如何使正则表达式变量捕获路由 - How to make regex variables to capture of routes

How to make regex variables to capture of routes

任何人都可以帮我解决这个问题。

example $uri = '/username/carlos'; => $routes[] = '/username/@name';

@name转换变量$name捕获字符串"carlos"

$routes[] = '/list/edit/@id:[0-9]{3}';
$routes[] = '/username/@name';
$routes[] = '/archive/*';
$routes[] = '/';

$uri = '/username/carlos';
foreach ( $routes as $pattern )
{
    if ( preg_match( '#^' . preg_replace( '#(?:{{)?@('w+'b)(?:}})?#i', '(?P<'1>['w'-'.!~'*''"(),'s]+)',
                            str_replace( ''*', '(.*)', preg_quote( $pattern, '/' ) ) ) . ''/?$#i', $uri, $matchs ) )
    {
        //how to make regex for this to work :

        echo $name; // carlos =>$uri = '/username/carlos'; or matt => $uri = '/username/matt';
    }
}

感谢阅读

你需要做更多的工作来让它工作，主要是将路由转换成正则表达式，然后你可以在给定的URI上使用。我的做法如下

$routes = $params = array();
$routes[] = '/list/edit/@id:[0-9]{3}';
$routes[] = '/username/@name';
$routes[] = '/archive/*';
$routes[] = '/';
$uri = '/username/carlos';
foreach($routes as $pattern) {
    // Convert all characters into safe characters
    $pattern = preg_quote($pattern, '~');
    // Convert * into .*?
    $pattern = preg_replace('/(?<!'''')'''''*/', '.*?', $pattern);
    #echo '<pre>'.print_r($pattern, true).'</pre>';
    // Convert any @name:expression into their expressions for capture
    $pattern = preg_replace_callback('%@([a-zA-Z]+)('''':([^/]+))?%', 'regex_callback', $pattern);
    $pattern = '~^' . $pattern . '$~';
    if(preg_match($pattern, $uri, $matches)) {
        $params = $matches;
        break;
    }
}
echo '<pre>'.print_r($params, true).'</pre>';
function regex_callback($data) {
    #echo '<pre>'.print_r($data, true).'</pre>';
    $pattern = '[^/]+';
    if(!empty($data[3])) {
        $pattern = stripslashes($data[3]);
    }
    return '(?P<' . $data[1] . '>' . $pattern . ')';
}

将*转换为.*?，再将@param转换为(?P<param>[^/]+)，或将:后的表达式替换为[^/]+

如果找到匹配项，则将匹配项设置为$params，并退出foreach循环。要查找名称，只需使用

echo $param['name'];