echo ("<img src=IMAGE.png>");

I use this code to call an image from the appropriate folder and the image shows up just fine.

My question is:

How do I determine the image's dimensions within the parameters of my script?

I want to center it on the page, and I want to link the image to a webaddress.

Thank you

try this code

PHP Code:

<?php

echo '<a href="'.$url.'"><img src="'.$image.'.png" class="center" /></a>';

?>

$url and $image are variables so set them higher up.

and in your css put

.center {
text-align: center;}

this code should work I think.

the image was called just fine, but it wouldn't link to an external address.

I am trying to place that linkable image in this particular script:

PHP Code:

<?
  session_start();
  include "../affconfig.php";
  include "./lang/$language";
  if(!aff_check_security())
  {
    aff_redirect('index.php');
    exit;
  }
 
  include "header.php"; 
 
  print "<p><br>".AFF_C_CONTACT." <a href=mailto:";
  print $emailinfo;
  print ">";
  print $emailinfo;
  print "</a>.";
  include "footer.php"; 
?>

I'm posting it to see if that gives you guys more insight as to why your recommendation wasn't working for me.

it works when I call the image with:

PHP Code:

<?php
 
echo ("<img src=IMAGE.png>");

 
?php>

but it doesn't allow me to link that image to an external website, or position it upon the page to my desired dimension.

Any ideas?