Correct regex to detect font family names from google font link src

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I’ve been trying to get array of the fonts that I’m enqeueing on my wordpress theme. This is just for testing.

On input:

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http://fonts.googleapis.com/css?family=Arimo:400,700|Quicksand:400,700|Cantarell:400,700,400italic,700italic|Muli:300,400,300italic,400italic|Roboto+Slab:400,700|Share:400,700,400italic,700italic|Inconsolata:400,700|Karla:400,700,400italic,700italic|Maven+Pro:400,500,700,900|Roboto+Slab:400,700|Open+Sans:400italic,600italic,700italic,400,600,700

What I need on output is like this: 

array(
[0] => 'Arimo',
[1] => 'Quicksand',
[2] => 'Cantarell',
... so on
)

Till now, I have done almost everything but one little problem.

My code: 

$input = 'http://fonts.googleapis.com/css?family=Arimo:400,700|Quicksand:400,700|Cantarell:400,700,400italic,700italic|Muli:300,400,300italic,400italic|Roboto+Slab:400,700|Share:400,700,400italic,700italic|Inconsolata:400,700|Karla:400,700,400italic,700italic|Maven+Pro:400,500,700,900|Roboto+Slab:400,700|Open+Sans:400italic,600italic,700italic,400,600,700';

$against = "/[A-Z][a-z]+[+][A-Z][a-z]+|[A-Z][a-z]+/";

$matches = array()

preg_match_all( $against, $input, $matches );

print_r($matches);

From this, the output is like this: 

array(
0   =>  Arimo
1   =>  Quicksand
2   =>  Cantarell
3   =>  Muli
4   =>  Roboto+Slab
5   =>  Share
6   =>  Inconsolata
7   =>  Karla
8   =>  Maven+Pro
9   =>  Roboto+Slab
10  =>  Open+Sans
)

There’s the + sign where the font name has spaces. I want to get rid of that.

I’m not a regex expert. So, couldn’t manage to do that.

Note: I know I could do it with str_replace() but don’t want to go through that long process. I want to know if it’s possible to escape the + sign through and leave an empty space there when we are collecting matched expressions.

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4 comments

  2. Without regex:

    $query = strtr(substr(parse_url($url, PHP_URL_QUERY),7), '+', ' ');

$result = array_map(function ($i) { return explode(':', $i)[0]; }, explode('|', $query));

    With regex:

    if (preg_match_all('~(?:G(?!A)|[^?&]+[?&]family=)([^:|&]+):[^:|&]*(?:[|&#]|z)~', strtr($url, '+', ' '), $m))
   $result2 = $m[1];

  3. From your code, output is given me something like this.

    array([0] => array([0]   =>  Arimo[1]   =>  Quicksand[2]   =>  Cantarell[3]   =>  Muli[4]   =>  Roboto+Slab[5]   =>  Share[6]   =>  Inconsolata[7]   =>  Karla[8]   =>  Maven+Pro[9]   =>  Roboto+Slab[10]  =>  Open+Sans))

    if is correct, then i was solve this issue ‘+’. here is the solution.

    $input = 'http://fonts.googleapis.com/css?family=Arimo:400,700|Quicksand:400,700|Cantarell:400,700,400italic,700italic|Muli:300,400,300italic,400italic|Roboto+Slab:400,700|Share:400,700,400italic,700italic|Inconsolata:400,700|Karla:400,700,400italic,700italic|Maven+Pro:400,500,700,900|Roboto+Slab:400,700|Open+Sans:400italic,600italic,700italic,400,600,700';

$against = "/[A-Z][a-z]+[+][A-Z][a-z]+|[A-Z][a-z]+/";

$matches = array();
$newArr=array();
preg_match_all( $against, $input, $matches );

for($i=0;$i< count($matches);$i++){
    for($j=0;$j< count($matches[$i]);$j++){
        $string=preg_replace('/[^A-Za-z0-9-]/', ' ', $matches[$j]);
        if($string!=""){
            $newArr[]=$string;
        }
    }    
}
print_r($newArr);

  4. In general, you have more than + characters to worry about.

    Special characters, such as the ampersand (&), and non-ASCII characters in URL query parameters have to be escaped using percent-encoding (%xx). In addition, when an HTML form is submitted, spaces are encoded using the + character.

    For example:

    • The font family “Jacques & Gilles” would be escaped as:

      Jacques+%26+Gilles

    • The Unicode character U+1E99 (LATIN SMALL LETTER Y WITH RING ABOVE), serialized into octets as UTF-8 (E1 BA 99), would be escaped as:

      %e1%ba%99

    To do what you want properly, you have to extract the query string from the URL, and use parse_str() to extract the name=value pairs. The parse_str() function will automatically urldecode() the names and values including the + characters.

    First, split the URL on the ? character to extract the query string:

    $url = 'http://fonts.googleapis.com/css?family=Arimo:400,700|...|Maven+Pro:400,500,700,900|Roboto+Slab:400,700|...';

$a = explode ('?', $url, 2);
if (isset ($a[1])) {
  $query = $a[1];
}

    You can also use parse_url ($url, PHP_URL_QUERY), but it doesn’t buy you much in this case.

    Then extract all the parameters:

    if (isset ($query)) {
  parse_str ($query, $params);

  if (isset ($params['family'])) {
    /* OK: Extract family names. */
  } else {
    /* Error: No family parameter found. */
  }
} else {
  /* Error: No query string found. */
}

    Note: You should always specify the second parameter of parse_str() to avoid clobbering existing variables.