PHP preg_replace limit not working?

Question

I have a file that contains several colors as hexadecimals (i.e. #000 or #ffffff) and I'm trying to replace each of them with @varN where N is a number that increments with each replacement done. I believed my code does that, but $count always returns 196 after doing the placement even though I put the limit to 1 so count should never goes past 1. This results an endless loop. Why isn't it the limit working and what I can do to get the desired output?

$list = file($filepath.$file, FILE_USE_INCLUDE_PATH);
$pattern = "/#((([a-f]|[A-F]|[0-9]){6})|(([a-f]|[A-F]|[0-9]){3}))/";
$replaceVar = "@var";
$replaceNum = 0;
$count = 1;
while($count != 0){ //never ends
    $replacement = $replaceVar.$replaceNum;
    $output = preg_replace($pattern,$replacement,$list,1,$count);
    $replaceNum++;
    echo $replaceNum." ".$count."\n"; //returns $replaceNum and 196
}
file_put_contents($filepath.'/new'.$file,$output,FILE_USE_INCLUDE_PATH);

Example input file:

#000000 #111111 #123456 #abcdef #123 #abc

Example output file:

@var1 @var2 @var3 @var4 @var5 @var6

file put all contents to an array, you'd better use file_get_contents, here your $count wouldn't decrease so it's normal to have an infinite loop and what is $count? — revo
– revo, Commented Apr 29, 2014 at 19:00
file_get_contents gives me a string though, meaning I would have to some how turn the string back into a file with all the newlines in the right spot. I understand that $count is the number of times a replacement has been done to my $list. And with $limit = 1, $count should never go above 1. — user2361174
– user2361174, Commented Apr 29, 2014 at 19:15

Casimir et Hippolyte · Accepted Answer · 2014-04-29 20:59:10Z

1

You don't use the good function to do that for several reasons, I suggest you to use preg_replace_callback instead. Example:

$source = file_get_contents($yourfile);

$pattern = '~#[[:xdigit:]]{3}(?:[[:xdigit:]]{3})?~';
$count = 0;

$output = preg_replace_callback($pattern,
                       function ($m) use (&$count) { return '@var' . ++$count; },
                                $source);

edited Apr 29, 2014 at 20:59

answered Apr 29, 2014 at 19:27

Casimir et Hippolyte

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3 Comments

user2361174 Over a year ago

It works, however your regex gave me some unexpected results.

Casimir et Hippolyte Over a year ago

@user2361174: What kind of unexpected results?

Casimir et Hippolyte Over a year ago

@user2361174: Sorry my mistake! It is not [:alnum:] but [:xdigit:] for hex numbers!

dognose · Accepted Answer · 2014-04-29 19:43:35Z

1

You are always replacing on $list, but writing the result of the replacement to $output:

$output = preg_replace($pattern,$replacement,$list,1,$count);

This means, your $list will ALWAYS contain 1 match, when it starts with 1 - its never modified. Therefore your while will run without an end. You have to REPLACE it inside the same string you are scanning for more hits:

$list = preg_replace($pattern,$replacement,$list,1,$count);

now - at one point - $count should become zero, and $list contain no more unwanted matches.

edited Apr 29, 2014 at 19:43

answered Apr 29, 2014 at 19:35

dognose

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1 Comment

user2361174 Over a year ago

The loop stopped, but it would always used @var0 as the replacement.

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