How to show image using php?

Question

I am just starting to use html, php, and mysql. I've successfully logged into my database using php, and formed a query. I now want to go a step further and show a picture instead of just strings or numbers.

The variable 'result' will be returning a string that has a url to an image i want to display on this webpage. How would I do this?

<html>
<head>
<title>My First Question</title>
</head>
<body>

<?php

$dbhost = 'someURL.com';
$dbname = 'user';
$dbuser = 'user';
$dbpass = 'password';

$mysql_handle = mysql_connect($dbhost, $dbuser, $dbpass)
    or die("Error Connecting To Database Server");

mysql_select_db($dbname, $mysql_handle)
    or die("Error selecting database: $dbname");

echo 'Successfully connected to database!';

$first = 'bobbie';

$query = sprintf("SELECT image FROM Player 
    p WHERE name='%s'", mysql_real_escape_string($first));

$result = mysql_query($query);

mysql_close($mysql_handle);     

?>

</body>
</html>

You are using an obsolete database API and should use a modern replacement. — Quentin
– Quentin, Commented Nov 1, 2012 at 7:22

Lix · Accepted Answer · 2012-11-01 08:19:58Z

10

Inside PHP, This will turn your SQL response into a usable variable.

$result = mysql_fetch_assoc(mysql_query($query));

Outside of your PHP tags, Echo the URL from the table into the SRC of an IMG element.

<img src="<?= $result['url_column_name'] ?>"/>

This will create a new IMG element with the source being the URL that you have fetched from your SQL query.

Short tags are also a way of echoing PHP variables in HTML.

<?= $var1, $var2 ?>

is the equivalent of using

<?php echo $var; echo $var2; ?>

edited Nov 1, 2012 at 8:19

Lix

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answered Nov 1, 2012 at 7:24

Matt Clark

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Comments

Lix · Accepted Answer · 2012-11-01 07:30:18Z

3

This is a simple case of echoing the relevant HTML. You'll also have to fetch the results after you execute the query -

$result = mysql_query($query);     
$data = mysql_fetch_assoc($result);
echo '<img src="'.$data['image'].'" />;

For added security, a good practice would be to escape any possible unwanted HTML content in your images path - htmlspecialchars($data['image']).

It should also be noted here that you are using a very old deprecated method to access your database. You might want to think about updating your code to use more modern PDO methods.

edited Nov 1, 2012 at 7:30

answered Nov 1, 2012 at 7:21

Lix

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3 Comments

Quentin Over a year ago

Don't forget htmlspecialchars

Robert Bain Over a year ago

I put it in the code you suggested all within my php tag, and it only writes: "Successfully connected to database!." Any idea why?

Lix Over a year ago

After you execute mysql_query(), try and see if there were any errors. You can use mysql_error() to see if there were any errors on the last query executed.

Mr. Alien · Accepted Answer · 2012-11-01 07:23:00Z

2

So what? simply use it as a source to your image

<?php $imgname = mysqli_fetch_array($connection, $result); ?>
<img src="<?php echo $imgname['image_column_name']; ?>" />

And btw use mysqli_() or PDO instead of using mysql_() as community is not maintaining it anymore

answered Nov 1, 2012 at 7:23

Mr. Alien

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Comments

Robert Bain · Accepted Answer · 2012-11-27 07:35:56Z

1

Once you update your mysql to mysqli you can echo the image's url in an html img tag as so:

echo '<img src="'.$result['image'].'"/>';

answered Nov 27, 2012 at 7:35

Robert Bain

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Comments

Ed Heal · Accepted Answer · 2012-11-01 07:24:30Z

0

<?php

$dbhost = 'someURL.com';
$dbname = 'user';
$dbuser = 'user';
$dbpass = 'password';

$mysql_handle = mysql_connect($dbhost, $dbuser, $dbpass)
    or die("Error Connecting To Database Server");

mysql_select_db($dbname, $mysql_handle)
    or die("Error selecting database: $dbname");

$first = 'bobbie';

$query = sprintf("SELECT image FROM Player 
    p WHERE name='%s'", mysql_real_escape_string($first));

$result = mysql_query($query);

mysql_close($mysql_handle);     
header("Location: $result");
?>

should work

answered Nov 1, 2012 at 7:24

Ed Heal

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