PHP code generator uses wrong quotes for $substr in substitution

[ismael-miguel]

Code Generator Language

PHP

Code Snippet

Just use the $group syntax (to use a value from a named group) for the substitution value, in substitution name.

Example: https://regex101.com/r/Iqww6d/1
In this example, I use the named group a

The generated code will contain this:

<?php

// ...

$substr = "$a"; // <-- the wrong quotes

// ...

This will try to use string interpolation to show the contents of the variable $a as the replacement value.
This can cause fatal errors or surprising results.

Alternatively, escape the $ with \, like this: "\$a".

---

Additinally, using the syntax ${a} for the replacement value will be worse, since that is the deprecated in PHP 8.2.
And has the same problems as before.

[working-name]

Hi @ismael-miguel,

The replacement notation you're using is regex101-specific, as stated in the quick reference entry for it. PHP, as far as I know, does not support calling capture groups by name, only by numerical ID.