Use variable as ID and get table from mysql (edited) - php

Question

I'm using a multiple select option form to get a table of venues. Each venue has an ID and this is what I used:
<?php
require("db_access.php");
if(isset($_POST['select3']))
{
$aVenues = $_POST['select3'];
if(!isset($aVenues))
{
echo("<p>You didn't select any venues!</p>\n");
}
else
{
$nVenues = count($aVenues);
echo("<p>You selected $nVenues venues: ");
for($i=0; $i < $nVenues; $i++)
{
echo($aVenues[$i] . " ");
}
echo("</p>");
$sql = "SELECT * FROM venues WHERE id IN (" . implode(",",$aVenues) . ")";
$comma_separated = implode(",", $aVenues);
echo $comma_separated;
}
}
?>
It results in this:
However I thought that the code would use those two numbers below and draw out a table with those id's I used :/ ? Am I missing something?

Answer 1

$array is used in implode(",", $array); but is not defined anywhere else that we can see. It is perhaps intended to be:
implode(",", $aVenues);
UPDATE
Per comments, it does not draw a table because you never actually query your database.
You build your SQL statement, but you need to execute it and fetch the result set.
// Make sure you actually have a database connection
$conn = mysql_connect('localhost', $username, $password);
mysql_select_db($database);
$sql = "SELECT * FROM venues WHERE id IN (" . implode(",",$aVenues) . ")";
$comma_separated = implode(",", $array);
echo $comma_separated;
// Execute query and fetch result rowset
$result = mysql_query($sql);
if ($result) {
$rowset = array();
while ($row = mysql_fetch_array($result)) {
$rowset[] = $row;
}
var_dump($rowset);
}
else echo mysql_error();