java - DB 列上的 Spring JPA 错误返回与列名称相关的错误

Question

尝试使用 JPA 查询 Postgres 数据库时收到错误。

学生.java:

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;

@Entity
public class Student {
    @Id
    @GeneratedValue
    private Long id;
    private String name;
    private String passportNumber;

    public Student() {
        super();
    }

    public Student(Long id, String name, String passportNumber) {
        super();
        this.id = id;
        this.name = name;
        this.passportNumber = passportNumber;
    }
    public Long getId() {
        return id;
    }
    public void setId(Long id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getPassportNumber() {
        return passportNumber;
    }
    public void setPassportNumber(String passportNumber) {
        this.passportNumber = passportNumber;
    }

}

应用程序属性:

spring.jpa.properties.hibernate.dialect = org.hibernate.dialect.PostgreSQLDialect

spring.jpa.hibernate.ddl-auto=none

spring.jpa.hibernate.show-sql=true

spring.datasource.url=jdbc:postgresql://localhost:5432/shorten-db

spring.datasource.username=my_user

spring.datasource.password=my_password

spring.datasource.initialization-mode=always

spring.datasource.initialize=true

spring.datasource.schema=classpath:/schema.sql

spring.datasource.continue-on-error=true

应用程序.yml:

spring:
  datasource:
    url: jdbc:postgresql://localhost:5432/shorten-db
    username: my_user
    password: my_password
    driverClassName: org.postgresql.Driver

 spring.datasource.schema=classpath:/schema.sql

引用此架构:

DROP TABLE student;

CREATE TABLE student

(

 id varchar(100) NOT NULL,

 name varchar(100) DEFAULT NULL,

 passportNumber varchar(100) DEFAULT NULL,

 PRIMARY KEY (id)

);

当我调用服务“/students”时:

@PostMapping("/students")
public ResponseEntity<Object> createStudent(@RequestBody Student student) {
    Student savedStudent = studentRepository.save(student);

    URI location = ServletUriComponentsBuilder.fromCurrentRequest().path("/{id}")
            .buildAndExpand(savedStudent.getId()).toUri();

    return ResponseEntity.created(location).build();

}

我收到以下错误:

postgres-db    | 2020-08-28 21:46:28.108 UTC [257] HINT:  Perhaps you meant to reference the column "student0_.passportnumber".
postgres-db    | 2020-08-28 21:46:28.108 UTC [257] STATEMENT:  select student0_.id as id1_1_, student0_.name as name2_1_, student0_.passport_number as passport3_1_ from student student0_

如何实现消息中指定的提示也许您想引用列“student0_.passportnumber”我没有明确指定 SQL 并尝试仅使用 JPA。我需要自定义查询还是应该修改学生实体？

Answer 1

JPA 期望列名称采用小写形式，并且每个单词用下划线分隔。这将转换为驼峰式 Java 变量名称。在你的例子中，你有java变量名称作为passportNumber，它转换为passport_number。可以通过三种方式解决这个问题:

1. 将java变量重命名为passportnumber，意思是它是一个单词。不幸的是，它看起来不太好，因为它破坏了 java 的变量命名约定。

2. 将表列重命名为passport_number，从而建立正确的JPA 连接。

3. 使用 @Column(name="passportNumber") 注释您的 java 变量。