我可以添加一个优先级来执行方法级别@bean吗？

Question

在我的测试应用程序中，我有我的安全配置类，如下所示。

package myProject.security;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.context.annotation.DependsOn;
import org.springframework.context.annotation.Primary;
import org.springframework.security.config.annotation.authentication.builders.AuthenticationManagerBuilder;
import org.springframework.security.config.annotation.web.builders.HttpSecurity;
import org.springframework.security.config.annotation.web.configuration.EnableWebSecurity;
import org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.crypto.password.PasswordEncoder;
import org.springframework.security.crypto.password.StandardPasswordEncoder;

import javax.annotation.PostConstruct;
import javax.annotation.PreDestroy;

@SuppressWarnings("deprecation")
@Configuration
@EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {

    @Autowired
    private UserDetailsService userService;

    @Override
    public void configure(HttpSecurity theHttp) throws Exception {
        theHttp.authorizeRequests()
                .antMatchers("/design", "/order")
                    .access("hasRole('ROLE_USER')")
                .antMatchers("/", "/**")
                    .access("permitAll")
                .and()
                .formLogin()
                    .loginPage("/login")
                    .defaultSuccessUrl("/", true) 
                .and()
                .logout()
                    .logoutSuccessUrl("/login")
                .and()
                .csrf()
                    .ignoringAntMatchers("/h2-console/**")
                .and()
                .headers()
                    .frameOptions()
                    .sameOrigin();
    }

    @Bean
    public PasswordEncoder encoder() {
        return new StandardPasswordEncoder("53cr3t");
    }

    @Override
    public void configure(AuthenticationManagerBuilder theAuth) throws Exception {
        theAuth.userDetailsService(userService)
                .passwordEncoder(encoder());
    }

}

在那里我定义了encoder（）并将其注释为@beans。这意味着该方法生成一个bean由spring容器管理。同样，我需要通过构造函数访问编码器，如下所示。

package myProject.security;

import org.springframework.context.annotation.DependsOn;
import org.springframework.security.crypto.password.PasswordEncoder;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.bind.annotation.RequestMapping;
import myProject.data.UserRepository;

@Controller
@RequestMapping("/register")
public class RegistrationController {

    private final UserRepository userRepo;
    private final PasswordEncoder passwordEncoder;

    public RegistrationController(UserRepository theUserRepo, PasswordEncoder thePasswordEncoder) {
        this.userRepo = theUserRepo;
        this.passwordEncoder = thePasswordEncoder;
    }

    @GetMapping
    public String registrationForm() {
        return "userRegistry";
    }

    @PostMapping
    public String processRegistration(RegistrationForm theUserRegistration) {
        userRepo.save(theUserRegistration.tranformUser(passwordEncoder));
        return "redirect:/loginPage";
    }

}

在上面的示例中，我使用密码编码器作为构造函数参数。问题是@Bean方法不是由调用RegistrationController的时间构造函数执行的。在将encoder（）添加到bootstrap类之后，我可以克服这个问题，但我认为这不是一个解决方案。谁能帮我想办法解决我的问题。事先谢谢。

Answer 1

请将@bean方法的名称更改为passwordencoder()。

@Bean
    public PasswordEncoder passwordEncoder () {
        return new StandardPasswordEncoder("53cr3t");
}

默认情况下，配置类使用@bean方法的名称作为生成bean的名称。但是，可以使用name属性重写此功能。

调用顺序是可以的，但是您没有名为passwordEncoder的bean