I need help with PHP include() echo and variables?

Question

I am having trouble taking a variable from my include and echoing it.
It's a nav bar that determines what css class to use based on the file name.

template.php:
$nav_bar =
"

>Home

>Forum

";
?>
index.php:
include("template.php");
$page = basename($_SERVER['SCRIPT_NAME']);
echo "$nav_bar";
?>
The problem is that when I echo "$nav_bar" it takes it literally and doesn't execute any of the php. How can i make it execute? Using it like this give me the undefined variable error ($page).
How can i make this work?
-Thank you!
-Aus

Answer 1

template.php:


$nav_bar =

"

Home

Forum

";

?>

index.php:


$page = basename($_SERVER['SCRIPT_NAME']);

include("template.php");

echo "$nav_bar";

?>



I used the ternary operator instead of an if...else to shorten the code.



Also, in index.php, you need to define $page before including template.php, because you use it in template.php.



PHP doesn't give undefined variable errors, because the first time you use it, if it is not already defined, PHP creates it for you. (It is null at that point.)

Answer 2

1) $nav_bar =

"

>Home

is all part of $nav_bar. You have to close the quote.



$nav_bar =

"

Home

etc.



2) echo "$nav_bar" will echo the contents of $nav_bar (echo '$nav_bar' will echo the string $nav_bar) but it's a waste of processing power, since you aren't echoing any literals. echo $nav_bar will give you the same output without making the PHP parser figure out ewhat's inside the quotes.



For a small site that gets a few hits, it makes very little difference. On a very busy site a few dozen uses of unneeded quotes could make the difference between the site working and the site crashing.

Answer 3

The php code is part of the string literal and will therefor be treated as such.



Try this:


$nav_bar = '

$nav_bar .= '">Home

$nav_bar .= '">Forum

';

?>

Answer 4

Well, I put it on my server and it executed properly, however, I believe that you do not need to put an opening php tag in your variable. i.e. "

>" should not have because you already are using php to echo the variable.

Answer 5

Simple answer!



When you echo a variable, don't use double quotes. Saying echo "$nav_bar" will give you output of "$nav_bar"



Change that line to:



echo $nav_bar;



and it should work!

Answer 6

i had this problem and i found a very simple fix after doing hours of research. so i went to command prompt, stood up infront of my computer and started dancing infront of it then it worked