MySQL/PHP check if data is in database table, if not insert it

I'm having an issue checking if the data I entered is already in the database table and if it's not then inserting it into the table. What I can't figure out is how to get the id (it is primary and auto increment) of the data just inserted into the database.

I don't think it's possible to insert it if it is not already in the database and then get the id of that data to be inserted into another table all from the same script. I could be wrong though. So what I have now is one script to check if it's inserted already and inserting it if it's not, then redirecting to another php script to get the id of the information.

I'm having many errors with this, it inserts it even if it's in the database and when it gets redirected to the second script I am receiving a server error. Any help would be great, thanks in advance!

EDIT: I replaced the two files with one, the new code is below. I am still getting problems though, it inserts duplicate interests in the first table with my interests list

<?php

require_once '../scripts2/app_config.php';
require_once '../scripts2/database_connection.php';
require_once '../scripts2/authorize.php';

session_start();

// Authorize any user, as long as they're logged in
authorize_user();

$user_id = $_SESSION['user_id'];

$interest = trim($_REQUEST['interest']);


$get_interests = "SELECT COUNT(*) AS cnt FROM interests WHERE name = " . $interest;
$query_interests = mysql_query($get_interests);
if($query_interests) {
$hjk = mysql_fetch_array($query_interests);
$cnt = $hjk['cnt'];
}
if($cnt < 1) {
// $num_interests = mysql_num_rows($query_interests);

// insert the interest if it is not already in the database
$insert_sql = sprintf("INSERT INTO interests " .
                              "(name) " .
    "VALUES ('%s');",
         mysql_real_escape_string($interest));


//insert the user into the database
$insert_query = mysql_query($insert_sql);
$id = mysql_insert_id();

// From the newly inserted interest get the id of the interest
$insert_user_interests = sprintf("INSERT INTO user_interests " .
                                                 "(user_profile_id, interest_id) " .
                            "VALUES (%d, %d);",
                             mysql_real_escape_string($user_id),
                             mysql_real_escape_string($id));

mysql_query($insert_user_interests);
//Redirect this user to the page that displays user information
$url = 'show_profile_user_interests.php';
header('Location: '. $url);
exit();
} else {
/* If the interest is already in the table, just insert the id of the interest and
user_profile_id */

$get_interests2 = "SELECT id FROM interests WHERE name = " . $interest;
$query_interests2 = mysql_query($get_interests2);
if($query_interests2) {
$hye = mysql_fetch_array($query_interests2);
$interest_id = $hye['id'];
}
$insert_user_interests = sprintf("INSERT INTO user_interests " .
                                                 "(user_profile_id, interest_id) " .
                            "VALUES (%d, %d);",
                             mysql_real_escape_string($user_id),
                             mysql_real_escape_string($interest_id));

mysql_query($insert_user_interests);

//Redirect this user to the page that displays user information
$url = 'show_profile_user_interests.php';
header('Location: '. $url);
exit();
}
?>

What I think is causing the main problem is below, But I could be wrong...

$get_interests = "SELECT COUNT(id) AS cnt FROM interests WHERE name = " . $interest;
$query_interests = mysql_query($get_interests);
if($query_interests) {
$hjk = mysql_fetch_array($query_interests);
$cnt = $hjk['cnt'];
}
if($cnt > 0) {