require/include into variable

Question

I want to require/include a file and retrieve its contents into a variable.

test.php

<?php
echo "seconds passed since 01-01-1970 00:00 GMT is ".time();
?>

index.php

<?php
$test=require("test.php");
echo "the content of test.php is:<hr>".$test;
?>

Like file_get_contents() but than it should still execute the PHP code. Is this possible?

Answer 1

If your included file returned a variable...

include.php

<?php
return 'abc';

...then you can assign it to a variable like so...

$abc = include 'include.php';

Otherwise, use output buffering.

ob_start();
include 'include.php';
$buffer = ob_get_clean();

Answer 2

I've also had this issue once, try something like

<?php
function requireToVar($file){
    ob_start();
    require($file);
    return ob_get_clean();
}
$test=requireToVar($test);
?>

Answer 3

You can write in the included file:

<?php
    return 'seconds etc.';

And in the file from which you are including:

<?php
    $text = include('file.php'); // just assigns value returned in file

Answer 4

In PHP/7 you can use a self-invoking anonymous function to accomplish simple encapsulation and prevent global scope from polluting with random global variables:

return (function () {
    // Local variables (not exported)
    $current_time = time();
    $reference_time = '01-01-1970 00:00';

    return "seconds passed since $reference_time GMT is $current_time";
})();

An alternative syntax for PHP/5.3+ would be:

return call_user_func(function(){
    // Local variables (not exported)
    $current_time = time();
    $reference_time = '01-01-1970 00:00';

    return "seconds passed since $reference_time GMT is $current_time";
});

You can then choose the variable name as usual:

$banner = require 'test.php';