Thanks for the help!;
It was fairly easy to write.
Here is the code if someone browses through the archives. It does
nothing special yet prints to console the uploaded resources if you use
{$files.filePrint} in your template. You can modify it however you like:
public class UploadedFilesModel implements Model {
private static Log log =
LogFactory.getLog(UploadedFilesModel.class);
private WeblogPageRequest weblogPageRequest = null;
public String getModelName() {
return "files";
}
public void init(Map params) throws RollerException {
this.weblogPageRequest =
(WeblogPageRequest)params.get("weblogRequest"); }
public Object getFilePrint(){
try {
FileManager fileManager =
RollerFactory.getRoller().getFileManager();
WebsiteData webSiteData =
weblogPageRequest.getWeblog();
WeblogResource[] resources =
fileManager.getFiles(webSiteData,null);
System.out.println("upload
files="+resources.length);
for (WeblogResource resource : resources) {
System.out.println(resource.getName()+"-"+resource.getPath());
}
} catch (RollerException e) {
log.warn("ERROR: FileManager failure", e);
}
return new Object();
}
}
-----Original Message-----
From: Dave [mailto:[EMAIL PROTECTED]
Sent: Monday, August 20, 2007 6:13 PM
To: [email protected]
Subject: Re: expose weblog uploaded files model?
On 8/20/07, Guy Katz <[EMAIL PROTECTED]> wrote:
> How do I make the new model accessible (UploadedFilesModel)?
> Do I need to configure it somewhere?
The README.html explains that http://tinyurl.com/22r2to
>What does the getModelName method do?
That is the name by which you will refer to the model in templates. For
example:
#set($myfiles = $mymodel.getMyFiles())
Where the model name is "mymodel"
- Dave
