Jasha,
Sorry to bother you again.
My code is:
OAuthProblemException ex =
OAuthProblemException.error(OAuthError.TokenResponse.UNAUTHORIZED_CLIENT).uri("
https://google.com";).setParameter("1", "2");
OAuthASResponse.errorResponse(HttpServletResponse.SC_FOUND).location(redirectURI).error(ex).buildQueryMessage();
But I still get: https://redirect.uri/oauth/callback?
error=unauthorized_client&error_uri=https%3A%2F%2Fgoogle.com
Please let me know if I'm doing wrong.
Thank you.
On Thu, Dec 17, 2015 at 4:52 PM, Jasha Joachimsthal <ja...@apache.org>
wrote:
>
>
> On 17 December 2015 at 09:13, Anders <innocentl...@gmail.com> wrote:
>
>> Jasha,
>>
>> I checked OAuthASResponse you mentioned and found:
>>
>> @Test
>> public void testAuthzImplicitResponseWithState() throws Exception {
>> HttpServletRequest request = createMock(HttpServletRequest.class);
>>
>> expect(request.getParameter(OAuth.OAUTH_STATE)).andStubReturn("ok");
>> replay(request);
>> OAuthResponse oAuthResponse =
>> OAuthASResponse.authorizationResponse(request,200)
>> .location("http://www.example.com";)
>> .setAccessToken("access_111")
>> .setExpiresIn("400")
>> .setParam("testValue", "value2")
>> .buildQueryMessage();
>>
>> String url = oAuthResponse.getLocationUri();
>> Assert.assertEquals("
>> http://www.example.com#testValue=value2&state=ok&expires_in=400&access_token=access_111";,
>> url);
>> Assert.assertEquals(200, oAuthResponse.getResponseStatus());
>> }
>>
>> Then I wrote my code as below:
>>
>> OAuthProblemException ex =
>> OAuthProblemException.error(OAuthError.TokenResponse.UNAUTHORIZED_CLIENT);
>> return
>> OAuthASResponse.errorResponse(HttpServletResponse.SC_BAD_REQUEST)
>> .error(ex)
>> .location(oauthReq.getRedirectURI())
>> .buildQueryMessage();
>>
>> I got this:
>> https://redirect.uri/oauth/callback?error_description=Not+allowed+to+go+IMPLICIT+grant+flow&error=unauthorized_client
>> But I expect this one: https://redirect.uri/oauth/callback#
>> error_description=Not+allowed+to+go+IMPLICIT+grant+flow&error=unauthorized_client
>>
>> I can't use OAuthASResponse.authorizationResponse(), because it doesn't
>> accept OAuthProblemException as argument.
>> DoI miss anything?
>>
>
>
> You are using a success method to return an error. See the
> testErrorResponse method for the example with the error response.
>
> OAuthASResponse.errorResponse(HttpServletResponse.SC_BAD_REQUEST).error(ex)...
>
>
>
>>
>> Thank you very much.
>>
>> On Thu, Dec 17, 2015 at 2:20 PM, Jasha Joachimsthal <ja...@apache.org>
>> wrote:
>>
>>> Hi Anderson,
>>>
>>> On 17 December 2015 at 07:00, Anders <innocentl...@gmail.com> wrote:
>>>
>>>> Hi,
>>>>
>>>> I'm using Oltu version 1.0.1.
>>>> According to OAuth 2.0 spec, I need to put error parameter in HTTP
>>>> fragment, like below:
>>>>
>>>> HTTP/1.1 302 Found
>>>> Location: https://client.example.com/cb#error=access_denied&state=xyz
>>>>
>>>> But I can't use OAuthASResponse.errorResponse() to put error parameter in
>>>> fragment.
>>>>
>>>> OAuthASResponse.errorResponse(HttpServletResponse.SC_FOUND)
>>>> .location(oauthReq.getRedirectURI())
>>>>
>>>> .setError(OAuthError.CodeResponse.ACCESS_DENIED)
>>>> .setState(oauthReq.getState())
>>>> .buildQueryMessage();
>>>>
>>>> Is there any way to do this?
>>>> Thank you for any comments.
>>>> --
>>>>
>>>> Anderson
>>>>
>>>
>>> First create an OAuthProblemException with the error and pass this
>>> exception to the OAuthASResponse. You can find examples in the test class
>>> of OAuthASResponse:
>>>
>>> https://svn.apache.org/repos/asf/oltu/trunk/oauth-2.0/authzserver/src/test/java/org/apache/oltu/oauth2/as/response/OAuthASResponseTest.java
>>>
>>> Regards,
>>>
>>> Jasha
>>>
>>>
>>
>>
>> --
>> －－－－－－－－－－－－－－－－－－
>> ~Mia は 最高！~
>> －－－－－－－－－－－－－－－－－－
>>
>
>
--
－－－－－－－－－－－－－－－－－－
~Mia は 最高！~
－－－－－－－－－－－－－－－－－－
