【发布时间】:2016-09-28 11:50:42
【问题描述】:
我正在尝试将字符串转换为 Zend Framework 格式的 URL。
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
我想接收一个包含 Zend_Controller_Request_Http 对象的列表,其中将识别控制器、动作、参数等参数。
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();
它适用于第一个 url,但对于其余部分,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
看起来我根本无法更改参数...
也许还有其他方法可以将字符串转换为 zf URL。
提前谢谢你!
【问题讨论】:
标签: php url zend-framework