A dispatcher is even easier; just think of it as a servlet that
returns true if it "handles the request". Get a response stream and
write XML to it.
On Thu, Apr 23, 2009 at 3:05 PM, Borut Bolčina <borut.bolc...@gmail.com> wrote:
> Thanks, but I am not sure how to do that in a dispatcher.
>
> -Borut
>
> 2009/4/23 Howard Lewis Ship <hls...@gmail.com>:
>> You can return a StreamResponse whose job is to generate an
>> InputStream of XML to be sent to the client.
>>
>> On Thu, Apr 23, 2009 at 6:12 AM, Borut Bolčina <borut.bolc...@gmail.com>
>> wrote:
>>> Hello,
>>>
>>> I have a dispatcher which matches some URL pattern like this:
>>>
>>> public boolean dispatch(Request request, Response response) throws
>>> IOException {
>>> String path = request.getPath();
>>> // just imagine this works ok
>>> String uglyWord = request.getContextPath() == "" ?
>>> StringUtils.substringAfterLast(path, "/") : path.substring(1);
>>>
>>> if (uglyWordSet.contains(uglyWord)) {
>>> // return some XML if url contains some ugly word
>>> // WHAT IS THE BEST APPROACH TO RETURN XML (STREAM) ?
>>> return true;
>>> }
>>>
>>> return false;
>>> }
>>>
>>> I know how to create a page which returns xml stream response (I
>>> actually have one). Can I just reuse this page?
>>>
>>> Using T 5.0.18.
>>>
>>> Regards,
>>> Borut
>>>
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>>>
>>
>>
>>
>> --
>> Howard M. Lewis Ship
>>
>> Creator of Apache Tapestry
>> Director of Open Source Technology at Formos
>>
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>>
>>
>
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>
--
Howard M. Lewis Ship
Creator of Apache Tapestry
Director of Open Source Technology at Formos
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