PHP preg_replace for hex

Question

I'm trying to use preg_replace() to remove all hexadecimal characters in a string (with only lower case letters):

$line = "sjdivfriyaaqa\xd2vkmpcuyyuen";
$line = preg_replace('/\\x[0-9a-f]{2}/', '', $line);
echo($line);

This should, as I understand it, echo sjdivfriyaaqavkmpcuyyuen ($line with \xd2 removed) but it echoes the original $line. Why?

Antonio Smoljan · Accepted Answer · 2015-12-08 12:35:46Z

1

This will work

$line = "sjdivfriyaaqa\\xd2vkmpcuyyuen";
$new = preg_replace('/\\\x[0-9a-f]{2}/', '', $line);
echo($new);

You need another blackslash to escape it in the string, if you try echoing it right away you will see that it's empty.

answered Dec 8, 2015 at 12:30

Antonio Smoljan

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Comments

Patrick Portal · Accepted Answer · 2015-12-08 12:33:03Z

1

In fact the double quotes "" interpretes escaped char \ meaning "\xd2" is 'Ò' for instance.

Using simple quote '' your code is ok:

$line = 'sjdivfriyaaqa\xd2vkmpcuyyuen';
$line = preg_replace('#\\\x[0-9a-f]{2}#', '', $line);
echo($line);

answered Dec 8, 2015 at 12:33

Patrick Portal

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btw why three backslashes ? \\\ => two \\ results to the real \ char, the third means the char is escaped, four \ will be ok too.