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我试图将字符串转换为Zend Framework格式的URL。从字符串获取Zend Framework URLs
例如,我有一个字符串列表
http://example.com/products/category/books
http://example.com/products/category/computers
http://example.com/contact
我想收到的Zend_Controller_Request_Http对象,其中类似的控制器,动作,参数,可以等参数将被承认的名单。
Zend_Controller_Request_Http Object (... [_params:protected] => Array ([controller] => index [action] => products [category] => books [module] => default) ...)
Zend_Controller_Request_Http Object (... [_params:protected] => Array ([controller] => index [action] => products [category] => computers [module] => default) ...)
Zend_Controller_Request_Http Object (... [_params:protected] => Array ([controller] => index [action] => contact ...)
我找到了一些解决方案here(感谢Willy Barro)
$url = 'http://example.com/module/controller/action/param1/test';
$request = new Zend_Controller_Request_Http($url);
Zend_Controller_Front::getInstance()->getRouter()->route($request);
$request->getParams();
,它的第一个URL工作正常,但其余我收到相同的参数:
[controller]=>index, [action]=>products, [category]=>books
[controller]=>index, [action]=>products, [category]=>books
[controller]=>index, [action]=>contact, [category]=>books
外貌像所有我无法改变参数......
也许有另一种方法将字符串转换为zf URL。
预先感谢您!