dropdown populated with prepared statement - php

Question

Hello and thank you for your time in advance.
I am looking to populate a dropdown menu from a mysql database using a prepared statement in PHP. I am just looking for any entries that have a matching "client_id". I have seen this addressed many ways (some of them out dated, some of them crazy complicated) but have not been able to find an example that is using php in a way similar enough to what I have been using (see below) so that I can cobble together something that works.
Here is the code I would like to work with.
<?php
include ('connect.php');
if ($_POST) {
if (!empty($_POST['client_id'])) {
$client_id = htmlspecialchars($_POST['client_id']);
$stmt = $link->prepare("SELECT project_id, nickname FROM project WHERE client_id=?");
$stmt->bind_param('s', $client_id);
$stmt->execute();
$row = $stmt->fetch();
while ($stmt->fetch()) {
echo ("<option value='$row['project_id']'>$row['nickname']</option>");
}
$stmt->close();
}
}
include ('disconnect.php');
?>
I know that the echo part is not correct, but I wanted to be able to show what I was thinking at the very least. In general I am not 100% I am using the row variable correctly either. I appreciate any assistance here.
EDIT:
the above file is called "fetchProjects.php"
Here is the code where this PHP file included.
<div class="twelve columns">
<p>select an existing project below</p>
<select size="10" name="selectedProject">
<?php include('fetchProjects.php')?>
</select>
</div>

Answer 1

You can use bind_result, on each iteration of the while loop $project_id and $nickname get updated with the current database row values
$stmt = $link->prepare("SELECT project_id, nickname FROM project WHERE client_id=?");
$stmt->bind_param('s', $client_id);
$stmt->bind_result($project_id, $nickname);
$stmt->execute();
$stmt->store_result();
while ($stmt->fetch()) {
echo '<option value="'.$project_id.'">'.$nickname.'</option>';
}
$stmt->close();
Not tested but it should work

Answer 2

You forgot to assign your fetch here:
while ($stmt->fetch()) {...
It should be:
while ($row = $stmt->fetch()) {...
Now the array elements you try to access with $row will work. You should delete your first fetch, because that will consume the first row of data. One more thing: you do not need parentheses areound your echo.

Answer 3

Please review:
http://php.net/manual/en/mysqli-result.fetch-assoc.php
http://php.net/manual/en/mysqli-stmt.fetch.php
Would advise:
<?php
include ('connect.php');
if (isset($_POST['client_id'])) {
if (!empty($_POST['client_id'])) {
$client_id = htmlspecialchars($_POST['client_id']);
$stmt = $link->prepare("SELECT project_id, nickname FROM project WHERE client_id=?");
$stmt->bind_param('s', $client_id);
$stmt->execute();
$stmt->bind_result($pid, $nick);
while ($stmt->fetch()) {
echo ("<option value='$pid'>$nick</option>");
}
$stmt->close();
}
}
include ('disconnect.php');
?>