CODEWITHSUNDEEP
javascriptphpcaching
Albin wärme
Albin wärme

Reputation: 261

Cache AND display images PHP

I am using a PHP script to display images stored as blobs in my database.

I used to display them with the following script.

<?php

if(!empty($_GET['user_id'])){//script to display an image from the database you basically just get the id from the picture in matter and fucking access it
include_once "DBH.php";
//Get image data from database

$id = mysqli_real_escape_string($conn, $_GET['user_id']);
$result = $conn->query("SELECT image FROM profilePictures WHERE user_id = $id");
if($result->num_rows > 0){
    $imgData = $result->fetch_assoc();
    header("Content-type: image"); 
    echo $imgData['image']; 
}else{
    echo 'Image not found...';
}
}
?>

In the context where

<img src = 'displayProfilePicture.php?user_id=n'>

The div containing the divs are updated frequently and to update the users image seems like a lot of unnecessary processing. I want to cache the profilepictures in the webpage so that I don't have to query them from the database every time. I started reading a lot about how you could cache the images but could not find any content on how to display the cached images.

This is a problem for me as the images flicker for a bit every time the img is updated with the PHP script. In an optimal world I see that the img load one time and then after that it does not have to load.

The context which I use the display img script is in a chat that is updated with a timer-interval within an ajax-request

 $("#chatlogs").load("logs.php");

logs.php

 if(isset($_SESSION['chatRoomId'])){
 while ($extract = mysqli_fetch_array($result1))
 {
   $from = $extract['from'];

   //make an object to echo.
   if($from == $id){
   echo "<div class='chatContainer self'>
    <div class = 'imgContainer'>
     <img src='displayProfilePicture.php?user_id=$selfId'>
    </div>
   <div class='content'>
    <div class = 'message'>
        ". $extract['msg'] ."
    </div>
</div>
</div>";
}else{
echo "<div class='chatContainer friend'>
<div class = 'imgContainer'>
  <img src='displayProfilePicture.php?user_id=$guestId'>
</div>
<div class='content'>
    <div class = 'message'>
        ". $extract['msg'] ."
    </div>
</div>
</div>";
}
}
}

Upvotes: 0

Views: 1571

Answers (1)

Kerkouch
Kerkouch

Reputation: 1456

I think this is what you looking for:

<?php

if(empty($_GET['user_id']) || !preg_match( '/^[0-9]+$/' , $_GET['user_id'])){
    header( '400 Bad Request' );
    exit(1);
}

include_once "DBH.php";


$user_id = intval( $_GET['user_id'] );

$result = $conn->query("SELECT image FROM profilePictures WHERE user_id = $user_id");
if($result->num_rows == 0){
    // Not Found
    header('404 Not Found');
    exit(1);
}

$imgData = $result->fetch_assoc();
header("Content-type: image"); 
$cache_for = 3600; // One hour in seconds 
$cache_until = gmdate("D, d M Y H:i:s", time() + $cache_for) . " GMT";
header("Expires: $cache_until");
header("Pragma: cache");
header("Cache-Control: max-age=$cache_for");
echo $imgData['image'];
exit(0);

Comments

First I checked if the user_id is supplied in the request, if so then check if it was a valid number if it doesn't then respond with a 400 error.

And also I have removed a SQLi in your code src='displayProfilePicture.php?user_id=-1 or 1=1.

And I have set the caching headers, so the browser will cache the image for an hour.

Upvotes: 1

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