处理 Spring Security 中基本身份验证的未授权错误消息 [英] Handle unauthorized error message for Basic Authentication in Spring Security

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问题描述

我正在尝试在我的 Web 应用程序中使用 Spring Security 3.0.5.基本上,我想要一个通过 HTTP GET 以 json 格式返回数据的 Web 服务.

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.

我已经实现了一个 RESTful 服务,它在请求 url http://localhost:8080/webapp/json 时返回数据.这适用于以下 curl 命令

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"}

使用 spring security 添加基本身份验证后,可以使用以下命令获取数据

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"}

前一个命令返回标准的 tomcat 错误页面,因为现在它需要用户名和密码.我的问题是是否可以通过打印出我自己的错误消息的方式来处理拒绝访问?

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"}

这是我的 spring 安全配置和 AccessDeniedHandler.如您所见,我正在尝试添加 access-denied-handler,它只是通过 servlet 响应打印出一个字符串,但它仍然没有在命令行上打印我自己的消息.

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"}");
    }

}

推荐答案

我已经解决了我的问题,所以我想我应该在这里分享它.此配置允许服务器根据请求的软件发送不同的错误消息.如果请求来自 Web 浏览器,它将检查 User-Agent 标头并在必要时重定向到表单登录.如果请求来自例如curl,则在认证失败时会打印出纯文本错误信息.

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 通过扩展 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());
    }
}

这篇关于处理 Spring Security 中基本身份验证的未授权错误消息的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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