php回显函数内部的变量

Question

I have a php function which displays a rating bar with the arguments. I have a variable called itemID inside my php page which holds the unique item number. I need to send this value to my function and also echo command must stay. Is there a way to achieve this?

Here is the code, which does not work. When I try it on the server, it does not show the id of item, it prints the variable name as it is.

<?php echo rating_bar('$as',5) ?>

What I get at html file:

<div id="unit_long$as"> 

instead of the item id in place of $as.

Answer 1

Single Quotes do not support variable replace,

$as = "test";
echo '$as'; //$as in your end result
echo "$as"; // test in your end result

echo $as; // test in your end result

//For proper use
echo " ".$as." "; // test in your end result

Update for newer PHP versions you should not use Template Syntax

echo "{$as}"