Question

This is what my table looks like in my database. I’m trying to display an image I stored it’s a mimetype (longblob) . When I run the code it gives me a small box with a ? , no error just that box. Does anyone know what the error is and how I can fix it?

Display
+-------+------------+----------+
| Index | Display_ID | Picture  |
+-------+------------+----------+
|     1 |         12 | longblob |
+-------+------------+----------+


<?php
    $mysqli=mysqli_connect('localhost','root','','draftdb');


    if (!$mysqli)
        die("Can't connect to MySQL: ".mysqli_connect_error());

    $imageid= 12;

    $stmt = $mysqli->prepare("SELECT PICTURE FROM display WHERE DISPLAY_ID=$imageid"); 
    $stmt->bind_param("i", $imageid);

    $stmt->execute();
    $stmt->store_result();

    $stmt->bind_result($image);
    $stmt->fetch();

    header("Content-Type: image/jpeg");
    echo $image; 
?>

Answer 1

This:

$stmt = $mysqli->prepare("SELECT PICTURE FROM display WHERE DISPLAY_ID=$imageid");

Should be:

$stmt = $mysqli->prepare('SELECT PICTURE FROM display WHERE DISPLAY_ID=?');

You were directly embedding the variable in the query instead of actually using the bound variables like you intended to.