PHP - str_replace returns original string - php

Question

I'm building a simple JS terminal shell emulator which posts its commands via AJAX to PHP.
Please leave security aside, this is only for learning and demo purposes.
Now my problem is, str_replace() won't work as expected, in fact, it returns the unchanged input string. It should work like this:
The name of this host is $hostname --> Yes, this string contains a variable --> Replace $hostname with testserver --> return The name of this host is testserver
What am I doing wrong?
This is my respond script for echo and export:
<?
// get environment variables from JSON
$vars = json_decode(file_get_contents('environment.json'), true);
// get request params
$method = $_SERVER['REQUEST_METHOD'];
$action = $_POST['action'];
$data = $_POST['data'];
switch ($action) {
case 'echo':
$cmd = $data;
// if the string in question contains a variable, eg. "the time is $time"
if (strpos($cmd,'$')) {
$output = '';
// for each environment variable as variable => value
foreach ($vars as $var => $val) {
// replace every variable in the string with its value in the command
$output = str_replace($var,$val,$cmd);
}
echo $output;
} else {
// if it does not contain a variable, answer back the query string
// ("echo " gets stripped off in JS)
echo $cmd;
}
break;
case 'export':
// separate a variable declaration by delimiter "="
$cmd = explode('=',$data);
// add a $-sign to the first word which will be our new variable
$var = '$' . array_shift($cmd);
// grab our variable value from the array
$val = array_shift($cmd);
// now append everything to the $vars-array and save it to the JSON-file
$vars[$var] = $val;
file_put_contents("environment.json",json_encode($vars));
break;
}

Answer 1

Better using :
if (strpos($cmd,'$') !== false) {
Then, every single replace will take the "first" data as its input data. You should proceed like this :
$output = $cmd;
// for each environment variable as variable => value
foreach ($vars as $var => $val) {
// replace every variable in the string with its value in the command
$output = str_replace($var, $val, $output);
}