Php - Display image from Database using path (PDO)

Question

I tried to display image from database,

image in directory

path in database

require_once "Connection.php";
class DisplayDataImageProfile {
function showImageProfile(){
    $connection = new Connection();
    $conn = $connection->getConnection();

    $id = $_GET['id'];

    try{
        $sqlDisplay = "SELECT photo FROM frofile WHERE id =$id";
        $getImage = $conn->prepare($sqlDisplay);
        $getImage->execute();
        $getImage->fetchAll(PDO::FETCH_ASSOC);

        foreach($getImage as $data){
            header('Content-type: image/jpg');
            // echo "<img src='$data'>";
            echo $data;
        }

    }catch (PDOException $e){
        echo "Error : " + $e->getMessage();
    }
}}

after that i call in html page like this :

<img src="DisplayDataImageProfile .php?id=3" align="center" />

i got problem that image cannot retrieve from database using path. other case on webpage the image something like broken image displayed.

for currently i just display one image.

@red Your code is open to SQL injection, you're not preparing correctly. — Matt
– Matt, Commented Mar 7, 2016 at 6:24

Your Common Sense · Accepted Answer · 2016-03-07 07:13:13Z

1

You should use the proper method.

Instead of fetchAll() which returns a nested array, you have to use fetchColumn() which returns single value. And you should be using prepared statements properly:

$sql = "SELECT photo FROM frofile WHERE id = ?";
$getImage = $conn->prepare($sqlDisplay);
$getImage->execute([$_GET['id']]);
header('Content-type: image/jpg');
echo $getImage->fetchColumn();

Edit: If you don't have an image itself in database but only path to the image, then you don't need this code at all. Get rid of Display.php and just select your path in the script that is echoing

<img src="Display.php?id=3" align="center" />

and echo selected path instead of Display.php?id=3.

edited Mar 7, 2016 at 7:13

answered Mar 7, 2016 at 6:32

Your Common Sense

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5 Comments

red Over a year ago

now it gave me broken image. on web page like display an image, but broken image.

Your Common Sense Over a year ago

I added header to my code. And obviously you have to have an image in your database stored properly.

red Over a year ago

in my database i just store image path, and the image in the folder still gave me broken image.

Your Common Sense Over a year ago

so you don't need no Display.php script at all!

red Over a year ago

sorry sir, once again my question, after i followed your code, i got error that id is Undefined. when should i define the variable id ?

devpro · Accepted Answer · 2016-03-07 06:25:54Z

0

You need to use $data as:

$data['photo']

instead of just

$data

Because $getImage consist on associative array, so you can not use like your example, you need to use index for print data.

answered Mar 7, 2016 at 6:25

devpro

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6 Comments

red Over a year ago

now it gave me broken image. on web page like display an image, but broken image.

devpro Over a year ago

@red: check url in view source

red Over a year ago

something like this : localhost:49185/insertData/AdminPage.php where image will be displayed

devpro Over a year ago

@red: and what is the result of print_r($getImage);

red Over a year ago

still broken image sir.

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