Keeping variables in an echo statement

[Anon]

Hi everyone.
This may seem like and odd thing I am trying to accomplish but trust me I am not crazy.

My problem is that this:

echo "<? include($includeDir$footer_filename); ?>";

Prints out in the document source when loaded in a browser as this:

<? include("/path/to/includesDir/footer.php"); ?>

But what I want it to do is keep the $varibales as $variables when it outputs. So I want it to show up in the source of the document like this:

echo "<? include($includeDir$footer_filename); ?>";

so 'as is'
thanks

[Anon]

Let me get this straight... you basically want to just dump the PHP code directly to the output?? If this is what you're looking to do, try using readfile() instead of include().

-Andy

[Anon]

Heh, ok..

I think you could do this..

print(" \$variable "); That should tell the dollar sign to be a dollar sign. Or you could also do this:

<?php
#your PHP code
?>

<!--escape back to HTML -->
include($includeDir$footer_filename);

<?php
#more php code
?>

[Anon]

Yes! I forgot about that... I believe it is actually $$variable. where the double dollar sign tells it to print a dollar sign.

[Anon]

No - $$variable is interpreted as the variable whose name is the value of $variable. For instance:

<?php
$foo = 'bar';
$variable = 'foo';
print $$variable; //prints 'bar'
?>

Use the \ to escape the dollar sign. Or, use single quotes if you don't want any of the variables to be interpreted as variables:

<?php
echo '<? include ($includeDir$footer_filename); ?>';
?>

-Keegan