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Special .ss Template for XML Version of Page

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3 Posts   1529 Views


Community Member, 197 Posts

24 August 2010 at 1:27pm

HTML versions of page on my website can be accessed at URLs like this:

I also want each page to be available in XML format at URLs like this:

Now "" is used as the HTML template.

I want "" should be used as the template for the XML version of pages.


Community Member, 904 Posts

25 August 2010 at 6:48pm

Shouldn't be too difficult.
First you need a special rewrite-rule:

<IfModule mod_rewrite.c>
	RewriteEngine On
	RewriteCond %{REQUEST_FILENAME} !-f
	RewriteRule (.*)\.xml$ sapphire/main.php?url=$1/xml [L,QSA]
	RewriteCond %{REQUEST_URI} ^(.*)$
	RewriteCond %{REQUEST_FILENAME} !-f
	RewriteRule .* sapphire/main.php?url=%1&%{QUERY_STRING} [L]

The red part is what's newly added.

Then add the following function to your Page_Controller:

public function xml(){
	return array();

And now all that is left to do is to create a special template for that. You have to name it and put it in your templates folder.

Calling /page.xml?flush=1 should now render your page with the XML template.


Community Member, 78 Posts

14 June 2011 at 12:37am

Hi Bana,

I am sorry to disturb you, I was trying to generate the record from DataObject into XML format for using on my flash page.
I have NewsItem as DataObject with following code:

class NewsItem extends DataObject {

static $db = array(
'Title' => 'Varchar(255)',
'Details' => 'HTMLText',
'Source' => 'Varchar(255)',
'Date' => 'Date'
static $has_one = array(
'NewsPage' => 'NewsPage',
'NewsPhoto' => 'Image'
static $summary_fields = array(
'Title' => 'Title',
'Source' => 'Source',
'Date' => 'Date'

public function getCMSFields() {
return new FieldSet(
new TextField('Title'),
new SimpleTinyMCEField('Details', 'News Details'),
new TextField('Source', 'News Source'),
new DatePickerField('Date', 'Release Date'),
new ImageField('NewsPhoto', 'Photo', Null, Null, Null, 'Uploads/latest-news/photos/')

function caView() {
return true;

//Return the Name as a menu title
public function MenuTitle() {
return $this->Title;

public function Link() {
if ($NewsPage = $this->NewsPage()) {
return $NewsPage->Link('show/') . $this->ID;

public function LinkingMode() {
if ($Controller = Controller::CurrentPage() && Controller::CurrentPage()->ClassName == 'NewsPage') {
if (Controller::CurrentPage()->getAction() == 'show' && $NewsItem = Controller::CurrentPage()->getNewsItem()) {
return ($NewsItem->ID == $this->ID) ? 'current' : 'link';


my purpose i would like to generate all record from this DataObject into XML Format, you have any example, I am trying many way but it return 404 error.

Your replies I really appreciated.