Showing Image From database in html using php

Question

I am having trouble displaying an image in my mysql database uploaded as a blob.

I have tried echoing the image. All other categories are displayed except the image.

$query = $handler->query('SELECT * FROM articles');
$results = $query->fetchAll(PDO::FETCH_ASSOC);

for ($i=0; $i < count($results); $i++) { 
  echo '<div class="col-md-4 col-xs-12 col-sm-12 height-news">';
  echo '<p class="news-title">'.$results[$i]['headline'].'<br>'.'</p>';
  echo '<img class="news-img" '.'src="'.$results[$i]['logo'].'">'.'</img>'.'<br>';
  echo '<p class="news-time">'.$results[$i]['date'].'<br>'.'</p>';
  echo '<p class="news-body">'.$results[$i]['text'].'<br>'.'</p>'.'</div>';
}

?>

Using this code I get this error:

Notice: Undefined variable: result in C:\xampp\htdocs\deeplake6\news.php on line 129

Using this code I get this error: Notice: Undefined variable: result in C:\xampp\htdocs\deeplake6\news.php on line 129 — Diced Mango, Aug 17 '17 at 14:47
@Rbaskam With 0 information about the database structure you have to assume those table names are correct — IsThisJavascript, Aug 17 '17 at 14:50
Quite possibly you have stored the image incorrectly in the first place. ___If you also show us the code you used to insert the image to the database it will help us help you.___ — RiggsFolly, Aug 17 '17 at 14:51
I do not see a variable called `result` in the code you show us. Can you identify `line 129` in the code you show us above — RiggsFolly, Aug 17 '17 at 15:21

score -1 · Answer 1 · edited Aug 18 '17 at 14:12

-1

try this ..

echo '<img src="data:image/jpeg;base64,'.base64_encode( $results[$i]['logo'] ).'" />';

the mysql link file :: https://pastebin.com/0bdCCw8P

and this my own test Code :

$dsn = 'mysql:host=localhost;dbname=stack1'; 
    $username = 'root'; 
    $pwd=''; 
    try {
        $db = new PDO($dsn, $username, $pwd);
    }
    catch (PDOException $e) {
        $error_message = $e->getMessage();
        echo "this is displayed because an error was found";
        exit();
}




$query = $db->prepare("SELECT * FROM articles");
$query->execute();
$result = $query->fetchall();


//echo $result['0']['2'] ;


echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['0']['2'] ).'"/>';

die ;

edited Aug 18 '17 at 14:12

Funk Forty Niner

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answered Aug 17 '17 at 14:51

Houssem Cherif

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This did not work thanks though! – Diced Mango Aug 17 '17 at 14:52
As the OP did not say they stored the image as a base64 encoding, this can only be a guess – RiggsFolly Aug 17 '17 at 14:53
1

@Houssem CH you can't give an answer based on your opinion.You don't have information about base64 encoding. – Mr Alb Aug 17 '17 at 14:56
i'm trying this example in my own laptop – Houssem Cherif Aug 17 '17 at 16:07

Showing Image From database in html using php

1 Answers1