Evaluate a string as PHP code without using eval

Question

Seeing how eval should be avoided, how do you evaluate a string as PHP code without using eval? For example consider this code:

<?php

$string = 'cup';
$name = 'coffee';
$str = 'This is a $string with my $name in it.'; 

$str = "\"$str\""; // Now I have a string with double quotes around it. 

// how to get the contents of $str evaluated without using eval()

?>

I can get the desired result by using eval like so - eval("echo $str;");, but eval is exactly what I am looking to avoid.

You might view this as a question of removing the double quotes. But it isn't about that. As @AmalMurali points out, what I am asking here is how to get the contents of $str evaluated without using eval()

possible duplicate of [Removing double quotes](http://stackoverflow.com/questions/10262951/removing-double-quotes) — Mihai Iorga, Nov 24 '13 at 07:34
@MihaiIorga: How is this question a duplicate of that? This one asks how to get the contents of `$str` evaluated without using `eval()`. — Amal Murali, Nov 24 '13 at 07:44
@MihaiIorga I have edited my question to better reflect my exact query. I dont feel its a duplicate anymore. — user1720897, Nov 24 '13 at 08:07

Giacomo1968 · Answer 1 · 2013-11-24T15:09:17.190

2

You can layout your code like this:

$string = 'cup';
$name = 'coffee';
$str = 'This is a ' . $string . ' with my ' . $name . ' in it.';

Or perhaps like this, using sprintf which is my personal favorite method of handling cases like this:

$string = 'cup';
$name = 'coffee';
$str = sprintf('This is a %s with my %s in it.', $string, $name);

There are different options based on personal style & preference.

edited Nov 24 '13 at 15:09

answered Nov 24 '13 at 07:46

Giacomo1968

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score 1 · Answer 2 · answered Nov 24 '13 at 07:42

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What you really want here is to use double quotes for $str so that variable replacement can take place.

See also the docs: http://www.php.net/manual/it/language.types.string.php#language.types.string.syntax.double

answered Nov 24 '13 at 07:42

Matteo Tassinari

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score 1 · Answer 3 · answered Nov 24 '13 at 07:42

1

Why are you adding double quotes then removing them? For a simple string variable inclusion you just use double quotes like

$string = 'cup';
$name = 'coffee';
$str = "This is a $string with my $name in it."; 

echo $str;

answered Nov 24 '13 at 07:42

WebChemist

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I have edited my question to better reflect my needs. – user1720897 Nov 24 '13 at 08:19

Evaluate a string as PHP code without using eval

3 Answers3