Hi,

my client wants me to build a page template which displays the content of a asset folder and its subfolders. Can someone please give me a tip how to display the content of a asset folder with some kind of tree?

Thanks much in advance!

patte

you are a lucky boy, when you use ss2.3.1

http://doc.silverstripe.com/doku.php?id=simpletreedropdownfield

Hi SalvaStripe,

thanks for that hint - i think i AM a lucky boy, using ss2.3.1 ;-)

The problem is that my client wants to display about six or seven hundred image files.... hm, too much for a simple tree, i think.

Do you know a way to generate a kind of unordered list of those files, like a sitemap?

patte

hm i looked into the DB tabel: FILE

i think all entries there are entries what are in "assets", sooo..

there are some different "ClassName" like

Folder
Image
File

###############################

when i want to make a page, that returns all my uploaded googleanalytics from the "googleanalyticsfolder" in my assets, i have to look up, what ID this folder has. then i use this code:

mysite/code/GoogleAnalytics.php
<?php

class GoogleAnalytics extends Page {
static $db = array(
);
static $has_one = array(
);
}

class GoogleAnalytics_Controller extends Page_Controller {
function GoogleAnalyticsList() {

$GoogleAnalyticsList = new DataObjectSet();

$result = DB::query("SELECT * FROM File WHERE ParentID = '293' ORDER BY Name DESC");
if($result) {
foreach($result as $value) {
$eintraege = array(
'ID' => $value['ID'],
'Name' => $value['Name'],
'Title' => $value['Title'],
'Filename' => $value['Filename']
);
$GoogleAnalyticsList->push(new ArrayData($eintraege));
}
}
return $GoogleAnalyticsList;
}
}

?>

themes/MYTHEME/templates/Layout/GoogleAnalytics.ss

...
...
<center>
<% control GoogleAnalyticsList %>
<a href="$Filename" target="_blank">$Name</a>
<% if Last %><% else %><hr /><% end_if %>
<% end_control %>
</center>
...
...
###############################

so, back to problem:

you can get with that code all "Images" or all files except folders (ClassName != 'Folder'), or you can get at first all Folders with ID, then you can fill between the folders an array with the files, who has ParantID = current Folder ID, after this you can output it nice..

hm, but just for outputting all Images or all Files, you can use this code i think..

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+
+ I know, there is a better way with "dataobject::get(..." or stuff like this.. but i did not yet used it.. but soon ;) i promise!
+
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

i hope you need some stuff from here^^

OKAY here is the better way: (check the syntax please :D )

PHP FILE
function ShowTheStuff() {
$obj = "File";
$filter = "ClassName != 'Folder'";
$sort = "ID ASC";
$join = "";
$limit = "";

$records = DataObject::get($obj, $filter, $sort, $join, $limit);
return $records;
}

SS FILE
<% if ShowTheStuff %>

<% control ShowTheStuff %>

<a href="$FileName" target="_blank">$Name</a>

<% end_control %>

<% else %>

Nothing FOUND!

<% end_if %>

it is NOT testet, nothing from this here. but this was in my head.. maybe is does work!

I will give it a try. Thanks for that & greetings from Cologne!