generating a css file incude using $_SERVER['SERVER_NAME'];

[junk]

Hi
can anyone tell me why this doesn't work:
<link rel="stylsheet" href="http://<?php echo $_SERVER['SERVER_NAME'];
?>/includes/test3.css" type="text/css" />

there is no wrap in the actual file.

when I view the source the code is printing the correct information.

I think I picked a good one for my first attempt at php. I am trying to
create a generic template.

cheers

Jim

[Janwillem Borleffs]

junk wrote:

can anyone tell me why this doesn't work:
<link rel="stylsheet" href="http://<?php echo $_SERVER['SERVER_NAME'];

/includes/test3.css" type="text/css" />

When you paste the printed stylesheet href in your browser and are able to
retrieve it from there, there's probably an error in the css file when it
isn't rendered the way you expected it to.

It might also be useful to use HTTP_HOST rather then SERVER_NAME (which can
be 2 different things).
JW

[John Dunlop]

junk wrote:

can anyone tell me why this doesn't work:
<link rel="stylsheet" href="http://<?php echo $_SERVER['SERVER_NAME'];
?>/includes/test3.css" type="text/css" />
A missing 'e' in the value of the rel attribute; it should
read 'stylesheet', I suppose, not 'stylsheet'.
there is no wrap in the actual file.

That makes no difference because white space between your
PHP tags isn't outputted.

--
Jock

[junk]

junk wrote:

Hi
can anyone tell me why this doesn't work:
<link rel="stylsheet" href="http://<?php echo $_SERVER['SERVER_NAME'];
?>/includes/test3.css" type="text/css" />

ROFLMAO.. It was the missing "e" in "stylesheet"
Now I better go read the manual about HTTP_HOST

Thanks people.

cheers

Jim