variables in strings: where's the magic?

$items = 5;
echo "you have $itemss items";

in a similar situation for me, that prints
"you have items"

where as something like this even
$items = 5;
echo "12k5j;12lk5j2050298$items)sjf0laskdfj;kj";

will find the variable $items and show it just fine.

why doesn't it work when i put an 's' at the end? is this a parser bug on the last letter or what?

You havn't set a var $itemss so why should it show it?

Nick

It's not a parser bug
you need somekind of non alpha delimiter so the parser knows what variable you want.

say

$foo = "hello";
$foobar = "bye";

echo "greeting: $foobar";
you are sayint that the program should print
greeting: hellobar

thats why you need somekind of delimiter like a space or a ')' in your second example if you really want to avoid this just do this

echo "you have " . $items . "s items";

that is how i do it, but i was just curious because i didn't once and it didn't work.

echo "you have " . $items . "s items";

No need to parse a literal string, this would be better...

echo 'you have ' . $items . 's items';