Error with custom PHP function: no variable being returned

Question

Alright, so here is a look at my PHP function. I can confirm I AM connected to the database, as I can make updates to it with mysqli_query functions directly in the file.

<?php
function username_from_id($id) {
$id = mysqli_real_escape_string($id);
$query = mysqli_query($d,"SELECT `username` FROM `users` WHERE `id` = '$id'");
$result = mysqli_fetch_array($query);
$res = $result['username'];
return $res;
}
?>

The purpose of the function is to select the username of a user if their ID equals what is put into the query, then return it. In the file, it looks like this

<?php
include 'file_where_function_is.php';
$id = '1';
echo username_from_id($id);
?>

Nothing shows up. Any ideas?

Add error reporting to the top of your file(s) error_reporting(E_ALL); ini_set('display_errors', 1); mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); if you're not already doing so. — Funk Forty Niner
– Funk Forty Niner, Commented May 16, 2014 at 4:31
$d should still be brought into the scope of the function if referenced. It must be passed into the function or retrieved from the global scope at the beginning of the function by using the global keyword. — Scopey
– Scopey, Commented May 16, 2014 at 4:33
There are 2 things that can be done: 1) pass $d as a function argument 2) omit $d and the mysqli_query() will use last used connection (which I find a great example of OOP fail in PHP) UPDATE: actually it has to be checked, as documentation describe that behavior for mysql_query() but there's nothing like that about mysqli_query(). — pawel7318
– pawel7318, Commented May 16, 2014 at 4:33
He "can make updates to it with mysqli_query functions directly in the file" which probably means he's testing that right after connecting to the database, where $d is initialized. His function has never heard of $d — ElefantPhace
– ElefantPhace, Commented May 16, 2014 at 4:40

Phil · Accepted Answer · 2014-05-16 05:05:05Z

2

As pointed out in comments, this is a scoping issue. Your $d variable (mysqli instance) is not in scope within username_from_id. Here's how to fix it...

function username_from_id(mysqli $d, $id) {
    if (!$stmt = $d->prepare('SELECT username FROM users WHERE id = ? LIMIT 1')) {
        throw new Exception($d->error, $d->errno);
    }
    $stmt->bind_param('i', $id);

    if (!$stmt->execute()) {
        throw new Exception($stmt->error, $stmt->errno);
    }

    $stmt->bind_result($username);

    if ($stmt->fetch()) {
        return $username;
    }
    return null;
}

and call it like this

include 'file_where_function_is.php';
$id = 1;
echo username_from_id($d, $id); // assuming $d exists in this scope

answered May 16, 2014 at 5:05

Phil

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