Reputation: 3
This is an HTML scrip that I created. I am trying to include a php script that contains a variable named $url which has a picture url in it. When I echo the var out, it displays nothing. What am I doing wrong?
<html>
<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<?php
include("script.php");
?>
<img src= "<?php echo $url; ?>">
</body>
</html>
Upvotes: 0
Views: 125
Reputation: 82804
Usually your server is not configured to execute PHP code in HTML files.
The files with name .php
are handled specially by your web server, i.e., executed as PHP programs.
It goes like this:
Browser asks for file
|
v
Server looks at found file
| |
v v
.html ? .php ?
| |
v v
send it to give it to
the browser PHP and send,
as-is what returns
You must rename your .html file to .php in order for PHP to work in the server's default configuration.
Upvotes: 2
Reputation: 1914
Since in the script.php you have the var $url equal to a url. I guess you dont support php. You should install xamp or wamp for localhost.
To test it change the code to this
<img src= "<?php echo "http://www.google.com//images/srpr/logo3w.png"; ?>">
and check if this is working, if not simple install xamp it will install php, apache and mysql :)
Upvotes: 0
Reputation: 44464
Provide the full URL.
In the included file, change to:
$url = "http://www.google.com//images/srpr/logo3w.png";
Upvotes: 0