PHP preg_replace with variable value using found expression as key

Question

I have a string $src with two variable placeholders %%itemA%% and %%itemB%%. The idea is for the regular expression to match any term listed ITEMA|ITEMB|ITEMC, case insensitively, and replace the term with the value of the array item using the term as the array key.

$replacements = [
    'itemA' => 'replacedItemA',
    'itemB' => 'replacedItemB',
    'itemC' => 'replacedItemC'
];

$src = "A B C %%itemA%% D E F %%itemB%% G H I";

$src = preg_replace('/\%\%(ITEMA|ITEMB|ITEMC)%%/i', $replacements['\1'], $src);

echo $src;

In this example the expected result is for %%itemA%% to be replaced with $replacements['itemA']'s value and %%itemB%% to be replaced with $replacements['itemB']'s value

Expected output from the echo was

A B C replacedItemA D E F replacedItemB G H I

Actual output, oddly, simply replaced the found terms with nothing (note the double spaces where the variable was removed)

A B C D E F G H I

Why is the term's string not being used in the $replacements['key'] to use the value of the array variable?

Casimir et Hippolyte · Accepted Answer · 2016-02-10 14:53:00Z

3

Your approach is wrong. Since you are dealing with literal strings, you can avoid the regex and use a faster way:

$replacements = [
    '%%itemA%%' => 'replacedItemA',
    '%%itemB%%' => 'replacedItemB',
    '%%itemC%%' => 'replacedItemC'
];

$str = strtr($str, $replacements);

answered Feb 10, 2016 at 14:53

Casimir et Hippolyte

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4 Comments

Craig Over a year ago

Hi, my understanding is that the regex is more efficient, as there will be a lot of strings filtered through? I'll do some benchmarking to test if I can find a solution to make my approach work too

Casimir et Hippolyte Over a year ago

@CraigMcArthur: "my understanding is that the regex is more efficient": it's a totally false idea!

Wiktor Stribiżew Over a year ago

Unless there is a need to also modify the $replacements, and specify some additional context for %%...%% strings, this can be used, too. Regex allows some more things that mere strtr does not.

Wiktor Stribiżew Over a year ago

The $replacements should be also modified to include delimiters to use this approach. If it is not possible, it not usable.

Wiktor Stribiżew · Accepted Answer · 2016-02-10 15:03:43Z

0

You just need to replace the preg_replace with preg_replace_callback like this:

$src = preg_replace_callback('/%%(ITEM[ABC])%%/i', function ($m) use ($replacements) {
 return isset($replacements[$m[1]]) ? $replacements[$m[1]] : $m[0];
}
, $src);

See IDEONE demo

The issue is that \1 backreference can only be used inside the pattern to refer to the text captured with the first capturing group or inside a replacement pattern, but only as a literal part of it. The \1 is not automatically evaluated to be used as an argument.

I also contracted the pattern a bit, but I guess it is just a sample.

edited Feb 10, 2016 at 15:03

answered Feb 10, 2016 at 14:54

Wiktor Stribiżew

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