Hi,

So I have noticed the following, if I place a form in a layout and will call that page directly then the form will be rendered. If I call it from another page then the form will not be displayed.

To be specific:

<?php
class ContactPage extends Page {
}
class ContactPage_Controller extends Page_Controller {
    private static $allowed_actions = array('Form');
    public function Form() { 
        $fields = new FieldList( 
            new TextField('Name'), 
            new EmailField('Email'), 
            new TextareaField('Message')
        ); 
        $actions = new FieldList( 
            new FormAction('submit', 'Submit') 
        ); 
        return new Form($this, 'Form', $fields, $actions); 
    }
}

Create ContactPage.ss with $Form entry only. From a different page call $renderWith(ContactPage) will result in no table output.

you can call 'ContactPage_Controller' Form to other page like below.

public function Form() {
$contact_page = ContactPage::get()->First();
$controller = new ContactPage_Controller($contact_page);
return $controller->Form();
}

also check the http://www.silverstripe.org/community/forums/form-questions/show/7334

Actually I cant. I use the following structure Page->Content->CustomPages and any php code that I add to either Content.php or Page.php and call it from the template Content.s does not get executed at all ! Yes I do add it to controler section of the page, yes the Content class extends Page class.

Hello Liutauras,

I does work well, I am attaching a screenshot.

Do you mind sharing php code, template file via http://www.sspaste.com/

Updated

Sure thing:

Page.php http://www.sspaste.com/paste/show/54eef35e9b275
Content.php http://www.sspaste.com/paste/show/54eef5bf53188
Content.ss http://www.sspaste.com/paste/show/54eef5903dbd5

More or less everything from the bootstrap template.

I have disabled rendering of other two tabs.

The only template file able to access any children of Page class is Page.ss. The only option I have is to cut out parts that require access to the base template and execute them in Page.ss file. If you have any more elegant ways of working around this let me know, that would be great.

Place the $Form outside loop.
<div id="content">
$Form
</div>

01 <div id="content">
02 <% loop Menu(1) %>
03 $Form
04 $testContent
05 $testPage
06 $testUser
07 <section id="$URLSegment">
08 $renderWith($ClassName)
09 </section>
10 <% end_loop %>
11 </div>

Initially reading the first post, it looked as though the ContactPage template was being used to render a different page directly. Which would make sense that no form would render, as it's simply a template, and another page does not have the contact for on it.
ie. the $Form variable template would refer to the current page's Form method, of which there is presumably none.

But now having seen the template, it makes more sense.

It looks like you're trying a single-page style layout for your website.
With that in mind, when you loop the Menu to get each page's data, you're getting the page model, where as a form is part of the controller.
So putting the Form outside of the loop is not the answer, as it will just print the form from the active page (the one loaded in the URL) - which is probably none (unless it's the contact page).

To get around this, you'd have to create a Form() method on the model, that returns

ModelAsController::controller_for($this)->Form();

Thanks for suggestion, will give a shot!