How do I replace the words of a text?

replied 10 years ago

$values = ['name' => 'Ricardo', 'drink' => 'coffee'];
$str = "Hello " . $values['name'] . ".. Do you like " . $values['drink'] . "?";

echo $str;

Output: "Hello Ricardo.. Do you like coffee?"

Your helpers function can do likewise as what that $str assignment is doing - learn about string concatenation in PHP.

Last updated 3 years ago.

0

aheinzm

replied 10 years ago

$str = 'Hello :name. Do you like :drink?';
echo str_replace([':name', ':drink'], ['Ricardo', 'Coffee'], $str);

Last updated 3 years ago.

0

ricardocanelas

replied 10 years ago

Solution

Good.. But it has to be dynamic .. So.. My solution is..

function vnsprintf( $format, array $data)
{
    preg_match_all( '/ (?<!%) % ( (?: [[:alpha:]_-][[:alnum:]_-]* | ([-+])? [0-9]+ (?(2) (?:\.[0-9]+)? | \.[0-9]+ ) ) ) \$ [-+]? \'? .? -? [0-9]* (\.[0-9]+)? \w/x', $format, $match, PREG_SET_ORDER | PREG_OFFSET_CAPTURE);
    $offset = 0;
    $keys = array_keys($data);
    foreach ( $match as &$value )
    {
        if ( ( $key = array_search( $value[1][0], $keys) ) !== FALSE || ( is_numeric( $value[1][0]) && ( $key = array_search( (int)$value[1][0], $keys) ) !== FALSE ) ) {
            $len = strlen( $value[1][0]);
            $format = substr_replace( $format, 1 + $key, $offset + $value[1][1], $len);
            $offset -= $len - strlen( $key);
        }
    }
    return vsprintf( $format, $data);
}


$str = "Hello %name\$s..Do you like %drink\$s?"; 
$values = array('name' => 'Ricardo', 'drink' => 'coffee');

echo vnsprintf( $str, $values);
// output: Hello Ricardo ..Do you like coffee?

Last updated 3 years ago.

0

aheinzm

replied 10 years ago

This should allow your replaced text in your string and the replacing text in your arrays to be in any order.

$replace = [':drink' => 'coffee', ':name' => 'Ricardo'];
$str = 'Hi my name is :name. I like to drink :drink.';

echo str_replace_dynamic($replace, $str);

function str_replace_dynamic(array $replace, $string) { 
   return str_replace(array_keys($replace), array_values($replace), $string);    
}

Last updated 3 years ago.

0

the94air

replied 7 years ago

@aheinzm Your answer is even more clean than the correct answer. Looks so nice.

Last updated 7 years ago.

0

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