我刚开始通过阅读Spring Boot in Action这本书来学习 Spring Boot，我正在学习这本书的示例，尝试自己运行它们，但我有一个使用 JpaRepository.findOne() 时出现问题。

我已经遍历了整章来寻找我可能的不匹配之处。但是，它就是行不通。

该项目应该是一个简单的阅读列表。

代码如下:

读者@Entity:

package com.lixin.readinglist;

import org.springframework.data.annotation.Id;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.UserDetails;

import javax.persistence.Entity;
import java.util.Collection;
import java.util.Collections;

/**
 * @author lixin
 */
@Entity
public class Reader implements UserDetails {

    private static final long serialVersionUID = 1L;

    @Id
    private String username;
    private String fullname;
    private String password;

    @Override
    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getFullname() {
        return fullname;
    }

    public void setFullname(String fullname) {
        this.fullname = fullname;
    }

    @Override
    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    @Override
    public Collection<? extends GrantedAuthority> getAuthorities() {
        return Collections.singletonList(new SimpleGrantedAuthority("READER"));
    }

    @Override
    public boolean isAccountNonExpired() {
        return true;
    }

    @Override
    public boolean isAccountNonLocked() {
        return true;
    }

    @Override
    public boolean isCredentialsNonExpired() {
        return true;
    }

    @Override
    public boolean isEnabled() {
        return true;
    }
}

Jpa 接口(interface):

package com.lixin.readinglist;

import org.springframework.data.jpa.repository.JpaRepository;

/**
 * @author lixin
 */
public interface ReaderRepository extends JpaRepository<Reader, String> {
}

安全配置:

package com.lixin.readinglist;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.context.annotation.Configuration;
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;

/**
 * @author lixin
 */
@Configuration
@EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {

    private final ReaderRepository readerRepository;

    @Autowired
    public SecurityConfig(ReaderRepository readerRepository) {
        this.readerRepository = readerRepository;
    }

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http
                .authorizeRequests()
                .antMatchers("/").access("hasRole('READER')")
                .antMatchers("/**").permitAll()
                .and()
                .formLogin()
                .loginPage("/login")
                .failureUrl("/login?error=true");
    }

    @Override
    protected void configure(AuthenticationManagerBuilder auth) throws Exception {
        auth
                .userDetailsService((UserDetailsService) username -> readerRepository.findOne(username));
    }
}

我一直收到这个错误:

Error:(40, 86) java: method findOne in interface org.springframework.data.repository.query.QueryByExampleExecutor<T> cannot be applied to given types;
  required: org.springframework.data.domain.Example<S>
  found: java.lang.String
  reason: cannot infer type-variable(s) S
    (argument mismatch; java.lang.String cannot be converted to org.springframework.data.domain.Example<S>)

最佳答案

findOne()定义为 <S extends T> Optional<S> findOne(Example<S> example); .
这意味着在您的情况下它接受 Example<Reader>并返回 Optional<Reader> .
你传递给它一个 String ，这是错误的，您将其用作 AuthenticationManagerBuilder.userDetailsService() 中的 lambda 返回值, 这也是错误的 因为UserDetailsService是定义为

UserDetails loadUserByUsername(String username) throws UsernameNotFoundException;

所以你需要返回一个UserDetails实例不是 Optional它或扔掉UsernameNotFoundException如果与用户名不匹配 to be compliant with the javadoc :

Returns:

a fully populated user record (never null)

Throws:

UsernameNotFoundException - if the user could not be found or the user has no GrantedAuthority

此外你不需要使用findOne()这是一个示例查询。按 ID 查询就足够了。

所以你可以这样写:

@Override
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
   auth.userDetailsService(username -> readerRepository.findById(username)
                                                       .orElseThrow( () -> new UsernameNotFoundException("user with username " + username + " not found"));
}

作为旁注，getOne()非常棘手，因为它依赖于延迟加载，在某些情况下可能会带来意外。
JB Nizet 的话很有趣。 所以我现在测试了。当实体(即 isAccountNonLocked() )被 Spring Security 类访问时，JPA session 还没有打开。
所以一个LazyInitializationException在任何情况下都会抛出(用户名正确或不正确):

org.hibernate.LazyInitializationException: could not initialize proxy - no Session
        at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:155)
        at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:268)
        at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:73)
        at davidhxxx.example.angularsboot.model.db.User_$$_jvstd90_5.isAccountNonLocked(User_$$_jvstd90_5.java)
        at org.springframework.security.authentication.dao.AbstractUserDetailsAuthenticationProvider$DefaultPreAuthenticationChecks.check(AbstractUserDetailsAuthenticationProvider.java:352)
        at org.springframework.security.authentication.dao.AbstractUserDetailsAuthenticationProvider.authenticate(AbstractUserDetailsAuthenticationProvider.java:165)

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关于java - 我应该如何将 JpaRepository.findOne() 与 SpringBoot 一起使用？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/54050430/