请有人可以帮助我从基于xml的spring配置迁移到基于java的吗？
这是我的xml配置：
<!--suppress SpringFacetInspection, SpringSecurityFiltersConfiguredInspection -->
<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"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.2.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">
<global-method-security pre-post-annotations="enabled"/>
<context:annotation-config/>
<context:spring-configured/>
<beans:bean name="userLoginService" class="service.UserLoginService"/>
<beans:bean name="standardPasswordEncoder"
class="org.springframework.security.crypto.password.StandardPasswordEncoder">
<beans:constructor-arg name="secret" value="supersecret"/>
</beans:bean>
<http auto-config="true" use-expressions="true">
<intercept-url pattern="/javax.faces.resources/**" access="permitAll"/>
<intercept-url pattern="/view/unsecured/**" access="permitAll"/>
<intercept-url pattern="/view/secured/**" access="isAuthenticated()" />
<intercept-url pattern="/view/admin/**" access="hasRole('ROLE_SUPERUSER')"/>
<intercept-url pattern="/admin/**" access="hasRole('ROLE_SUPERUSER')"/>
<form-login login-page="/view/unsecured/login.xhtml"/>
<logout logout-success-url="/index.xhtml" invalidate-session="true" delete-cookies="true"/>
</http>
<authentication-manager alias="authenticationManager">
<authentication-provider user-service-ref="userLoginService">
<password-encoder ref="standardPasswordEncoder"/>
</authentication-provider>
</authentication-manager>
</beans:beans>
这是我的java尝试：
@Configuration
@EnableWebSecurity
@EnableGlobalMethodSecurity(prePostEnabled = true)
public class WebSecurityConfig extends WebSecurityConfigurerAdapter {
@Bean(name = "standardPasswordEncoder")
public PasswordEncoder standardPasswordEncoder() {
return new StandardPasswordEncoder("supersecret");
}
@Bean(name = "userDetailService")
@Override
public UserDetailsService userDetailsServiceBean() {
return new UserLoginService();
}
@Override
protected void configure(HttpSecurity http) throws Exception {
http
//.userDetailsService(userDetailsService())
.authorizeRequests()
.antMatchers("/view/secured/**").fullyAuthenticated()
.antMatchers("/admin/**", "/view/admin/**").access("hasRole('ROLE_SUPERUSER')")
.antMatchers("/index.xhtml", "/view/unsecured/**", "/javax.faces.resources/**").permitAll()
.and()
.formLogin().loginPage("/view/unsecured/login.xhtml")
.usernameParameter("email").passwordParameter("password")
.and()
.logout().logoutSuccessUrl("/index.xhtml").invalidateHttpSession(true).deleteCookies("JSESSIONID")
.and()
.exceptionHandling().accessDeniedPage("/error.xhtml")
.and()
.csrf().disable();
}
@Autowired
public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
auth.userDetailsService(userDetailsService()).passwordEncoder(standardPasswordEncoder());
}
}
当我尝试登录时出现异常
Caused by: java.lang.StackOverflowError: null
at org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter$UserDetailsServiceDelegator.loadUserByUsername(WebSecurityConfigurerAdapter.java:386)
at org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter$UserDetailsServiceDelegator.loadUserByUsername(WebSecurityConfigurerAdapter.java:387)
...
我想念什么，或者怎么了？谢谢
覆盖userDetailsService()方法，而不是userDetailsServiceBean()方法。这就是导致您无限递归的原因。无需将其声明为@Bean那时。
应该只是：
@Override
public UserDetailsService userDetailsService() {
return new UserLoginService();
}
或者，如果您@Service在自己的注释上UserLoginService（以及将要拾取该类的组件扫描），则可以避免手动创建bean，而直接将其注入UserLoginService配置方法中：
@Autowired
public void configureGlobal(AuthenticationManagerBuilder auth, UserLoginService userDetailsService) throws Exception {
auth.userDetailsService(userDetailsService).passwordEncoder(standardPasswordEncoder());
}
然后，您将不需要重写userDetailsServiceBean或userDetailsService：，因为它们所做的只是创建该bean的实例。
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