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PHP:MVC路由函数解析请求视图的好解决方案?_Php_Model View Controller_Design Patterns - Fatal编程技术网

PHP:MVC路由函数解析请求视图的好解决方案?

PHP:MVC路由函数解析请求视图的好解决方案?,php,model-view-controller,design-patterns,Php,Model View Controller,Design Patterns,可能重复: 我想实现MVC设计结构,目前正在努力寻找一个好的解决方案来解析请求的视图 在我的路由文件中,我有以下代码: public function parseRequestedView() { $this->ressource_requested = explode('/', trim($_GET['view'], '/')); // e.g.: http://www.foo.com/article/{id}/comments/show if (!empty($

可能重复:

我想实现MVC设计结构,目前正在努力寻找一个好的解决方案来解析请求的视图

在我的路由文件中,我有以下代码:

public function parseRequestedView() {

   $this->ressource_requested = explode('/', trim($_GET['view'], '/'));

   // e.g.: http://www.foo.com/article/{id}/comments/show
   if (!empty($this->ressource_requested[3])) {

      // Format: [0] viewpoint (article), [1] child (comments), [2] action (show), [3] reference ({id}),
      //         [4] additional information (from $_POST)
      return array($this->ressource_requested[0], $this->ressource_requested[2], $this->ressource_requested[3],
                   $this->ressource_requested[1], $_POST);

   // e.g.: http://www.foo.com/article/{id}/show
   } elseif (!empty($this->ressource_requested[2])) {

      return array($this->ressource_requested[0], NULL, $this->ressource_requested[2], $this->ressource_requested[1],
                   $_POST);

   // e.g.: http://www.foo.com/archive/show
   } else {

      return array($this->ressource_requested[0], NULL, $this->ressource_requested[1], NULL, NULL);

   }

}
其思想是,无论访问者在浏览器中键入什么,函数都会解析请求并始终返回相同的格式化数组/输出。主机名后面的URL的第一段始终是主视点,例如:article。最后,我通过另一个名为includeTemplateFile的函数包含视图。这些文件具有以下命名约定:

viewpoint.child.action.template.php
e.g.: article.comments.show.template.php
我现在的问题是:有没有更优雅的解决方案?我读了一些关于这个主题的文章,例如:但我不喜欢大多数解决方案,因为它们设计得不好

以下是.htaccess文件的内容:

RewriteEngine on

RewriteBase /
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d

RewriteRule ^(.*)$ index.php?view=$1 [L,QSA]

提前感谢。

好的,我首先要说我不是PHP专家,我建议使用symphony之类的框架来完成路由,但这里有一个可能的解决方案,您可以使用

function regexPath($path)
{
    return '#' . str_replace([":int:", ":string:"], ["\d+", ".+"], $path) . '#';
}

function parseRequestedView($url)
{
    $ressource_requested = explode('/', trim($url, '/'));


    // define our routes, and the indices that each route will use (from the exploded url)
    // this could be defined as another parameter or as a member of the class
    $routes = [
        regexPath("article/:int:/comments/show") => [0,  2, 3,  1], // will return array(resource[0], resource[2], resource[3], resource[1]), etc
        regexPath("article/:int:/show")          => [0, -1, 2,  1], // -1 will return a null
        regexPath("archive/show")                => [0, -1, 1, -1]
    ];


    // go through each route, checking to see if we have a match
    foreach ($routes as $regex => $indices)
    {
        if (preg_match($regex, $url))
        {
            // it matched, so go over the index's provided and put that data into our route array to be returned
            foreach ($indices as $index)
            {
                $route[] = $index > -1 ? $ressource_requested[$index] : null;
            }

            // include the post data (not really nessesary)
            $route[] = $_POST; // unnessesary to pass $_POST data through function, because it is global

            return $route;
        }
    }

    return null; // or some default route maybe?
}

$route = parseRequestedView("article/13/comments/show");

echo '<pre>';
print_r($route);
echo '</pre>';

/* returns:
Array
(
    [0] => article
    [1] => comments
    [2] => show
    [3] => 13
    [4] => Array // this is our $_POST data
        (
        )

)
*/

首先,非常感谢你的回答。但让我指出,我的问题不是重复的,因为我在问我的解决方案设计得好还是有更好的。我对路由库没有任何问题。还是我看错了张贴的问题?非常感谢。你的解决方案比我的好得多。遗憾的是,我不能使用任何第三方框架规则。还有一件事:您能解释一下str_中的[:int:,:string:]和[\d+,.+]替换[:int:,:string:,[\d+,.+],$path到底是做什么的吗?非常感谢。这基本上只是为了方便。它将可读性更强的字符串转换为正则表达式模式。网上有很多关于它的好资源。