or even easier (if you dont need a bookmarkable url)
abstract class ByteArrayLink extends Link {
protected abstract byte[] getbytes();
protected abstract String getfilename();
public void onclick() {
getRequestCycle().setRequestTarget(new IRequestTarget() {
void detach(RequestCycle rc) {}
Object getLock(RequestCycle rc) { return ByteArrayLink.this; }
void respond(RequestCycle rc) {
WebResponse r = (WebResponse)requestCycle.getResponse();
r.setAttachmentHeader(getfilename());
response.getOuptutStream().write(getbytes());
}}
}
-igor
On 1/3/07, Janos Cserep <[EMAIL PROTECTED]> wrote:
Very short, very quick DynamicWebResource tutorial:)
1. Subclass DynamicWebResource
public class MyResource extends DynamicWebResource {
@Override
protected DynamicWebResource.ResourceState getResourceState() {
return new ResourceState() {
public byte[] getData() {
return "Example".toBytes("UTF-8");
}
public String getContentType() {
return "text/plain";
}
};
}
}
2. Open your Application class and append your init() method with the
following two lines:
getSharedResources().add("myResource", new MyResource( );
mountSharedResource("/my/resource/url", new
ResourceReference("myResource").getSharedResourceKey());
3. Browse to the http://server/context/.../my/resource/url URL and it
should show the "Example" string
4. modify getContentType() and getData() to return your objects (you can
get any HTTP parameter with the getParameter() call of the Resource class in
getData() so you could pass arguments to theURL like
http://server/context/.../my/resource/url?id=XkdfG12
Janos
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