Preg_replace regex doesn't display anything

Question

    $str = '<div class="hello">{author}</div><div id="post_{id}"><div class="content">{content}</div></div>';
    $pattern = '/\{([a-z _0-9]+)\}/i';
    $subst= array('author'=>'Mr.Google','id'=>'1239124587123','content'=>'This is some simple text');
    $str = preg_replace($pattern,$subst['$1'],$str);
    echo $str;

Each instance {text}turns out to be " ", is there something I am doing wrong with the capture group? I did this with help preg_match_all, and it comes back like {author}that author. Is this the problem here?

Answer 1

In this case, you need preg_replace_callback :

$str = preg_replace_callback($pattern, function ($matches) use ($subst) {
    return $subst[$matches[1]];
}, $str);

Another solution uses strtr :

// $subst needs to be changed a bit.
$subst= array('{author}'=>'Mr.Google','{id}'=>'1239124587123','{content}'=>'This is some simple text');
echo strtr($str, $subst);

Answer 2

The second parameter preg_match_allis evaluated before the function starts. (Like all options, there are always.)

error_reporting, PHP : : Undefined index: $1 [...]

, $1 , , - preg_replace_callback:

$str = preg_replace_callback(
    $pattern,
    function($matches) use ($subst) {
        return $subst[$matches[1]];
    },
    $str
);

: , PHP < 5.3.0:

$str = preg_replace_callback($pattern, 'my_custom_replacement_function', $str);

function my_custom_replacement_function($matches) {
    global $subst;
    return $subst[$matches[1]];
}