How can I use a variable variable in PHP?

Question

Here is what I am trying to do:

<?php
    $my_str = "My String";
    $str = "%my_str";

    $str = str_replace("%", "$", $str);

    echo $str;
?>

The above code prints '$my_str' to the screen. But I want it to print 'My String', as in the actual value of the variable $my_str

Anyone know how to do this?

The reason I want this is because I'm in the process of writing my own, very basic, parsing language so I kinda need this to be functional before I can continue.

Answer 1

  1 <?php
  2     $my_str = "My String";
  3     $str = "%my_str";
  4
  5     $str = str_replace("%", "$", $str);
  6
  7         eval("echo $str;");
  8 ?>
~

Answer 2

$my_str = 'My String';
$str    = 'my_str';

echo $$str;

This construction is called a variable variable.

Answer 3

eval is not necessary. Just get rid of the % and use $$str

<?php
    $my_str = "My String";
    $str = "%my_str";

    $str = str_replace("%", "", $str);

    echo $$str;
?>

Answer 4

The behaviour you see is not surprising - expansion of variables ("interpolation") is performed on only string literals, rather than variables. If it wasn't, then it would be a large security hole (any instance of a string with $ in it that your code used would suddenly be revealing variables).

You can try and fix this using eval, as in

eval ("\$str = \"" . str_replace("%", "\$", $str) . "\"");

but this pretty dangerous - if your string is from the user, then can make $str something like

"; system("rm -rf /"); $x = "

and suddenly, you're in trouble. The best solution, I believe, will be to parse out variables using your favourite methods (stripos and substring, or something more), then replace each one by hand.

Answer 5

Try:

echo $$str;

Answer 6

You could search and replace the %var patterns using preg_replace and the e modifier, which makes the replacement being evaluated:

<?php
 $my_str = "My String";
 $str = "Foo %my_str bar";

 $str = preg_replace("/%([A-Za-z_]+)/e", "$\\1", $str);

 echo $str;
?>

Here preg_replace will find %my_str, \1 contains my_str and "$\\1" (the backslash needs to be escaped) becomes the value of $my_str.

However, it would maybe be cleaner to store your replacement strings in an associative array:

<?php
 $replacements = array(
  "my_str" => "My String",
 );
 $str = "Foo %my_str bar";

 $str = preg_replace("/%([A-Za-z_]+)/e", '$replacements["\\1"]', $str);

 echo $str;
?>

Answer 7

PHP automatically parses and expands variables within double-quoted strings:

<?php
    $my_str = "My String";
    $str = "{$my_str}";

    echo $str;
?>

Outputs:

My String

Answer 8

How about using a regex with the e modifier

$str = preg_replace("/%([a-zA-Z0-9_]+)/e", '$\1', $str);

(This is untested, but should work)