referencing an image from php - php

Question

I have a function on a website where I have a img reference leading to a php page
<img src="selectimage.php?uid=21312412">
which I need to return an image and update a database file
<?php
$servername = "localhost";
$username = "webz";
$password = "BLAH";
$dbname = "contactlist";
$readid = $_GET["uid"];
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "UPDATE emails SET `read`='1' WHERE `id`='" .$readid. "'";
if ($conn->query($sql) === TRUE) {
echo "<img src='logo11w.png'>";
}
$conn->close();
?>
I figured this would not work as I am referencing a php page in which is referencing the HTML tag
While this is updating my database it is not displaying the image.
how do I make this php file send the image logo11w.png to that img tag
Further note : include or other methods of including this code on the page loading are out of the question because this procedure may be called from a off-server source.

Answer 1

Store user id in one hidden field
You can use ajax for get respone back from php page after updating value in database
<input type="hidden" id="userid" value="user_id_value">
<input type="button" id="getimage" value="GET IMAGE">
<div id="u_img"></div>
$(document).ready(function(){
$('#getimage').click(function() {
var uid = $('#userid').val();
var datastring = "uid="+uid;
$.ajax({
type: "POST",
url: "php_page.php",
data: datastring,
cache: false,
success: function(result) {
$('#u_img').append(result);
}
});
});
});

Answer 2

The most effective method for what I am trying to do turns out to be:
...
if ($conn->query($sql) === TRUE) {
header('Content-Type: image/jpeg');
echo file_get_contents('https://www.google.com/images/srpr/logo11w.png');
} else {
echo "Error updating record: " . $conn->error;
}
...