Displaying a date specific image from a databse - php

Question

The aim is for a customer to upload their advert to the database, that triggers and email for approval. Once approved ad will display in location between the set dates.
I have written the code correctly to display by date and approval from an admin area. I just cannot get the image to display.
The code below details the process so far.
<?php
include('mysql_connect.php');{
$location='1';
}
$resultSet = $mysqli->query("SELECT * FROM adverts WHERE adloc = '$location' AND approval ='Y' ");
if($resultSet->num_rows > 0){
while($rows = $resultSet->fetch_assoc())
{
$id = $rows ['id'];
$start = $rows ['start'];
$end = $rows ['enddate'];
$business = $rows ['business'];
$email = $rows ['email'];
$tel = $rows ['tel'];
$web = $rows ['web'];
$advert = $rows ['image'];
$Date = date('Y-m-d');
$Date=date('Y-m-d', strtotime($Date));;
$DateBegin = date('Y-m-d', strtotime("$start"));
$DateEnd = date('Y-m-d', strtotime("$end"));
}
if (($Date > $DateBegin) && ($Date < $DateEnd))
{
echo "ADVERT";
echo '<img src="getad.php?id=$id">';
}
else
{
echo "FILLER IMAGE";
}
}
?>
The code for the getad.php file is as follows
if(isset($_GET['id']))
{
$id = mysqli_real_escape_string($_GET['id']);
$query = mysqli_query("SELECT * FROM adverts WHERE id= '$id'");
while ($row = mysqli_fetch_assoc($query))
{
$image = $row['image'];
}
header("content-type: image/jpeg");
echo $image;
}
else
{
echo "Error!";
}
?>
I am sure I can't be far off. The image is a JPEG LONGBLOB in the database.
Thank you.

Answer 1

Firstly, this is failing you:
$id = mysqli_real_escape_string($_GET['id']);
$query = mysqli_query("SELECT * FROM adverts WHERE id= '$id'");
Both of those functions require a database connection be passed and as the first parameter:
$id = mysqli_real_escape_string($mysqli, $_GET['id']);
$query = mysqli_query($mysqli, "SELECT * FROM adverts WHERE id= '$id'");
References:
http://php.net/manual/en/mysqli.real-escape-string.php
http://php.net/manual/en/mysqli.query.php
However, may need to use base64_decode:
header("content-type: image/jpeg");
echo base64_decode($image);
or
echo '<img src="data:image/gif;base64,' . $image . '" />';
Here is an example borrowed from https://stackoverflow.com/a/20564797/
$db = mysqli_connect("localhost","root","","DbName"); //keep your db name
$sql = "SELECT * FROM products WHERE id = $id";
$sth = $db->query($sql);
$result=mysqli_fetch_array($sth);
echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image'] ).'"/>';
For debugging purposes:
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote: Displaying errors should only be done in staging, and never production.
As well as or die(mysqli_error($mysqli)) to mysqli_query().