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Java Spring安全问题-404错误

java spring spring-mvc spring-security

Java Spring安全问题-404错误,java,spring,spring-mvc,spring-security,Java,Spring,Spring Mvc,Spring Security,我只是试图执行一个非常简单的Spring安全示例项目，但我得到了404错误。请在此处帮助查找问题 控制器： package mypack; import org.springframework.stereotype.Controller; import org.springframework.web.bind.annotation.RequestMapping; import org.springframework.web.bind.annotation.RequestMethod; imp

我只是试图执行一个非常简单的Spring安全示例项目，但我得到了404错误。请在此处帮助查找问题

控制器：

package mypack;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.ModelAndView;

@Controller
public class SecurityController {

@RequestMapping(value = { "/", "/welcome**" }, method = RequestMethod.GET)
public ModelAndView welcomePage() {

    ModelAndView model = new ModelAndView();
    model.addObject("title", "Spring Security Hello World");
    model.addObject("message", "This is welcome page!");
    model.setViewName("hello");
    return model;

}

@RequestMapping(value = "/admin**", method = RequestMethod.GET)
public ModelAndView adminPage() {

    ModelAndView model = new ModelAndView();
    model.addObject("title", "Spring Security Hello World");
    model.addObject("message", "This is protected page!");
    model.setViewName("admin");

    return model;

}

}
spring-servlet.xml

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

<context:component-scan base-package="mypack" />

<bean
  class="org.springframework.web.servlet.view.InternalResourceViewResolver">
  <property name="prefix">
    <value>/WEB-INF/views/</value>
  </property>
  <property name="suffix">
    <value>.jsp</value>
  </property>
</bean>


/WEB-INF/views/
.jsp

spring-security.xml

<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"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.2.xsd">

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

<authentication-manager>
  <authentication-provider>
    <user-service>
    <user name="user" password="123456" authorities="ROLE_USER" />
    </user-service>
  </authentication-provider>
</authentication-manager>

hello.jsp

<%@page session="false"%>
<html>
<body>
<h1>Title : ${title}</h1>   
<h1>Message : ${message}</h1>   


标题：${Title}
消息：${Message}

在您的spring security xml文件中尝试以下代码。还根据您的要求更改了角色

<http auto-config="true" use-expressions="true">
        <intercept-url pattern="/signin" access="permitAll"></intercept-url>
        <intercept-url pattern="/logout" access="permitAll"></intercept-url>
        <intercept-url pattern="/accessdenied" access="permitAll"></intercept-url>          
        <intercept-url method="GET"   pattern="/**"
            access="hasRole('USER') 
            or hasRole('ADMIN')"></intercept-url>   

        <form-login login-page="/signin" default-target-url="/index"
            authentication-failure-url="/accessdenied" always-use-default-target="true"
            username-parameter="username" password-parameter="password"></form-login>
        <logout logout-success-url="/logout"></logout>
    </http>
尝试spring security xml文件中的以下代码。还可以根据您的要求更改角色

<http auto-config="true" use-expressions="true">
        <intercept-url pattern="/signin" access="permitAll"></intercept-url>
        <intercept-url pattern="/logout" access="permitAll"></intercept-url>
        <intercept-url pattern="/accessdenied" access="permitAll"></intercept-url>          
        <intercept-url method="GET"   pattern="/**"
            access="hasRole('USER') 
            or hasRole('ADMIN')"></intercept-url>   

        <form-login login-page="/signin" default-target-url="/index"
            authentication-failure-url="/accessdenied" always-use-default-target="true"
            username-parameter="username" password-parameter="password"></form-login>
        <logout logout-success-url="/logout"></logout>
    </http>
因为您已经在web.xml到url模式中将servlet映射配置为servlet映射中的“/”。因此，DispatcherServlet将查找名为index.html/jsp的文件

要配置控制器，必须将组件扫描添加到servlet.xml

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

<context:component-scan base-package="mypack" />

<bean
  class="org.springframework.web.servlet.view.InternalResourceViewResolver">
  <property name="prefix">
    <value>/WEB-INF/views/</value>
  </property>
  <property name="suffix">
    <value>.jsp</value>
  </property>
</bean>
添加下面的行

<context:component-scan base-package="mypack" />
因为您已经在web.xml到url模式中将servlet映射配置为servlet映射中的“/”。因此，DispatcherServlet将查找名为index.html/jsp的文件

要配置控制器，必须将组件扫描添加到servlet.xml

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

<context:component-scan base-package="mypack" />

<bean
  class="org.springframework.web.servlet.view.InternalResourceViewResolver">
  <property name="prefix">
    <value>/WEB-INF/views/</value>
  </property>
  <property name="suffix">
    <value>.jsp</value>
  </property>
</bean>
添加下面的行

<context:component-scan base-package="mypack" />
您必须定义路径的过滤器：

<intercept-url pattern="/" access="permitAll" />
<intercept-url pattern="/welcome" access="permitAll" />
因此，在这种情况下，没有角色（匿名）的用户可以访问“/”和“/欢迎**”。

您必须定义路径的筛选器：

<intercept-url pattern="/" access="permitAll" />
<intercept-url pattern="/welcome" access="permitAll" />
因此，在这种情况下，没有角色（匿名）的用户可以访问“/”和“/欢迎**”。

组件扫描扫描控制器的注释组件将解决spring-servlet.xml中存在的问题。很抱歉，它在代码段中不可见。现在编辑它。我开始得到404错误，一旦我包括了spring安全配置。若我对web.xml中的spring安全配置进行注释，那个么该项目运行良好。请注意，我没有包括登录页面。我希望spring.component scan提供的默认登录页面能够扫描控制器的注释组件，以解决spring-servlet.xml中存在的问题。很抱歉，它在代码段中不可见。现在编辑它。我开始得到404错误，一旦我包括了spring安全配置。若我对web.xml中的spring安全配置进行注释，那个么该项目运行良好。请注意，我没有包括登录页面。我期待spring提供的默认登录页面。存在于spring-servlet.xml中。很抱歉，它在代码段中不可见。现在编辑它。存在于spring-servlet.xml中。很抱歉，它在代码段中不可见。现在编辑它。