CODEWITHSUNDEEP
phphtml
charlie
charlie

Reputation: 1384

show logo in html without showing url in source

i am trying to display an image in HTML using:

<img src="/logo.php?seq=5" />

then logo.php looks like:

<?php
$sql="SELECT * from reseller where sequence = '".$_GET["seq"]."' ";
$rs=mysql_query($sql,$conn);
$result=mysql_fetch_array($rs);

echo '<img src="http://www.integradigital.co.uk/customer/'.$result["logo"].'" />';
?>

but its not working - whats the best way to do this so the user seeing the image cannot look at the URL of the image. if they open the image in its own window i want them to see something like http://www.domain.com/logo.php?seq=5 ???

Upvotes: 0

Views: 160

Answers (1)

Rudolf
Rudolf

Reputation: 1886

Use readfile() to read the image in the image.php:

// Read URL from database
$sql    = "SELECT * from reseller where sequence = '" . $_GET["seq"] . "'";
$rs     = mysql_query($sql,$conn);
$result = mysql_fetch_array($rs);

// Generate path
$path = '/customer/' . $result["logo"];

// Set proper headers
$headers = get_headers( $path );

foreach( $headers as $h )
    if( strpos( $h, 'Content-Type:' ) !== false )
        header( $h );

// Send file to user
readfile( $path );

Then PHP reads the right logo and outputs it, the user won't be able to see the real path. You can link the logo like you proposed:

<img src="/logo.php?seq=5" alt="Logo">

Upvotes: 2

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