Re: Best practise? Creating a detailed data report from different linked models?

Michael,

You could certainly have a Reports controller. The expectation would
be that it is operating on the Report model. This seems pretty
unlikely given the information so far. More likely you are wanting a
report on data in some other model (like Project or Customer or ...)
and it's related models.

Starting with the assumption that you want to generate reports on the
model Project, you would want a report action in your Projects
controller. A simple example would be this:

function report() {
if(!empty($this->data)) {
$report = $this->Project->find('all', array('conditions'=>'your
conditions', 'contain'=>'fields you want'));
$this->set('report', $report);
}
}

In the folder app/views/projects/ you would store your file report.ctp
This file would output a form so the user can control and modify
reports. Also have this file output a formatted report if data is
available to do so. User the Form Helper to generate the form on your
page (http://book.cakephp.org/view/1383/Form).

Given this setup the user would get to the page at http://yoursite.com/projects/report

The report action in your Project controller looks to see if it was
being provided with posted data. On the first visit there would be
none so the logic would just fall out the bottom of the action and
report.ctp is rendered with only the form displayed. The user fills
the form, submits the data back to projects/report and this time the
if() is satisfied, the find() is done and data passed to the view in
$report. Now the page shows the form AND the report!

One action, one view, sweet!

Your various models would be designed with hasMany, belongsTo,
hasAndBelongsToMany, etc. as appropriate. Probably you would also want
your models to use the Containable behavior (http://book.cakephp.org/
view/1323/Containable) since you imply there is a lot of linking
involved.

I think you're fighting against the framework a bit. To become more
familiar with where to store your files and how to name things, look
these sections over again:
http://book.cakephp.org/view/899/CakePHP-Folder-Structure
http://book.cakephp.org/view/901/CakePHP-Conventions

Regards,
Don

On May 31, 1:19 pm, mivogtGermanyLU <miv...@mivogt.net> wrote:
> Don,
>
> yes I need a form to offer the user a way to select the criteria.
>
> As the SQL is a bit more complex, the use of find will not be my way.
> I know it is a dirty way of using cake but a long sqlquery-string
> works just better for me and offers me a nice result so far without
> learning the complex find (in my case I am using nearly all my models
> linked with each other)
>
> I am not sure about naming files and functions this for and placing
> them in the right directory.
>
> Is it ok to create a "reports_controller.php" containing an action
> "CreateReport($id,$date)" and 2 views "selectDataSet.ctp",
> "showReport.ctp"
>
> .. pleas feel free to correct my naming as it will not be within the
> convention so far ;)
>
> TIA
>
> michael
>
> On 31 Mai, 20:20, dreamingmind <dreamingmin...@gmail.com> wrote:
>
>
>
>
>
>
>
> > Michael,
>
> > I'm not sure I see what the problem is. Your link/button can certainly
> > pass any search parameters you want in to your action. It sounds like
> > you might be planning to offer a small form for the user to choose
> > search params? The Form Helper will let you put all that together and
> > the data will come back to you in $this->data in your action (http://
> > book.cakephp.org/view/1383/Form). The action can make use of the
> > parameters to build a $this->yourModel->find() call. The returned data
> > will be be passed along to the view because actions by default render
> > their associated views. All pretty standard stuff unless I'm
> > misreading your message.
>
> > If you're really planning on writing that sql by hand, you might want
> > to review thishttp://book.cakephp.org/view/1017/Retrieving-Your-Data
>
> > Regards,
> > Don
>
> > On May 31, 10:28 am, mivogtGermanyLU <miv...@mivogt.net> wrote:
>
> > > Hi there,
>
> > > my app has several models with defined relations.
> > > Entering Data works fine also all the CRUD stuff is fine.
>
> > > Now I need to collect data from the database ostly leftjoined and
> > > limited and sorted by some definitions... so far as first step I made
> > > me an SQL statement trying insidephpmyadminand it works fine.
>
> > > To get it more comfortable I would like to have a view offering to set
> > > the data-limiters.
> > > In my case it will be a month/year datafield and the id of one model
> > > all the other stuff is linked somehow.
>
> > > I am not sure about how to go on with this and would be happy to get a
> > > helping comment...
>
> > > I guess I will need a SelectWhatToReport-View to select month/year/
> > > model_id
> > > passing this to a controller-action CreateReport($month,$year,
> > > $model_id)
>
> > > in controller I might add the new action
> > > CreateReport($month,$year, $model_id)
> > > {sql='select ... where .. $month .. ,$year, .. $model_id'
> > > // . creating sql result array to be passed to a view
>
> > > }
>
> > > not sure about how to call a 2nd view from controller to display the
> > > result from SqlResultArray using some while statements ...
>
> > > ok maybe I better rewrite in short sentences what I want to know/need
> > > to do:
> > > - need to get a view offering to choose an entry of a model from
> > > database (sems to be easy I hope)
> > > - passing this selectioncriteria to controller/action (hopefully just
> > > by pressing a button)
> > > - calling a view from controller side to display the result (sql
> > > array)
>
> > > Thanks in advance
>
> > > Michael

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