[cas-user] Re: Cas 5.1 How to get client's service url.
Thank you Andy for your time, everything works as I wanted. -- - Website: https://apereo.github.io/cas - Gitter Chatroom: https://gitter.im/apereo/cas - List Guidelines: https://goo.gl/1VRrw7 - Contributions: https://goo.gl/mh7qDG --- You received this message because you are subscribed to the Google Groups "CAS Community" group. To unsubscribe from this group and stop receiving emails from it, send an email to cas-user+unsubscr...@apereo.org. To view this discussion on the web visit https://groups.google.com/a/apereo.org/d/msgid/cas-user/e6188091-57e9-4e1b-8363-16df90b2a5b2%40apereo.org.
[cas-user] Re: Cas 5.1 How to get client's service url.
Hi again,
I have just dig deeper in the source code, and I found the way to extract
the *service URL*, here how it can be done in CAS 5.2.0-RC4 (I am too lazy
to test it in other version, please change the code accordingly)
public class CustomAuthenticationHandler implements AuthenticationHandler {
protected PrincipalFactory principalFactory = new
DefaultPrincipalFactory();
@Autowired
PropertiesConfiguration propertiesConfiguration;
private transient Logger logger =
LoggerFactory.getLogger(this.getClass());
@Override
public HandlerResult authenticate(Credential credential) throws
GeneralSecurityException, PreventedException {
UsernamePasswordCredential usernamePasswordCredential =
(UsernamePasswordCredential) credential;
// Here I would like to have access to the encoded address
final RequestContext context =
RequestContextHolder.getRequestContext();
final WebApplicationService service = WebUtils.getService(context);
//WebUtils use the CAS one
*String theUrlThatYouNeed = service.getId();*
//Of course you need to extract the hostname from the url
}
*Additional Info:*
I will also shows you the steps I acquire this information, so you might
reference if you want.
1) I search all types of AuthenticationHandler and see if any of them is
using the serviceParam url
2) I sumble accross this class *YubiKeyAuthenticationHandler *(actually, I
don't know what this handler does... but it doesn't matter much)
3) I see that it uses RequestContext, so I immediate see what this class
does (Since I know request context -> GET parameter -> service parameter ->
goal)
4) I see that with RequestContext, I can get WebApplicationService and
Service
https://github.com/apereo/cas/blob/master/api/cas-server-core-api-authentication/src/main/java/org/apereo/cas/authentication/principal/Service.java
5) I see this line *final String serviceUrl =
URLDecoder.decode(service.getId(), StandardCharsets.UTF_8.name());*
6) hence I know the I can use service.getId() to get the service URL
Hope this helps you!
-Andy
On Monday, 27 November 2017 06:50:25 UTC+8, Andy Ng wrote:
>
> In that case, I would suggest using a custom template. And make all the
> username, password parameters to be hidden. Or maybe create a
> loading
> ... so user know they are being redirected.
>
> Here are how to set up custom template:
> https://groups.google.com/a/apereo.org/forum/m/#!topic/cas-user/k-yfoou7Zy0
>
> See if that helps you.
>
> - Andy
>
>
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[cas-user] Re: Cas 5.1 How to get client's service url.
In that case, I would suggest using a custom template. And make all the username, password parameters to be hidden. Or maybe create a loading ... so user know they are being redirected. Here are how to set up custom template: https://groups.google.com/a/apereo.org/forum/m/#!topic/cas-user/k-yfoou7Zy0 See if that helps you. - Andy -- - Website: https://apereo.github.io/cas - Gitter Chatroom: https://gitter.im/apereo/cas - List Guidelines: https://goo.gl/1VRrw7 - Contributions: https://goo.gl/mh7qDG --- You received this message because you are subscribed to the Google Groups "CAS Community" group. To unsubscribe from this group and stop receiving emails from it, send an email to cas-user+unsubscr...@apereo.org. To view this discussion on the web visit https://groups.google.com/a/apereo.org/d/msgid/cas-user/33484ff9-8186-42ff-a465-be23d6b5c240%40apereo.org.
[cas-user] Re: Cas 5.1 How to get client's service url.
On the issue of using a hostname, I need it to retrieve a user from the
database - the user is searched using his name and an additional parameter
that I can specify using the hostname.
As for your suggestions, they are also useful. This is the next step I am
going to take, namely:
I would like to users from the specified domain to be properly redirected.
If the user comes from https://localhost:8445/casclient2/ then let him show
standard cas login page with appropriate handler:
{
"@class" : "org.apereo.cas.services.RegexRegisteredService",
"serviceId" : "https://localhost:8445/casclient2/;,
"name" : "JDBC",
"id" : 1003,
"description" : "Sample 1",
"evaluationOrder" : 51,
"requiredHandlers" : [ "java.util.HashSet", [
"QueryAndEncodeDatabaseAuthenticationHandler" ] ]
}
- here's no problem everything works
However if the user comes from https://localhost:8445/casclient2/ then I
would like to have authenticated using SAML and external Idp - delegated
authentication.I have tested it using the automatically generated button -
in the second case I would like the users not even see the form from cas.
W dniu piątek, 24 listopada 2017 11:29:27 UTC+1 użytkownik Andy Ng napisał:
>
> I also don't know a way to retrive the hostname. However I would like to
> know what is your use case for the hostname
>
> If, for example, you have 3 hosts that need 3 different customized
> authentication
>
> 1) https://i-only-use-saml.example.com
> 2) https://i-only-use-oauth.example.com
> 3) https://i-only-use-usernamepassword.example.com
>
> Then you can just registered the above three services, and use the
> "requiredHandler" param to redirect the three use case to a specific
> authentication handler.
> https://apereo.github.io/cas/development/installation/Service-Management.html
> (if you are planning to use CAS 5.2.0)
>
> However, if you need to do much more with the hostname, that service
> management cannot help with your use case, then I might not be able to help
> you... Maybe you need to read the source code to see how it can be retrived.
>
> - Andy
>
>
> On Thursday, 23 November 2017 22:41:23 UTC+8, snaffy wrote:
>>
>> I would like to know how I can retrieve the encoded address of the
>> service parameter on the authentication step on the CAS serve side.
>> More precisely I have created CustomAuthenticationHandler and at this
>> stage, I need to know from what domain the authentication request came
>> from. For example from this address
>> https://localhost:8445/cas/login?service=https%3A%2F%2Flocalhost%3A8445%2Fcasclient%2F
>>
>> I need to extract "localhost"
>>
>>
>> public class CustomAuthenticationHandler implements AuthenticationHandler
>> {
>>
>> protected PrincipalFactory principalFactory = new
>> DefaultPrincipalFactory();
>>
>> @Autowired
>> PropertiesConfiguration propertiesConfiguration;
>> private transient Logger logger =
>> LoggerFactory.getLogger(this.getClass());
>>
>>
>> @Override
>> public HandlerResult authenticate(Credential credential) throws
>> GeneralSecurityException, PreventedException {
>>
>> UsernamePasswordCredential usernamePasswordCredential =
>> (UsernamePasswordCredential) credential;
>>// Here I would like to have access to the encoded address
>>}
>>
>> Thanks in advance for help.
>>
>
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- Gitter Chatroom: https://gitter.im/apereo/cas
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[cas-user] Re: Cas 5.1 How to get client's service url.
I also don't know a way to retrive the hostname. However I would like to
know what is your use case for the hostname
If, for example, you have 3 hosts that need 3 different customized
authentication
1) https://i-only-use-saml.example.com
2) https://i-only-use-oauth.example.com
3) https://i-only-use-usernamepassword.example.com
Then you can just registered the above three services, and use the
"requiredHandler" param to redirect the three use case to a specific
authentication
handler.
https://apereo.github.io/cas/development/installation/Service-Management.html
(if you are planning to use CAS 5.2.0)
However, if you need to do much more with the hostname, that service
management cannot help with your use case, then I might not be able to help
you... Maybe you need to read the source code to see how it can be retrived.
- Andy
On Thursday, 23 November 2017 22:41:23 UTC+8, snaffy wrote:
>
> I would like to know how I can retrieve the encoded address of the service
> parameter on the authentication step on the CAS serve side.
> More precisely I have created CustomAuthenticationHandler and at this
> stage, I need to know from what domain the authentication request came
> from. For example from this address
> https://localhost:8445/cas/login?service=https%3A%2F%2Flocalhost%3A8445%2Fcasclient%2F
>
> I need to extract "localhost"
>
>
> public class CustomAuthenticationHandler implements AuthenticationHandler {
>
> protected PrincipalFactory principalFactory = new
> DefaultPrincipalFactory();
>
> @Autowired
> PropertiesConfiguration propertiesConfiguration;
> private transient Logger logger =
> LoggerFactory.getLogger(this.getClass());
>
>
> @Override
> public HandlerResult authenticate(Credential credential) throws
> GeneralSecurityException, PreventedException {
>
> UsernamePasswordCredential usernamePasswordCredential =
> (UsernamePasswordCredential) credential;
>// Here I would like to have access to the encoded address
>}
>
> Thanks in advance for help.
>
--
- Website: https://apereo.github.io/cas
- Gitter Chatroom: https://gitter.im/apereo/cas
- List Guidelines: https://goo.gl/1VRrw7
- Contributions: https://goo.gl/mh7qDG
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