display image using php

Question

I have a the following code that i intend to should return a filepath name from mysql in the php section and then display the image in html. But i am only getting up a tiny thumbnail in my browser. By the way "username" and "imagefile" are the only columns in the table "images".

I'm sure there is just some silly mistake but i need a fresh pair of eyes to spot it. Thanks a lot in advance. P.S i know i should really me moving over to mysqli but i will simply translate at a later date. Cheers

<?php


session_start();
$username = $_SESSION['username'];

$con = mysql_connect('localhost','root','password');
mysql_select_db("db");


$profileimage = mysql_query("

SELECT * FROM images WHERE username='$username'

");

$row = mysql_fetch_array($profileimage);
$showimage = $row['imagefile'];

?>

<html>

<img src = "$showimage">


</html>

i have the following code that should return...sorry!

Ross
– Ross

2012-12-01 19:39:03 +00:00
Commented Dec 1, 2012 at 19:39 — Ross
– Ross, Commented Dec 1, 2012 at 19:39

rg88 · Accepted Answer · 2012-12-01 19:45:23Z

8

First off, HTML doesn't know what "$showimage" means. That is a PHP variable and HTML cannot interpret it. You need to output it so that HTML can just deal with the result.

So if the value for $showimage is "/images/foo.jpg" you would need something like:

<img src="<?php echo $showimage; ?>" />

which would give you

<img src="/images/foo.jpg" />

Now, switching things to mysqli is as simple as replacing mysql with mysqli. It's no more complicated than that. Since it looks like you are just starting to learn about these things you may as well, when you go to improve things, learn about PDO.

answered Dec 1, 2012 at 19:45

rg88

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2 Comments

Ross Over a year ago

I pretty new to web developing. What you says seems obvious now. Thanks a lot to both of you above. Really appreciate the quick reply.

rg88 Over a year ago

@Ross Don't forget to upvote comments that help you and "accept" the one you find best for you.

koopajah · Accepted Answer · 2012-12-01 19:44:13Z

2

Is this your current real code or is it a simplified version? If it is your real code the problem is in the HTML part where the PHP variable is unknown, you should do this:

<html>
<img src ="<?php echo $showimage; ?>" />
</html>

answered Dec 1, 2012 at 19:44

koopajah

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2 Comments

Ross Over a year ago

It was. I see what you are saying. Thanks a lot for the quick response.

koopajah Over a year ago

No problem! Don't forget to add one of the answers as "accepted" to show you had a proper answer.

Michael Dodd · Accepted Answer · 2017-09-27 16:47:04Z

-1

<?php  
    $db = mysqli_connect("localhost:3306","root","","databasename"); 
    $sql = "SELECT * FROM table_name ";
    $sth = $db->query($sql);
    while($result=mysqli_fetch_array($sth)){    
        echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image'] ).'" height="100" width="100"/>';
    }
?>

Works for me

edited Sep 27, 2017 at 16:47

Michael Dodd

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answered Sep 27, 2017 at 15:55

JHM16

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