Crudepository（spring boot）中自定义方法的java问题

1 年，11 月 Questions & Answers 86

我正在尝试使用spring boot为一所小型驾校开发一个web应用程序。此时，我正在尝试创建使用外键的表：一个学生将有多个评估，所以我正在进行1:N关联。问题是，当我试图运行项目进行测试时，我得到了这个错误（只想说明原因）：

Caused by: org.springframework.data.mapping.PropertyReferenceException: No property rut found for type ResultadosPsicotecnicos!

我做了一个自定义方法来获取一个学生的所有评估，但由于某种原因，它不起作用。我查看了数据库，看看这个列是否真的被创建了，下面是我看到的代码：

-- Table: public.resultados_psicotecnicos

-- DROP TABLE public.resultados_psicotecnicos;

CREATE TABLE public.resultados_psicotecnicos
(
    id bigint NOT NULL,
    fecha_evaluacion character varying(255) COLLATE pg_catalog."default",
    nota1 character varying(255) COLLATE pg_catalog."default",
    nota2 character varying(255) COLLATE pg_catalog."default",
    nota3 character varying(255) COLLATE pg_catalog."default",
    nota4 character varying(255) COLLATE pg_catalog."default",
    rut bigint NOT NULL,
    CONSTRAINT resultados_psicotecnicos_pkey PRIMARY KEY (id),
    CONSTRAINT fkf57g1daekqxergtw9kck81esq FOREIGN KEY (rut)
        REFERENCES public.students (id) MATCH SIMPLE
        ON UPDATE NO ACTION
        ON DELETE NO ACTION
)

TABLESPACE pg_default;

ALTER TABLE public.resultados_psicotecnicos
    OWNER to postgres;

我对这个框架有点陌生，所以可能错误很明显，但我看不到。现在我将展示类和存储库

这是该项目的结构：

申请。属性

spring.datasource.url=jdbc:postgresql://localhost:5432/postgres
spring.datasource.username=postgres
spring.datasource.password=password
spring.jpa.show-sql=true
spring.datasource.driver-class-name=org.postgresql.Driver

## Hibernate Propierties
spring.jpa.properties.hibernate.dialect = org.hibernate.dialect.PostgreSQL94Dialect

# Hibernate auto ddls
spring.jpa.hibernate.ddl-auto = create

学生。爪哇

package net.BBB.ProjectB.entity;


import java.sql.Date;
import java.util.Set;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.Table;
import javax.validation.constraints.NotBlank;
import javax.validation.constraints.NotNull;

@Entity
@Table(name = "students")
public class Student {

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

    @NotBlank(message = "Este campo es obligatorio")
    @Column(name = "nombre")
    private String nombre;

    @NotBlank(message = "Este campo es obligatorio")
    @Column(name = "nacionalidad")
    private String nacionalidad;

    @NotNull(message = "Este campo es obligatorio")
    @Column(name = "edad")
    private long edad;

    @NotBlank(message = "Este campo es obligatorio")
    @Column(name = "rut")
    private String rut;

    @NotBlank(message = "Este campo es obligatorio")
    @Column(name = "sexo")
    private String sexo;

    @NotNull(message = "Este campo es obligatorio")
    @Column(name = "fecha_nacimiento")
    private Date fecha_nacimiento;

    @NotBlank(message = "Este campo es obligatorio")
    @Column(name = "domicilio")
    private String domicilio;

    @NotBlank(message = "Este campo es obligatorio")
    @Column(name = "email")
    private String email;

    @NotNull(message = "Este campo es obligatorio")
    @Column(name = "telefono")
    private long telefono;

    @OneToMany(mappedBy = "student", cascade = CascadeType.ALL)
    private Set<ResultadosPsicotecnicos> resultadosPsicotecnicos;

   /* constructors, getters and setters ommited */

不要把Student中的rut和Resultadospicotecnicos中的混淆

结果球囊炎。爪哇

package net.BBB.ProjectB.entity;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;

@Entity
@Table(name = "Resultados_Psicotecnicos")
public class ResultadosPsicotecnicos {

    @Id
    @Column(name = "id")
    private long id;

    @ManyToOne
    @JoinColumn(name = "rut", nullable = false, referencedColumnName = "id")
    private Student student;

    private String fecha_evaluacion;

    private String nota1;

    private String nota2;

    private String nota3;

    private String nota4;

    public ResultadosPsicotecnicos() {}

    public ResultadosPsicotecnicos(Student student) {
        this.student = student;
    }

    /* getters and setters ommited */

结果：ospicotecnicose阳性。爪哇

package net.BBB.ProjectB.repository;

import java.util.List;

import org.springframework.data.jpa.repository.EntityGraph;
import org.springframework.data.jpa.repository.EntityGraph.EntityGraphType;
import org.springframework.data.repository.CrudRepository;
import org.springframework.stereotype.Repository;

import net.BBB.ProjectB.entity.ResultadosPsicotecnicos;

@Repository
public interface ResultadosPsicotecnicosRepository extends CrudRepository<ResultadosPsicotecnicos, Long> {

    @EntityGraph(value = "Student.resultadosPsicotecnicos", type = EntityGraphType.FETCH)
    List<ResultadosPsicotecnicos> findByRut(long rut);

}

最后是导致错误的自定义方法。现在我将使用学生的id，稍后如果我解决了这个问题，我将使用rut，但是现在外键的列名为rut，它使用学生的id

欢迎任何帮助

共 (1) 个答案

# 1 楼答案
实际上，JPA正试图在ResultadosPsicotecnicos中找到列rut，正如您在ResultadOSpicoteCnicosRespository中指定的findByRut一样
1. 按如下所示修改Resultadoscsicotecnicosrepository中的findByRut，因为您需要在jpa存储库中指定实体的对象名，而不是db列名列出findByStudent_id（long studentid）
2. 或者，您可以尝试以下解决方案。要获得特定学生的结果分类列表，您需要在StudentRepository中添加以下方法
@EntityGraph（value=“Student.resultadospicotecnicos”，type=EntityGraphType.FETCH）学生findById（长id）；请在学生课的最上面加上下面一行 @NamedEntityGraph（name=“Student.resultadospicotencicos”，attributeNodes={@NamedAttributeNode（“resultadospicotencos”）}）

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Crudepository（spring boot）中自定义方法的java问题

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