"If" ternary with echo(). Parser show me an error

Question

I'm trying to identify my pages (for the menu) using if ternary operator.

$page_name = "workers";

($page_name != "home") ? echo "home";

So I suppose "home" should appear, but no. Parser show me an error:

Parse error: syntax error, unexpected '?'

What I'm doing wrong??

Answer 1

You ternary syntax is wrong, try this:

$page_name = "workers";
echo ($page_name != "home" ? "home" : null);