Functions not able to find variable? - php

Question

I have the following variable $user_id being set by
//Check if user is logged in
session_start();
if (!isset ($_SESSION['user_id']))
{
header("location:login.php");
}
elseif(isset ($_SESSION['user_id']))
{
$user_id = $_SESSION['user_id'];
}
and then within the same function file I have the following:
function course_menu()
{
$sqlSubscription = "SELECT * FROM subscriptions WHERE `user_id` = '".$user_id."'";
$subscriptionResult = mysql_query($sqlSubscription);
while ($rows = mysql_fetch_assoc($subscriptionResult))
{
$user_id = $rows['user_id'];
$course_id = $rows['course_id'];
$course_title = $rows['course_title'];
if ($data_id == $rows['course_id'])
{
echo
'<li>
',$course_title,'
</li>';
}
else
{
echo
'<li>',$course_title,' </li>';
}
}
}
The problem is I keep getting undefined variable user_id every time I try to run the function. I can echo $user_id on another page lets say index.php by using require_once function.php and then echo $user_id, but for some reason the function itself can't access it?
I think it might be because it's outside its scope - but if so I'm not entirely sure what to do about it.
My question is, how can I get the function to be able to use the variable $user_id?
EDIT
So I've started doing
$user_id = $_SESSION['user_id'];
global $conn;
$sqlSubscription = "SELECT * FROM subscriptions WHERE `user_id` = '".$user_id."'";
$subscriptionResult = $conn->query($sqlSubscription);
while ($rows = mysqli_fetch_assoc($subscriptionResult))
{
$user_id = $rows['user_id'];
$course_id = $rows['course_id'];
$course_title = $rows['course_title'];
if ($data_id == $rows['course_id'])
{
echo
'<li>
',$course_title,'
</li>';
}
else
{
echo
'<li>',$course_title,' </li>';
}
}
which seems to work fine, but it's a bit tedious to add a new connection each time with a function or set the $user_id manually. Is there any way around this as I have several functions that require a connection to the db to pull data. Is there a better way to structure this type of stuff? I'm not very familiar with OOP but I can try it out if I can get some direction, here's another function that I use (and there are at least another 5-6 that require db connections)
function render_dashboard()
{
$user_id = $_SESSION['user_id'];
global $conn;
//Following brings up the number of subscription days left on the user dashboard
$sqlDate = "SELECT * FROM subscriptions WHERE `user_id` = '".$user_id."'" ;
$date = $conn->query($sqlDate);
while ($daterows = mysqli_fetch_assoc($date))
{
$course_registered = $daterows['course_title'];
$date_time = $daterows['end_date'];
$calculate_remaining = ((strtotime("$date_time")) - time())/86400;
$round_remaining = round("$calculate_remaining", 0, PHP_ROUND_HALF_UP);
// Here we assign the right term to the amount of time remaining I.E DAY/DAYS/EXPIRED
if($round_remaining > 1)
{
$remaining = $course_registered." ".$round_remaining." "."Days Remaining";
$subscriptionStatus = 2;
echo '<p>',$remaining,'</p>';
}
elseif ($round_remaining == 1)
{
$remaining = $course_registered." ".$round_remaining." "."Day Remaining";
$subscriptionStatus = 1;
echo '<p>',$remaining,'</p>';
}
elseif ($round_remaining <= 0)
{
$remaining = $course_registered." "."Expired"." ".$date_time;
$subscriptionStatus = 0;
echo '<p>',$remaining,'</p>';
}
}
//Check for most recent viewed video
$sqlVideo = "SELECT `last_video` FROM users WHERE `user_id` = '".$user_id."'" ;
$videoResult = $conn->query($sqlVideo);
if ($videoRows = mysqli_fetch_assoc($videoResult))
{
$last_video = $videoRows['last_video'];
$videoLink = "SELECT `chapter_id` FROM chapters WHERE `chapter_title` = '".$last_video."'";
if ($chapteridResult = mysql_fetch_assoc(mysql_query($videoLink)));
{
$chapter_id = $chapteridResult['chapter_id'];
}
$videoLink = "SELECT `course_id` FROM chapters WHERE `chapter_title` = '".$last_video."'";
if ($courseResult = mysql_fetch_assoc(mysql_query($videoLink)));
{
$course_id = $courseResult['course_id'];
}
}
}

Answer 1

The function course_menu() will not recognize your $user_id, Since it is outside its scope.
Make use of global keyword to solve this issue.
function course_menu()
{
global $user_id;
// your remaining code .........

Answer 2

The solution to getting around it without using global is to either DEFINE and pass it through ie - define ('var', '$var') then function x($var) or dependency injection as stated here How can I use "Dependency Injection" in simple php functions, and should I bother?