DB列上的Spring JPA错误返回与列名称相关的错误

如何解决DB列上的Spring JPA错误返回与列名称相关的错误

尝试使用JPA查询Postgres数据库时收到错误消息。

Student.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;
    }

}

appliction.properties：

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

appliction.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

是指以下架构：

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Perhaps you meant to reference the column "student0_.passportnumber"中指定的提示我没有明确指定SQL并仅尝试使用JPA。我需要自定义查询还是应该修改学生实体？

解决方法

JPA希望列名采用小写形式，并且每个单词都用下划线分隔。这将转换为驼峰式的Java变量名称。在您的情况下，您具有java变量名，即passportNumber，可转换为passport_number。有三种解决方法：

1. 将Java变量重命名为passwordnumber，这意味着它是一个单词。不幸的是，它看起来不太好，因为它违反了Java的变量命名约定。

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

3. 用@Column（name =“ passportNumber”）注释Java变量。