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。
我已经实现了RESTful服务,该服务在请求url http://localhost:8080/webapp/json
时返回数据。使用以下curl命令
> curl http://localhost:8080/webapp/json
{"key":"values"}
使用spring security添加基本身份验证后,我可以使用以下命令获取数据
> 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,因为它现在需要用户名和密码。 我的问题是,是否有可能以打印出我自己的错误信息的方式处理拒绝访问?即
> curl http://localhost:8080/webapp/json
{"error":"401", "message":"Username and password required"}
这是我的spring安全配置和AccessDeniedHandler
。正如您所看到的,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,但它仍然不会在命令行上打印我自己的消息。
<?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\"}");
}
}
答案 0 :(得分:15)
我已经解决了我的问题所以我想我应该在这里分享。此配置允许服务器根据请求软件以不同方式发送错误消息。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
,则在身份验证失败时将打印出纯文本错误消息。
<?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
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());
}
}
答案 1 :(得分:0)
我想知道它是否曾经用于AccessDeniedHandler。是因为身份验证还是授权而出错?是否为两种方案调用AccessDeniedHandeler?我现在正在解决这个问题。