Question

我在Spring-boot中有一个项目。对不起英国人不好。因此，问题是：当我通过＆＃34; bootRun＆＃34;运行我的项目时，仅在我的postgresql数据库中创建3个表中的2个。他们是＆＃34;角色＆＃34;和＆＃34; user_role＆＃34;。当Hibernate尝试创建第三个表＆＃34; user＆＃34;时，它会返回一条消息。这是消息：enter image description here

这是user.java代码

import com.fasterxml.jackson.annotation.JsonIgnore;
import org.hibernate.validator.constraints.NotEmpty;

import javax.persistence.*;
import java.util.HashSet;
import java.util.Set;

@Entity
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Integer id;

    @NotEmpty
    private String name;

    @NotEmpty
    @Column(unique = true, nullable = false)
    private String login;

    @NotEmpty
    private String password;

    @JsonIgnore
    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(name = "user_role", joinColumns = {@JoinColumn(name = "user_id")}, inverseJoinColumns = {@JoinColumn(name = "role_id")})
    private Set<Role> roles = new HashSet<Role>();

    public User() {
    }

    public User(User user) {
        super();
        this.id = user.getId();
        this.name = user.getName();
        this.login = user.getLogin();
        this.password = user.getPassword();
        this.roles = user.getRoles();
    }

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getLogin() {
        return login;
    }

    public void setLogin(String login) {
        this.login = login;
    }

    public String getPassword() {
        return password;
    }

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

    public Set<Role> getRoles() {
        return roles;
    }

    public void setRoles(Set<Role> roles) {
        this.roles = roles;
    }

}

这是role.java代码

import java.util.HashSet;
import java.util.Set;

import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.ManyToMany;

import org.hibernate.validator.constraints.NotEmpty;
import org.springframework.security.core.GrantedAuthority;

import com.fasterxml.jackson.annotation.JsonIgnore;

@Entity
public class Role implements GrantedAuthority {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Integer id;

    @NotEmpty
    private String name;

    @JsonIgnore
    @ManyToMany(fetch = FetchType.LAZY, mappedBy = "roles")
    private Set<User> users = new HashSet<User>();

    @Override
    public String getAuthority() {
        return name;
    }

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public Set<User> getUsers() {
        return users;
    }

    public void setUsers(Set<User> users) {
        this.users = users;
    }

    public Role(String name, Set<User> users) {
        this.name = name;
        this.users = users;
    }

    public Role() {
    }
}

application.properties文件

spring.profiles.active=dev

spring.datasource.url= jdbc:postgresql://localhost:5432/postgres
spring.datasource.username=postgres
spring.datasource.password=12345

spring.jpa.hibernate.ddl-auto=create

和build.gradle

buildscript {
    repositories {
        mavenCentral()
        maven { url "https://repo.spring.io/plugins-release" }
    }
    dependencies {
        classpath("org.springframework.boot:spring-boot-gradle-plugin:1.3.0.RC1")
    }
}

apply plugin: "java"
apply plugin: "eclipse"
apply plugin: "idea"
apply plugin: "spring-boot"
apply plugin: "war"

jar {
    baseName = "spring-rest-service-oauth"
    version =  "0.1.0"
}

repositories {
    mavenCentral()
    maven { url "https://repo.spring.io/libs-release" }
}

dependencies {
//  compile("org.springframework.boot:spring-boot-starter-data-mongodb")
    compile 'postgresql:postgresql:9.1-901-1.jdbc4'

    compile("org.springframework.boot:spring-boot-starter-web")
    compile("org.springframework.boot:spring-boot-starter-security")
    compile("org.springframework.boot:spring-boot-starter-data-jpa")
    compile("org.springframework.security.oauth:spring-security-oauth2")
    providedRuntime("org.springframework.boot:spring-boot-starter-tomcat")
    testCompile("org.springframework.boot:spring-boot-starter-test")
    testCompile("com.jayway.jsonpath:json-path")
    testCompile("com.jayway.jsonpath:json-path-assert")
}

task wrapper(type: Wrapper) {
    gradleVersion = "2.7"
}

感谢谁回复

Answer 1

因为User是PostgreSQL中的保留关键字。

Answer 2

你可以转义你的表名，以避免保留关键字，如下所示

import javax.persistence.Entity;
import javax.persistence.Table;

@Entity
@Table(name = "\"User\"")
public class User

Answer 3

检查PostgreSQL的SQL保留字：Appendix B. SQL Key Words

为什么表用户没有创建？我得到这个消息＆＃34;不成功：创建表用户＆＃34;

3 个答案: