Php Zend框架创建用户友好的url
我想创建用户友好的URL,如下所示:Php Zend框架创建用户友好的url,php,zend-framework,Php,Zend Framework,我想创建用户友好的URL,如下所示: mysite.com/flat-sale/london/1-room/1 当URL的某些部分为参数时: flat-sale is post/list/type/1 london is city/123 1-room is rooms/1 and 1 id page/1 为此,我在database-url\u别名中创建了表。此表有三列: 援助,网址,地址 我在此表中插入了下一行: 1 post/list/type/1 flat-sale 2
mysite.com/flat-sale/london/1-room/1
当URL的某些部分为参数时:
flat-sale is post/list/type/1
london is city/123
1-room is rooms/1
and 1 id page/1
为此,我在database-url\u别名中创建了表。此表有三列:
援助,网址,地址
我在此表中插入了下一行:
1 post/list/type/1 flat-sale
2 city/123 lonodon
3 1-room rooms/1
我正在使用Controller_插件解析URL:
class My_Controller_Plugin_UrlAlias extends Zend_Controller_Plugin_Abstract {
public function routeStartup(Zend_Controller_Request_Abstract $request) {
$alias = substr($request->getRequestUri(), 1);
$pattern = "([^/]+)";
//this model for CRUD from tables url_alias
$resources = new Admin_Resource_Materialalias();
$match = array();
if (preg_match_all($pattern, $alias, $match)) {
$url = array();
foreach($match['0'] as $m) {
//this is page
if (preg_match("#^[\d]+$#", $m)) {
$url[] = "page/$m";
} else {
$url[] = $resources->getUrl($m);
}
}
$url = implode("/", $url);
//echo $url;
}
if (isset($url) && strlen($url)) {
$request->setRequestUri($url);
}
}
}
这个插件是完美的工作
但我还需要创建这样的url:mysite.com/flat-sale/london/1-room/1。
为此,我创建了新的View_Helper:
class My_View_Helper_Alias extends Zend_View_Helper_Url {
public function alias(array $urlOptions = array(), $name = null, $reset = false, $encode = true) {
$url = $this->url($urlOptions, $name, $reset, $encode);
$pattern = "#([^/]+)\/([^/]+)\/([-a-zA-Z0-9_/.]+)#";
$params_pattern = "#([^/]+\/[\d]+)#";
if (preg_match($pattern, $url, $match)) {
$resources = new Admin_Resource_Materialalias();
if (preg_match_all($params_pattern, $match[3],$params)) {
$p_alias = array();
foreach($params[0] as $p) {
//add controller, action and first params
if (empty($p_alias)) {
$p = "/".$match[1].'/'.$match[2]."/".$p;
}
//this is page
if (preg_match("#page\/([\d]+)#", $p, $page)) {
$p_alias[] = $page[1];
continue;
}
//this model for CRUD from tables url_alias
$part = $resources->getAlias($p);
$p_alias[] = strlen($part)?$part:$p;
}
$alias = implode("/",$p_alias);
}
}
$alias = strlen($alias)? $alias : $url;
return $alias;
}
}
这个视图助手也可以工作,但我认为它不是最佳的。任何人都可以评论这个代码,或者可能有相同的任务吗?多谢各位
问题解决了,谢谢KA_lin:
routes.flat_sale_city_rooms.route = /:type/:city/:rooms/:page
routes.flat_sale_city_rooms.defaults.module = main
routes.flat_sale_city_rooms.defaults.controller = post
routes.flat_sale_city_rooms.defaults.action = list
routes.flat_sale_city_rooms.reqs.type = [^/]+
routes.flat_sale_city_rooms.reqs.city = [^/]+
routes.flat_sale_city_rooms.reqs.rooms = [^/]+
routes.flat_sale_city_rooms.defaults.page = 1
routes.flat_sale_city_rooms.reqs.page = \d+
routes.flat_sale_city.route = /:type/:city/:page
routes.flat_sale_city.defaults.module = main
routes.flat_sale_city.defaults.controller = post
routes.flat_sale_city.defaults.action = list
routes.flat_sale_city.reqs.type = [^/]+
routes.flat_sale_city.reqs.city = [^/]+
routes.flat_sale_city.defaults.page = 1
routes.flat_sale_city.reqs.page = \d+
为什么不使用Zend Framework路由?看看这个例子:您可以对用户友好的URL使用路由 例如,看一看
尝试在Bootstrap.php中添加如下内容:
$route = new Zend_Controller_Router_Route (
'user/summary/:id/connection',
array('controller' => 'user',
'action' => 'get-summary',
'id' => FALSE,
)
);
$router->addRoute('equity', $route);
这将创建您可以调用的自定义url:
user/my-personal-url,
user/my-personal-url/connection
假设Zend 1被使用您为我的目的提供如下:
$route=new Zend\u Controller\u Router\u route(':type/:city/:rooms/:page',数组('Controller'=>'post','action'=>'list','city'=>false',rooms'=>false',page'=>:page))$路由器->添加路由('equity',$route)代码>并在控制器中获取此参数的别名?是的…关于友好url,我也有此问题:)