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How do I replace characters not in range [0x5E10, 0x7F35] with '*' in PHP?

Tags:

regex

php

hex

I'm not familiar with the how regular expressions treat hexadecimal, anyone knows?

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openid Avatar asked Apr 28 '10 09:04

openid


1 Answers

The following does the trick:

$str = "some മനുഷ്യന്റെ";

echo preg_replace('/[\x{00ff}-\x{ffff}]/u', '*', $str);
// some **********

echo preg_replace('/[^\x{00ff}-\x{ffff}]/u', '*', $str);
// *****മനുഷ്യന്റെ

The important thing is the u-modifier (see here):

This modifier turns on additional functionality of PCRE that is incompatible with Perl. Pattern strings are treated as UTF-8. This modifier is available from PHP 4.1.0 or greater on Unix and from PHP 4.2.3 on win32. UTF-8 validity of the pattern is checked since PHP 4.3.5.

And here a short description why \uFFFF is not working in PHP:

Perl and PCRE do not support the \uFFFF syntax. They use \x{FFFF} instead. You can omit leading zeros in the hexadecimal number between the curly braces. Since \x by itself is not a valid regex token, \x{1234} can never be confused to match \x 1234 times. It always matches the Unicode code point U+1234. \x{1234}{5678} will try to match code point U+1234 exactly 5678 times.

like image 141
Stefan Gehrig Avatar answered Nov 09 '22 17:11

Stefan Gehrig