使用 Spring 保护 REST url

Question

我有一个正常工作的 Spring REST Web 服务，并希望向它添加基本身份验证。控制器方法附有普通的 url 和 HTTP 方法注释。为了增加安全性，我做了两件事

1) 将 spring-security.xml 添加到 WEB-INF 文件夹，如下所示：

<?xml version="1.0" encoding="UTF-8"?>
<beans  xmlns="http://www.springframework.org/schema/beans"
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
        xmlns:security="http://www.springframework.org/schema/security"
        xsi:schemaLocation="http://www.springframework.org/schema/beans 
            http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
            http://www.springframework.org/schema/security 
            http://www.springframework.org/schema/security/spring-security-3.1.xsd">


    <security:http auto-config="true">
            <security:http-basic />
            <security:intercept-url pattern="/*" access="ROLE_USER" />
    </security:http>

    <security:authentication-manager>
      <security:authentication-provider>
        <security:user-service>
        <security:user name="spring" password="spring" authorities="ROLE_USER" />
        </security:user-service>
      </security:authentication-provider>
    </security:authentication-manager>

</beans>

2）在web.xml中配置spring security为：

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4"
    xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">

    <servlet>
        <servlet-name>spring</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <filter>
        <filter-name>springSecurityFilterChain</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
    </filter>

    <servlet-mapping>
        <servlet-name>spring</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            /WEB-INF/spring-security.xml
        </param-value>
    </context-param>

    <filter-mapping>
        <filter-name>springSecurityFilterChain</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

</web-app>

3) 弹簧控制器：

package com.sample.main;

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.PathVariable;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.ResponseBody;

import com.sample.employee.Employee;

@Controller
public class EmpController 
{
    List<Employee> list = new ArrayList<Employee>();  

    @RequestMapping(value = "/getEmp/{emp}", method = RequestMethod.GET)
    public @ResponseBody Employee getEmployee(@PathVariable("emp") int empid) {
        System.out.println("meet getEmp");
        for (Iterator<Employee> iterator = list.iterator(); iterator.hasNext();) {
            Employee emp = (Employee) iterator.next();
            if(emp.getEmpId()==empid) {
                return emp;
            }
        }
        return new Employee();
    }
}

4) applicationContext.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:mvc="http://www.springframework.org/schema/mvc"
    xsi:schemaLocation="http://www.springframework.org/schema/beans 
            http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
            http://www.springframework.org/schema/context
            http://www.springframework.org/schema/context/spring-context-3.1.xsd
            http://www.springframework.org/schema/mvc 
            http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd">

    <!-- Activates various annotations to be detected in bean classes -->
    <context:annotation-config />

    <!-- Scans the classpath for annotated components that will be auto-registered as Spring beans.
     For example @Controller and @Service. Make sure to set the correct base-package-->
    <context:component-scan base-package="com.sample" />

    <!-- Configures the annotation-driven Spring MVC Controller programming model.
    Note that, with Spring 3.0, this tag works in Servlet MVC only!  -->
    <mvc:annotation-driven />

</beans>

5) spring-servlet.xml

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:p="http://www.springframework.org/schema/p"
        xmlns:mvc="http://www.springframework.org/schema/mvc"
    xmlns:context="http://www.springframework.org/schema/context"
    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/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">

</beans>

我已将完整的申请上传至： http://www.javaexperience.com/wp-content/uploads/2012/12/SpringWebServiceSecurity.zip

配置 springSecurityFilterChain 后，我开始在应用程序中收到 REST url 的 404 错误页面。任何想法...

Answer 1

尝试更改调度程序 servlet 的 url-mapping 以处理所有请求。

Web.xml

<servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>spring</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

同样在您的安全配置中更改安全过滤器intercept-url-pattern 以处理所有请求。这可以在工作后进行调整。

spring-security.xml

<security:http auto-config="true">
        <security:http-basic />
        <security:intercept-url pattern="/**" access="ROLE_USER" />
</security:http>