Replace variable with its value inside HTML without <?php ?> - php

Question

Here is the situation. I have 2 files
content.php
<?php $my_var = "this is a variable"; ?>
<h1> php{my_var} </h1>
index.php
<?php include "content.php" ?>
The result should be:
<h1>this is a variable</h1>
I know how to work with preg_replace_callback. But I don't know how can I change php{my_var} with the value of $my_var.
All the logic should happens inside the index.php.
Edit
index.php
function replace_pattern($match)
{
what should I write here
}
echo preg_replace_callback("/php\:\{(.*)\}/", "replace_pattern", $Content);
Edit 2
Variables are not declare in the global scope

Answer 1

Note the added question-mark in the regular expression to make it less greedy.
$my_var = 'Hello World!';
// Get all defined variables
$vars = get_defined_vars();
$callback = function($match) use ($vars)
{
$varname = $match[1];
if (isset($vars[$varname])) {
return $vars[$varname]; // or htmlspecialchars($vars[$varname]);
} else {
return $varname . ' (doesn\'t exists)';
}
};
echo preg_replace_callback("/php\:\{(.*?)\}/", $callback, $Content);
Demo: http://phpfiddle.org/main/code/ax15-bpyw

Answer 2

Try this:
$my_var = "this is a variable";
$string = '<h1> php{my_var} </h1>';
$pattern = '/php{(.*)}/i';
preg_match($pattern, $string, $match);
$varName = $match[1];
$newString= preg_replace($pattern, $$varName, $string);
echo $newString;
But, warning!
In this case, i assume, if the code is php{somethin}, then there should be a $something variable. I used dynamic variable name. If you want to use only $my_var then use like this:
$newString= preg_replace($pattern, $my_var, $string);