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Home/ Questions/Q 9062921
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T15:51:28+00:00 2026-06-16T15:51:28+00:00

Looking for the most efficient way to do the following: Details: 1 controller: Reports

  • 0

Looking for the most efficient way to do the following:

Details:

1 controller:

  • Reports

2 views:

  1. create_report
  2. view_report

create_report has a form where information is collected that will be used in the view_report view.
I know the data entered in the form is available within the create_report view via the $this->session->data element.

Looking for Help on This:

What I need to do is send that same info ($this->session->data) from the create_report view to the view_report view (through the controller, I assume).

What I have so far:

Just a link:

echo $this->Html->link('View Report', array('controller' => 'reports', 'action' => 'view_report'));

But this only takes the user to the view_report view. It doesn’t send the info in $this->session->data to that view.

I am thinking it has something to do with the Js helper, but I wasn’t able to find any posts dealing directly with this situation for CakePHP 2.x.

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  1. Editorial Team
    Editorial Team
    2026-06-16T15:51:30+00:00Added an answer on June 16, 2026 at 3:51 pm

    You could bypass data processing by controller action create_report and send them directly to report view and do validation there if needed. Just alter form creation in create_report.ctp view.

    $this->Form->create('Report', array('action' => 'view_report'));
    

    Data will be submited directly to view_report action and available in $this->request->data

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