PHP preg_replace - finding the replacement from an array using the match as the key

Question

I have a string which can contain multiple matches (any word surrounded by percentage marks) and an array of replacements - they key of each replacement being the match of the regex. Some code will probably explain that better...

$str = "PHP %foo% my %bar% in!";
$rep = array(
  'foo' => 'does',
  'bar' => 'head'
);

The desired result being:

$str = "PHP does my head in!"

I have tried the following, none of which work:

$res = preg_replace('/\%([a-z_]+)\%/', $rep[$1], $str);
$res = preg_replace('/\%([a-z_]+)\%/', $rep['$1'], $str);
$res = preg_replace('/\%([a-z_]+)\%/', $rep[\1], $str);
$res = preg_replace('/\%([a-z_]+)\%/', $rep['\1'], $str);

Thus I turn to Stack Overflow for help. Any takers?

Artefacto · Accepted Answer · 2010-07-13 12:48:53Z

7

echo preg_replace('/%([a-z_]+)%/e', '$rep["$1"]', $str);

gives:

PHP does my head in!

See the docs for the modifier 'e'.

answered Jul 13, 2010 at 12:48

Artefacto

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2 Comments

user9645 Over a year ago

The 'e' modifier is "deprecated and use is highly discouraged" since PHP 5.5. according to the docs cited in this answer. An alternative solution using preg_replace_callback() is presented here

Jpsy Over a year ago

And the 'e' modifier is REMOVED as of PHP 7.0.0.

Claudio · Accepted Answer · 2013-09-19 09:25:34Z

3

It seems that the modifier "e" is deprecated. There are security issues. Alternatively, you can use the preg_replace_callback.

$res = preg_replace_callback('/\%([a-z_]+)\%/', 
                             function($match) use ($rep) { return  $rep[$match[1]]; },
                             $str );

answered Sep 19, 2013 at 9:25

Claudio

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Comments

Rob · Accepted Answer · 2010-07-13 12:51:11Z

2

You could use the eval modifier...

$res = preg_replace('/\%([a-z_]+)\%/e', "\$rep['$1']", $str);

answered Jul 13, 2010 at 12:51

Rob

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Comments

Mark Baker · Accepted Answer · 2010-07-13 12:55:56Z

1

Just to provide an alternative to preg_replace():

$str = "PHP %foo% my %bar% in!";
$rep = array(
  'foo' => 'does',
  'bar' => 'head'
);


function maskit($val) {
    return '%'.$val.'%';
}

$result = str_replace(array_map('maskit',array_keys($rep)),array_values($rep),$str);
echo $result;

answered Jul 13, 2010 at 12:55

Mark Baker

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