Calling Functions

[Anon]

This is probably a very easy question, but being that I have little to no knowledge of PHP, I hope you can help me.

I have a function call name()
I want the function to go at the following
<?
if ($PIC=="2") {
echo"<img src=\"
http://domain.com/name()$PIC.jpg\">";
?>

This doesn't work, but I was wondering how it can be done.

Thanks.

[mattw]

Give this a try....

<?
if ($PIC=="2") {
echo "<img src=\"http://domain.com/" . name() . $PIC . ".jpg\">";
?>

Matt Wade
codewalkers.com

[Anon]

It doesn't seem to work. The name() output appears before the link.

Doe,_Doe<img src="http://domain.com/2.jpg">