Print a PHP script's echo in another PHP script without include - php

Question

I have a file, let's call it first.php. It echoes a lot of text. Inside that text I also want to include something another file, second.php echoes. I'd like to do this without including the code from second.php, because there are collisions.
TL;DR I want a PHP script to execute another one, wait for it to print, and then echo whatever it printed.
Tried
echo something
include second script
echo something else
But it doesn't work because of said collisions. The second script only prints a short plain text.
Better example
a.php prints "hello" and "world" and declares $i=1
b.php prints "beautiful" and declares $i=2
I want to print "hello beautiful world" and have $i=1

Answer 1

You could run the second script through the CLI with shell_exec
echo something
shell_exec("php second.php");
echo something else
However, this won't work if the second script needs to access variables that are set by the webserver, like $_GET or $_SERVER. If you need to pass certain variables along, you can send them as command line arguments. But the second script will then have to access them using $argv, not $_GET.
shell_exec("php second.php " . escapeshellarg($_GET['id']));
second.php can get the ID from $argv[1].

Answer 2

Imo you should really solve the actual problem by not re-using variables which are sharing scope. If you cannot do that it means you are putting too much in the current scope. However if you really want to do it you could wrap it in a function (which has its own scope).
<?php
$i = 1;
echo 'hello';
echo call_user_func(function() {
// in your case this would be a include statement
$i = 2;
return 'beautiful';
});
echo 'world';
var_dump($i); // 1
If you actually need to "import" variables from the current scope create a closure instead:
<?php
$i = 1;
echo 'hello';
echo call_user_func(function() use ($theVariable) {
// in your case this would be a include statement
$i = 2;
return 'beautiful';
});
echo 'world';
var_dump($i); // 1

Answer 3

You could include the second file in a function. Functions have their own variable scopes. So, variables of the second file can not overwrite variables with the same name of the first script file.
$myvar = 1;
include 'first.php';
callSecond ();
echo $myvar;
function callSecond () {
include 'second.php';
}
echo $myvar would give 1, even when 'second.php' assigns $myvar = 2;