);
return $query;
}
}
$db = new database();
$query = $db->query("SHOW TABLES FROM DATABASENAME");
$db->fetch_assoc($query);
Is the most simple SQL statement for doing that. You can also take a look at INFORMATION_SCHEMA.TABLES if you want to have more details or do some filtering or such.
SELECT * FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_SCHEMA LIKE 'your_database';
here is little example
There are many ways.
Is the most simple SQL statement for doing that. You can also take a look at
INFORMATION_SCHEMA.TABLES
if you want to have more details or do some filtering or such.Using PHP 5.5 or later, a simple solution is using PHP's built-in array_column() function.
I'd try something like:
You might also be interested in skimming this: https://devzone.zend.com/13/php-101-part-8-databases-and-other-animals_part-2/ (EDIT: this link refers to mysql API and not mysqli, but most calls do have a mysqli parallel).
HTH