替换为相同的命名PHP变量

Question

My Example:

$name = "Simon";
$string = "My name is [name].";
echo preg_replace("/\[(.*)]/", ${"$1"}, $string);
// Expected: My name is Simon.
// I get: My name is .
// ${"$1"} should be $name?
exit();

When I do only:

echo preg_replace("/\[(.*)]/", "$1", $string);
// I get: My name is name.
// $1 = name

What am i doing wrong? Why is PHP not using the generated $name var? This is only a example. I would like to work this with any replace:

[foo] --> $foo
[bar] --> $bar
...

Answer 1

As per OP's wish to post my comment as an answer:

In order for this to work, you will need to change [name] to [$name] and ${"$1"} to "$1"

$name = "Simon";
$string = "My name is [$name].";

echo preg_replace("/\[(.*?)]/", "$1", $string);

PHP needs a variable to go on, so using [name] isn't being populated.

---

As per another and earlier comment I made, alternatively you could very well do:

$name = "Simon";
$string = "My name is " .$name;

echo $string;

If you have a framework with existing brackets, then that is something you haven't told us, only that you said in comments:
" have much more Placeholders, not only [name]", whether it's part of something bigger, then stick to the accepted method.

---

As per in comments, you could alternatively use "/\[([^\]]+)\]/" instead of "/\[(.*)]/"
Or "/[(.*?)]/" :)

in regards to using for example: "[$foo]bar[$foo]"