zea*_*eal 5 java spring spring-security spring-boot
我想将基于XML的配置用于Spring Security.第一个想法是使用SHA-256或任何其他散列函数用于用户密码.我找不到用普通java解决这个问题的好方法.所以我开始在xml中配置东西.当它开始变得有趣时,这就是重点.
我的配置:
弹簧security.xml文件:
<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:jdbc="http://www.springframework.org/schema/jdbc"
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.xsd>
<http pattern="/css/**" security="none"/>
<http pattern="/login.html*" security="none"/>
<http>
<intercept-url pattern="/**" access="ROLE_USER" />
<form-login login-page='/login.html'/>
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="admin" password="admin"
authorities="ROLE_USER, ROLE_ADMIN"/>
<user name="bob" password="bob"
authorities="ROLE_USER"/>
</user-service>
</authentication-provider>
</authentication-manager>
</beans:beans>
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我在类中加载了xml文件,其中public static void main可以找到:
@Configuration
@ComponentScan
@EnableAutoConfiguration
@Order(HIGHEST_PRECEDENCE)
@ImportResource({
"/spring-security.xml"
})
public class PhrobeBootApplication extends SpringBootServletInitializer {
...
}
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但是我在任何pageload上都得到以下异常:
[ERROR] org.apache.catalina.core.ContainerBase.[Tomcat].[localhost].[/].[dispatcherServlet] - Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception
org.springframework.security.authentication.AuthenticationCredentialsNotFoundException: An Authentication object was not found in the SecurityContext
...
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所以看起来配置来自resources/WEB-INF/web.xml不加载,如果我从文档中有很好的理解,我应该在使用普通弹簧时使用它,而不需要启动.(应配置过滤器).我对吗?
为什么会发生这种错误?有没有更好的方法在spring-boot中使用基于xml的配置来实现spring-security?web.xml甚至可以通过tomcat加载吗?
根据Dave Syer在最新版本的 Spring Boot 中的说法,配置 Spring Security 的最佳方法是使用 Java 配置。
我需要一个 SHA-256 编码器,但我还没有找到任何简单且良好的解决方案来实现它。您只需使用passwordEncoder 配置jdbcAuthentication 即可。这真的非常简单:
@EnableWebSecurity
public class SpringSecurityConfigurer extends WebMvcConfigurerAdapter {
@Override
public void addViewControllers(ViewControllerRegistry registry) {
registry.addViewController("/login").setViewName("login");
}
@Bean
public ApplicationSecurity applicationSecurity() {
return new ApplicationSecurity();
}
@Order(SecurityProperties.ACCESS_OVERRIDE_ORDER)
protected static class ApplicationSecurity extends WebSecurityConfigurerAdapter {
@Autowired
private SecurityProperties security;
@Autowired
private DataSource dataSource;
@Override
protected void configure(HttpSecurity http) throws Exception {
http.authorizeRequests().antMatchers("/css/**").permitAll().anyRequest().fullyAuthenticated()
.and().formLogin().loginPage("/login").failureUrl("/login?error").permitAll()
.and().logout().logoutRequestMatcher(new AntPathRequestMatcher("/logout")).logoutSuccessUrl("/login");
}
PasswordEncoder sha256PasswordEncoder = new PasswordEncoder() {
@Override
public String encode(CharSequence rawPassword) {
return Hashing.sha256().hashString(rawPassword, Charsets.UTF_8).toString();
}
@Override
public boolean matches(CharSequence rawPassword, String encodedPassword) {
return encodedPassword.equals(Hashing.sha256().hashString(rawPassword, Charsets.UTF_8).toString());
}
};
@Override
public void configure(AuthenticationManagerBuilder auth) throws Exception {
auth.jdbcAuthentication()
.dataSource(this.dataSource)
.passwordEncoder(sha256PasswordEncoder);
}
}
}
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