PHP - Display image based on email - php

Question

I am looking for some help.
Need to create a branding page, that upon logging in would display an image based on user's email that's stored in the database.
Creating a cookie and session id works just fine. Just can't find a way to select image based on user's email.
Pasting just part of the code that actually handles user id, as anything else in my index.php file is pure html.
So far:
<?php
include 'function.php';
include 'sqlconnect.php';
//User Check
$adminid = "1";
$company1= "#gmail.com";
$company2= "#yahoo.com";
$gmail= mysqli_query($connection, "SELECT Email FROM Users WHERE Email='$company1'");
$yahoo= mysqli_query($connection, "SELECT Email FROM Users WHERE Email='$company2'");
if(loggedin()){
$checkid = $_SESSION['user_id'];
if ($adminid == $checkid){ //if user id is 1 then display img of administrator
echo '<img src="images/admin.png" title="Administrator" width:"75px" height="75px">';
echo "<li><a href='logout.php'>Logout</a></li>";
} elseif(filter_var($company1, FILTER_VALIDATE_EMAIL) && preg_match('/#.+\.gmail/', $company1)) {
echo "<li><a href='logout.php'>Logout</a></li>";
echo '<img src="images/google.png">';
} elseif (filter_var($company2, FILTER_VALIDATE_EMAIL) && preg_match('/#.+\.yahoo/',$company2)) {
echo "<li><a href='logout.php'>Logout</a></li>";
echo '<img src="images/yahoo.png>';
}
}
?>
Thank you in advance!

Answer 1

This should get your user based on the session's user_id then display the image as expected. Here are a few notes:
I wasn't sure what your id field was, so if it is different, replace UserId with the name of your column in the $results = line.
I did not run the query in the case that it is an admin, so if you need data for something else in that case, you'll have to move the $result = .. line to above the if ($userid == 1){ line.
I didn't check the regex for your preg_match conditions, so if the results are coming in, but the image isn't being displayed, you can see what the Email field is by adding echo <p>'.$result['Email'].'</p> right after the if(mysql_num_rows($result) == 1) { line.
<?php
include 'function.php';
include 'sqlconnect.php';
if(loggedin()){
$userid = $_SESSION['user_id'];
if ($userid == 1){ //if user id is 1 then display img of administrator
echo '<img src="images/admin.png" title="Administrator" width:"75px" height="75px">';
echo "<li><a href='logout.php'>Logout</a></li>";
} else {
$results = mysqli_query($connection, "SELECT Email FROM Users WHERE UserId='$userid'");
if(mysqli_num_rows($results) == 1) {
foreach(mysqli_fetch_assoc($results) as $result) {
echo "<li><a href='logout.php'>Logout</a></li>";
if (strpos($result,'yahoo')) {
echo '<img src="images/yahoo.png>';
} elseif (strpos($result,'gmail')) {
echo '<img src="images/google.png">';
} else {
// you could echo a default image here
}
}
}
}
}
?>