How do I remove only few url parameters in zend framework

Question

I am trying to develop an application using zend framework (v 1.11). Am totally new to this framework.

I have a URL like this.

http://xyz.local/client/feedback/index/username/abc/page/2

The above link points to client module, feedback controller's index action.

url parameter username along with a valid value is necessary for every request.

On this page I have paginated all the feedbacks. My probem is that when I navigate away from the index action to some other action in same or different controller the parameter name /page/2 still remains in the url.

For suppose if I navigate to new action of same controller then the new URL appears like this

http://xyz.local/client/feedback/new/username/newClient/page/2

whereas it should have been like this

http://xyz.local/client/feedback/new/username/newClient

Using the $this->url(array('module' => 'client', 'controller' => 'feedback', 'action' => 'new'), null, true) resets all the parameters including the username, which I do not want.

What's the standard Zend's way of doing this.

You could pass the username to the view and add it to the url's parameters: $this->url(array('module' => 'client', 'controller' => 'feedback', 'action' => 'new', 'username' => $username), null, true) or you just set the page-parameter to null and omit the final argument when calling the URL-ViewHelper. I think the former is the cleaner way... — dbrumann
– dbrumann, Commented Oct 3, 2012 at 7:30
cool, that worked (former approach).. thanks. :) Can you please post that reply as an answer to this question.? — Gaurav Sharma
– Gaurav Sharma, Commented Oct 3, 2012 at 7:50

dbrumann · Accepted Answer · 2012-10-03 08:28:18Z

5

You can pass the username to the view and add it to the url's parameters in the URL-ViewHelper:

$this->url(array('module' => 'client', 'controller' => 'feedback', 'action' => 'new', 'username' => $username), null, true)

answered Oct 3, 2012 at 8:28

dbrumann

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