preg replace with multiple curly brackets

Question

I want to replace what is between the }string{ with }string%{

I'm adding the the % to edit a search query.

string example - {a26f089ed1530be8648027b493a21bfd44c5fb15d632afc8f1c539c7c3da88a4}out{a26f089ed1530be8648027b493a21bfd44c5fb15d632afc8f1c539c7c3da88a4}

The 2 values between the pair of curlies changes each time {rAndows2ring}

I've tried a lot of variations and closest I've got to is;

$where = preg_replace("/\}([^}]+)\{/", "}$1%{", $where);

But that returns an extra % at the first {.

%{212190298d692385253efb2a1062006ddf3ea008e3cdc2f8b8f11884ec863202}out%{212190298d692385253efb2a1062006ddf3ea008e3cdc2f8b8f11884ec863202}

thanks in advance!

[It does not](https://regex101.com/r/Rzggzl/1), it works as expected. Show the code. — Wiktor Stribiżew, Jun 18 '20 at 17:45
You may just add `{` to the negated character class, `\}([^{}]+)\{`, to make sure there are no `{` and `}` between `}` and `{`. — Wiktor Stribiżew, Jun 18 '20 at 17:57

score -1 · Answer 1 · answered Jun 18 '20 at 17:49

-1

This works:

<?php

$oldString = '{a26f089ed1530be8648027b493a21bfd44c5fb15d632afc8f1c539c7c3da88a4}out{a26f089ed1530be8648027b493a21bfd44c5fb15d632afc8f1c539c7c3da88a4}';
echo '<p>oldString: ';
var_dump($oldString);
echo '</p>';


$newString = preg_replace('#(\{[\d\w]+\})([\d\w]+)(\{[\d\w]+\})#', '$1$2%$3', $oldString);

echo '<p>newString: ';
var_dump($newString);
echo '</p>';

?>

(\{[\d\w]+\}) captures first string of digits and word characters enclosed with curly brackets.
([\d\w]+) captures 'out', or any other string of digits and word characters in between
(\{[\d\w]+\}) same as above

answered Jun 18 '20 at 17:49

Adam Winter

973
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1

OP code works well and the original regex is much cleaner and better than what you suggest. – Wiktor Stribiżew Jun 18 '20 at 17:56
It doesn't work at all if it puts another '%' where he doesn't want it. – Adam Winter Jun 19 '20 at 19:44

preg replace with multiple curly brackets

1 Answers1