Now, if it's just for practice, then you need a little change in you
code, since strtoupper('\\2') will give '\\2' (i.e. does nothing) and
the letter will remain the same.
I meant, since *strtoupper('\\2')* is evaluated _before_ being included
in the string (used as replacement), i.e...
preg_replace('/(^)(.)(.*$)/', strtoupper('\\2') . '\\3', 'yikes!');
is evaluated as
preg_replace('/(^)(.)(.*$)/', '\\2\\3', 'yikes!');
since *strtoupper('\\2')* => *'\\2'*, that's why I said it does nothing.
In the code sent, *strtoupper()* is passed as a (literal) string which,
combined with the "e" modifier (at the end of the expression), gives the
effect you were looking for (i.e. the "replacement string" is evaluated
--as code-- before it actually replaces the string found)
--
Atentamente / Sincerely,
J. Rafael Salazar Magaña
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