Accessing a MySQL Link Identifier from within a Function - php

Question

I'm having some difficulty returning an array out of a while lopp which I have in a function. Here is the code I am using. I am meant to be able to return an array of results from the function which contains the id numbers of pictures associated with a particular user id - in this case I want to print_r the array for the user id of 17. When this code isn't in the function it works, but when I place it in the function, no luck. I presume its related to a mistake I am making in the returning of the array. Your help is greatly appreciated.
function picture($id)
{
$sql = "SELECT * FROM avatar WHERE user_id={$id}";
$result = $database->query($sql);
$results = array();
while ($row = mysql_fetch_assoc($result))
{
$results[] = $row;
}
return $results;
}
$results = picture(17);
print_r($results);

Answer 1

Your function can't access your MySQL link identifier
First of all, you're mixing object-oriented paradigm ($database->query($sql)) with procedural paradigm (mysql_fetch_assoc($result)) which will make your code a nightmare to maintain.
Assuming that $database is a mysql_ link identifier, you'll need to pass it into your function in order to access it there.
function getUserAvatar($database, $id){
$sql = 'SELECT * FROM `avatar` WHERE `user_id`=' . intval($id) . ' LIMIT 1;';
$result = mysql_query($database, $sql);
$row = mysql_fetch_assoc($result);
return $row;
}
$results = picture($database, 17);
Don't just copy-paste that, keep reading!
The above will probably work, but if you're allowing a user to pass that user ID into the function, it's quite possible that they'll be able to find a vulnerability to inject an SQL statement of their choice into your MySQL database.
mysql_ functions are deprecated, so you should ideally stop using them and switch to mysqli or PDO. You'll also want to get an understanding of prepared statements in order to prevent SQL injections. If you can't upgrade, look at the mysql_real_escape_string and intval functions and make sure you sanitize all user inputs before processing them.
The resulting code will look something like this, if you switch to mysqli and prepared statements:
function getUserAvatar($db, $userId) {
$stmt = $db->prepare("SELECT * FROM `avatar` WHERE `user_id`=? LIMIT 1;");
$stmt->bind_param("i", $userId);
$stmt->execute();
$res = $stmt->get_result();
return $res->fetch_assoc();
}
$db = new mysqli("localhost", "user", "password", "database");
$result = getUserAvatar($db, 17);

Answer 2

may be you should try this..
function picture($id)
{
$sql = "SELECT * FROM avatar WHERE user_id={$id}";
$result = $database->query($sql);
$row = mysql_fetch_assoc($result);
return $row;
}
$results = picture(17);
print_r($results);