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DataObject PAge
18 February 2009 at 11:47pm
Hi, I have read and understood the dataobject tutorials and how to display them on a page
what I would like now is to create a unique page for each dataobject row. Is this possible and how ?
Thanks
E.
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Re: DataObject PAge
24 February 2009 at 11:35am
Hiya,
I don't quite follow what you're trying to do here. Can you be a bit mroe detailed?
Cheers
Aaron -
Re: DataObject PAge
25 February 2009 at 6:41am Last edited: 25 February 2009 6:58am
Ron, I was enquiring about dataobjects : I know how to display a list of records on a page, but I don't know how to create a unique url for each record of a databobject
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Re: DataObject PAge
25 February 2009 at 8:08am
If the DataObject needs a URLSegment, it needs to be a Page, not a DataObject..
I think you're misunderstanding something.. ?
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Re: DataObject PAge
25 February 2009 at 3:53pm
but I don't know how to create a unique url for each record of a databobject
The forum module does something like this, You need to have 'Holder' page though. Eg a forum thread isnt a page in the site but by using the ID of the thread in the url and a function 'show' or something you can show your dataobject.
So if its not a page object you can do something like this in your controller class (mysite/code/page.php). It could be called anything
function view() {
if($id = Director::urlParam('ID')) { // check to see we have been passed an ID
return DataObject::get_by_id("SomeDataObject", $id);
}
}In the template you could do $view.Title or $view.Date to access the information in the DataObject.
Then you would have the url like yoursite.com/home/view/12 to view the 12th dataobject. Thats just a basic example which is easily expanded (eg if that object doesn't exist)
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Re: DataObject PAge
15 July 2009 at 7:50am
Hi, this is exactly what I was looking for, however I can't get it working.
First I did what willr wrote in his post and added this to PersonsPage.php:class PersonsPage_Controller extends Page_Controller {
function view(){
if($id = Director::urlParam('ID')){
return DataObject::get_by_id("PersonsObject", $id);
}
}
}
That returned a "Page not found" and no signs of my PersonsPage.ss-template.So I did some research in the forum-module, and added the following lines to PersonsObject.php:
function Link() {
return "show/" . $this->ID;
}
Right now I can call $Link in PersonsPage.ss instead of fx a hardcoded localhost/myss/show/1
That's nice, but I still get no signs of my PersonsPage.ss-template (and still the same "Page not found"-message.Does someone have any suggestions?
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Re: DataObject PAge
15 July 2009 at 8:02am
You need a url segment first so SS knows which controller to use.
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Re: DataObject PAge
16 July 2009 at 12:53am Last edited: 16 July 2009 1:02am
yeah, that makes sense.
But now I stumble upon two other learning-opportunities (problems don't exist, do they?):
1: how do I retrieve the URLSegment dynamically?
adding $this->PersonsPages()->URLSegment or $this->URLSegment to the function Link() rendered http://www.mysite.com/view/123.
In the forummodule the following string is used: $this->Forum()->Link(), however that gives an error: Uncaught Exception: Object->__call(): the method 'link' does not exist on 'ComponentSet'. Because the error didn't occur on the forummodule, I searched in the module for another place where a method Link() is defined, but couldn't find it.
2: hardcoding the urlsegment (http://www.mysite.com/the-persons-page-you-want-to-view/view/123) renders a white screen (nothing on it and no firebug-errors). I checked the templates and there is a PersonsObject.ss template with content in the mysite/template/Layout-folder so that shouldn't be the cause (I also tried moving it to mysite/template or renaming it to PersonsPage-view.ss (as is done in the forum-module) but that didn't work either)
BTW, I found some errors in my original code, so I'll post a better version here:
class PersonsPage_Controller extends Page_Controller {
function view($id = null) {
if($id == null)
$id = $this->urlParams['ID'];
if($id && is_numeric($id))
return DataObject::get_by_id("PersonsObject", $id);
}
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