问题[hibernate]

我正在使用 Hibernate 5.2.3.Final。我有一个看起来尽可能正常的实体ExternalGrade（这是所有实体的问题） 但是在HQL中编写查询时，idea看不到这些实体并指示查询中的错误。

我的配置是在xml中，也许这是因为它？

<hibernate-configuration>
<session-factory>
    
    <property name="hibernate.connection.driver_class">com.mysql.cj.jdbc.Driver</property>
    <property name="hibernate.connection.useUnicode">true</property>
    <property name="hibernate.connection.characterEncoding">UTF-8</property>
    <property name="hibernate.connection.charSet">UTF-8</property>
    
    <property name="hibernate.dialect">org.hibernate.dialect.MySQLDialect</property>
    <property name="show_sql">false</property>
    <property name="hibernate.c3p0.min_size">5</property> <!-- Используется с 03.11.2021 -->
    <property name="hibernate.c3p0.max_size">5000</property> <!-- Используется с 03.11.2021 -->
    <property name="hibernate.c3p0.acquire_increment">25</property> <!-- Используется с 03.11.2021 -->
    <property name="hibernate.c3p0.timeout">30</property> <!-- Используется с 03.11.2021 -->
    <property name="hibernate.c3p0.max_statements">50</property> <!-- Используется с 03.11.2021 -->
    <property name="hibernate.c3p0.idle_test_period">25</property> <!-- Используется с 03.11.2021 -->

</session-factory>

我尝试设置方面，看起来像这样

添加框架支持甚至不想打开，并且仅在显示禁用操作时才会显示在搜索中

当尝试从 org.hibernate 进行任何导入时，会抛出错误The import org.hibernate cannot be resolved

例如，以下所有导入都不起作用：

import org.hibernate.SessionFactory;
import org.hibernate.boot.Metadata;
import org.hibernate.boot.MetadataSources;
import org.hibernate.boot.registry.StandardServiceRegistry;
import org.hibernate.boot.registry.StandardServiceRegistryBuilder;

我正在使用 Hibernate 版本 6.1.3.FINAL。

我在 Eclipse 中工作。

我尝试下载 Hibernate 版本 6.0.0.CR2 （我找不到更新的版本）并将其添加到项目中Build Path > Configure Build Path > Libraries > Classpath > Add external JARs

我尝试将项目中的Java版本从20更改为17，调用Clean，调用Maven > Update Project- 没有任何效果。

pom：

<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 https://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>
    <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>3.1.4</version>
        <relativePath/> <!-- lookup parent from repository -->
    </parent>
    <groupId>ru.terentyev</groupId>
    <artifactId>playtoxapp</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <packaging>war</packaging>
    <name>playtoxapp</name>
    <description>Demo project for Spring Boot</description>
    <properties>
        <java.version>21</java.version>
    </properties>
    <dependencies>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-security</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-thymeleaf</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-web</artifactId>
        </dependency>
        <dependency>
            <groupId>org.thymeleaf.extras</groupId>
            <artifactId>thymeleaf-extras-springsecurity6</artifactId>
        </dependency>

        <dependency>
            <groupId>org.postgresql</groupId>
            <artifactId>postgresql</artifactId>
            <scope>runtime</scope>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-tomcat</artifactId>
            <scope>provided</scope>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-test</artifactId>
            <scope>test</scope>
        </dependency>
        <dependency>
            <groupId>org.springframework.security</groupId>
            <artifactId>spring-security-test</artifactId>
            <scope>test</scope>
        </dependency>
<!-- https://mvnrepository.com/artifact/org.hibernate/hibernate-core -->
<dependency>
    <groupId>org.hibernate</groupId>
    <artifactId>hibernate-core</artifactId>
    <version>6.3.1.Final</version>
    <type>pom</type>
</dependency>
<!-- https://mvnrepository.com/artifact/org.springframework/spring-orm -->


    </dependencies>

    <build>
        <plugins>
            <plugin>
                <groupId>org.springframework.boot</groupId>
                <artifactId>spring-boot-maven-plugin</artifactId>
            </plugin>
        </plugins>
    </build>

</project>

互联网上的建议要么行不通，要么写得非常混乱和难以理解。

我需要删除 User'a，但由于他有连接（订阅和订阅者），对象没有被删除，我该如何正确实现呢？

抛出的异常：

Cannot delete or update a parent row: a foreign key constraint fails (`sweater_db`.`sweater_user_subscribers`, CONSTRAINT `user_role_subscribers_subscriber_fk` FOREIGN KEY (`subscriber_id`) REFERENCES `sweater_user` (`id`))

用户

@Entity
@Table(name = "sweater_user")
public class User implements UserDetails{

    private final static long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @NotBlank(message = "Username should not be empty")
    private String username;

    @NotBlank(message = "Password should not be empty")
    private String password;

    @Email(message = "Email is not correct")
    @NotBlank(message = "Email should not be empty")
    private String email;

    private String activationCode;

    private boolean active;

    @ManyToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    @JoinTable(
            name = "sweater_user_role",
            joinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"),
            inverseJoinColumns = @JoinColumn(name = "role_id", referencedColumnName = "id")
    )
    private Collection<Role> roles;

    @OneToMany(mappedBy = "author", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    private Collection<Message> messages;

    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(
            name = "sweater_user_subscribers",
            joinColumns = @JoinColumn(name = "channel_id"),
            inverseJoinColumns = @JoinColumn(name = "subscriber_id")
    )
    private Set<User> subscribers = new HashSet<>();

    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(
            name = "sweater_user_subscriptions",
            joinColumns = @JoinColumn(name = "subscriber_id"),
            inverseJoinColumns = @JoinColumn(name = "channel_id")
    )
    private Set<User> subscriptions = new HashSet<>();

为什么我可以通过 hibernate.properties 文件设置属性：

通过 application.properties 是不可能的吗？

例如，有两个实体 Users 和 System 。用户具有角色 admin 和 user 。具有管理员角色的用户必须与系统多对多关联，而具有用户角色的用户必须一对多（系统）关联（用户）。hibernate 提供了任何标准方法吗？另外，这样通过spring security进行身份验证时就不会发生冲突