PDO issue with displaying appropriate data - php

Question

To begin with, I'm having this small PDO snippet of code which gets me all available databases on the server and another function for the tables:
I need whenever I select and submit an option from the db list to display its corresponding tables in the select menu below. I managed to display the databases with a foreach but I kinda miss the point why this doesn't work.
Thanks in advance for any answers/solutions :)

Answer 1

see below for a revision to your code in order to achieve what you're looking for.
<form method='post' action="<?php $_SERVER['PHP_SELF']; ?>">
<?php
define('PDO_USER', 'root');
define('PDO_PASS', '');
function getDatabases(PDO $pdo) {
$stmt = $pdo->query('SHOW DATABASES');
$result = $stmt->fetchAll();
$dbs = array();
foreach($result as $row) {
$dbs[] = $row['Database'];
}
return $dbs;
}
$pdo = new PDO('mysql:host=localhost', PDO_USER, PDO_PASS);
$pdo->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);
$databases = getDatabases($pdo);
// Table code
$selectedDB = (!empty($_POST['database'])) ? $_POST['database'] : null;
function getTables(PDO $pdo, $databaseName) {
if(!in_array($databaseName, getDatabases($pdo))) {
return array();
}
$stmt = $pdo->query('SHOW TABLES FROM '.$databaseName);
$result = $stmt->fetchAll();
$tables = array();
foreach($result as $row) {
$tables[] = $row['Tables_in_'.$databaseName];
}
return $tables;
}
$tables = array();
if(!empty($selectedDB)) {
$tables = getTables($pdo, $selectedDB);
}
?>
Database:
<select name='database'>
<?php foreach($databases as $row): ?>
<option value="<?php echo $row; ?>"><?php echo $row; ?></option>
<?php endforeach; ?>
</select> 
<input type='submit' name='formSubmit' value='Submit'></form>
UPDATE
<select name='formTable1'>
<?php foreach($tables as $tbName): ?>
<option value='<?php echo $tbName; ?>'><?php echo $tbName; ?></option>
<?php endforeach; ?>
</select>
A couple of notes:
You need to include the "action" attribute in the form tag in order to give the form a target URL for submission.
In your original select you had 2 name attributes
(name='formDatabases[]' name='database'). You should only have one
name per form element, which is what is determined to be used as the
key in the $_POST array for that particular field.
You should only use "name='var[]'" format in the HTML, if you expect
that particular POST item to be an array in PHP. Otherwise you can
just use "name='var'"
This may be out of the scope of your original question, but you may
want to consider separating your database business logic into a separate
class, to keep your HTML clean. For now at least, I updated the code to have
the logic at the top of the file so that in can be extracted into a
class more easily in the future. This avoids the perilous path into PHP Spaghetti string code!