Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Handle unauthorized error message for Basic Authentication in Spring Security

I am trying to use Spring Security 3.0.5 in my web application. Basically, I want to have a web service which return data in json format via HTTP GET.

I have implemented a RESTful service which returns data when the url http://localhost:8080/webapp/json is requested. This works fine with the following curl command

> curl http://localhost:8080/webapp/json
{"key":"values"}

After I added basic authentication using spring security, I can use following commands to get the data

> curl  http://localhost:8080/webapp/json
<html><head><title>Apache Tomcat/6.0.29 - Error report .....
> curl -u username:password http://localhost:8080/webapp/json
{"key":"values"}

The former command returns standard tomcat error page since now it requires username and password. My question is whether it is possible to handle access denied in such a way that it prints out my own error message? i.e.

> curl  http://localhost:8080/webapp/json
{"error":"401", "message":"Username and password required"}

Here is my spring security configuration and AccessDeniedHandler. As you can see, I am trying to add access-denied-handler which simply prints out a string through servlet response but it still does not print my own message on command line.

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/security"
       xmlns:beans="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:context="http://www.springframework.org/schema/context"
       xmlns:p="http://www.springframework.org/schema/p"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
          http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd
          http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.0.xsd">
    <global-method-security secured-annotations="enabled"/>

    <beans:bean name="access-denied" class="webapp.error.JSONAccessDeniedHandler"></beans:bean>

    <http auto-config="true">
        <access-denied-handler ref="access-denied"/>
        <intercept-url pattern="/json" access="ROLE_USER,ROLE_ADMIN"  />
        <http-basic />
    </http>

    <authentication-manager>
        <authentication-provider>
            <password-encoder hash="md5"/>
            <user-service>
            ...
            </user-service>
        </authentication-provider>
    </authentication-manager>

</beans:beans>

AccessDeniedHandler.java

package webapp.error;

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.security.access.AccessDeniedException;
import org.springframework.security.web.access.AccessDeniedHandler;

public class JSONAccessDeniedHandler implements AccessDeniedHandler  {

    @Override
    public void handle(HttpServletRequest request, HttpServletResponse response, AccessDeniedException accessDeniedException) throws IOException, ServletException {
        PrintWriter writer = response.getWriter();
        writer.print("{\"error\":\"401\", \"message\":\"Username and password required\"}");
    }

}
like image 737
gigadot Avatar asked Dec 09 '10 10:12

gigadot


People also ask

How do you handle security exceptions in Spring?

Spring security exceptions can be directly handled by adding custom filters and constructing the response body. To handle these exceptions at a global level via @ExceptionHandler and @ControllerAdvice, we need a custom implementation of AuthenticationEntryPoint.


1 Answers

I have solved my problem so I think I should share it here. This configuration allows server to send out error message differently depending on the requesting software. If the request comes from a web browser, it will check the User-Agent header and redirect to a form login if necessary. If the request comes from, for example, curl, it will print out plain text error message when the authentication fails.

<?xml version="1.0" encoding="UTF-8"?>
<beans
    xmlns="http://www.springframework.org/schema/beans"
    xmlns:sec="http://www.springframework.org/schema/security"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:p="http://www.springframework.org/schema/p"
    xsi:schemaLocation="
        http://www.springframework.org/schema/beans    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
        http://www.springframework.org/schema/context  http://www.springframework.org/schema/context/spring-context-3.0.xsd
        http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.0.xsd">

    <!-- AspectJ pointcut expression that locates our "post" method and applies security that way
    <protect-pointcut expression="execution(* bigbank.*Service.post*(..))" access="ROLE_TELLER"/>-->
    <sec:global-method-security secured-annotations="enabled"/>

    <bean id="basicAuthenticationFilter"
          class="org.springframework.security.web.authentication.www.BasicAuthenticationFilter"
          p:authenticationManager-ref="authenticationManager"
          p:authenticationEntryPoint-ref="basicAuthenticationEntryPoint" />

    <bean id="basicAuthenticationEntryPoint"
          class="webapp.PlainTextBasicAuthenticationEntryPoint"
          p:realmName="myWebapp"/>

    <bean id="formAuthenticationEntryPoint"
          class="org.springframework.security.web.authentication.LoginUrlAuthenticationEntryPoint"
          p:loginFormUrl="/login.jsp"/>

    <bean id="daep" class="org.springframework.security.web.authentication.DelegatingAuthenticationEntryPoint">
        <constructor-arg>
            <map>
                <entry key="hasHeader('User-Agent','Mozilla') or hasHeader('User-Agent','Opera') or hasHeader('User-Agent','Explorer')" value-ref="formAuthenticationEntryPoint" />
            </map>
        </constructor-arg>
        <property name="defaultEntryPoint" ref="basicAuthenticationEntryPoint"/>
    </bean>

    <sec:http entry-point-ref="daep">
        <sec:intercept-url pattern="/login.jsp*" filters="none"/>
        <sec:intercept-url pattern="/json" access="ROLE_USER,ROLE_ADMIN"  />
        <sec:intercept-url pattern="/json/*" access="ROLE_USER,ROLE_ADMIN"  />
        <sec:logout
            logout-url="/logout"
            logout-success-url="/home.jsp"/>
        <sec:form-login
            login-page="/login.jsp"
            login-processing-url="/login"
            authentication-failure-url="/login.jsp?login_error=1" default-target-url="/home.jsp"/>
        <sec:custom-filter position="BASIC_AUTH_FILTER" ref="basicAuthenticationFilter" />
    </sec:http>

    <sec:authentication-manager alias="authenticationManager">
        <sec:authentication-provider>
        ...
        </sec:authentication-provider>
    </sec:authentication-manager>

</beans>

PlainTextBasicAuthenticationEntryPoint by extending org.springframework.security.web.authentication.www.BasicAuthenticationEntryPoint

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.security.core.AuthenticationException;
import org.springframework.security.web.authentication.www.BasicAuthenticationEntryPoint;

public class PlainTextBasicAuthenticationEntryPoint extends BasicAuthenticationEntryPoint {

    @Override
    public void commence(HttpServletRequest request, HttpServletResponse response, AuthenticationException authException) throws IOException, ServletException {
        response.addHeader("WWW-Authenticate", "Basic realm=\"" + getRealmName() + "\"");
        response.setStatus(HttpServletResponse.SC_UNAUTHORIZED);
        PrintWriter writer = response.getWriter();
        writer.println("HTTP Status " + HttpServletResponse.SC_UNAUTHORIZED + " - " + authException.getMessage());
    }
}
like image 74
gigadot Avatar answered Sep 19 '22 05:09

gigadot