为什么 PSQL 会抱怨未知关系:org.postgresql.util.PSQLException:错误:关系“benannte_person”不存在?

如何解决为什么 PSQL 会抱怨未知关系:org.postgresql.util.PSQLException:错误:关系“benannte_person”不存在?

当我尝试使用 Testcontainers 运行 SpringBootTest 时,我得到一个

org.springframework.dao.InvalidDataAccessResourceUsageException: could not extract ResultSet; SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet
...
org.postgresql.util.PSQLException: ERROR: relation "benannte_person" does not exist

当我在 DAO 上运行 findAll() 并且我不知道原因时,因为在我的整个代码中找不到“benanntee_person”(甚至在注释中也找不到)。

SQL DDL:

CREATE TYPE Erklaerungstyp AS ENUM ('aaaa','bbbb','cccc','dddd');
CREATE TYPE Geschlecht AS ENUM ('D','F','M');

DROP TABLE IF EXISTS Anschrift;
CREATE TABLE Anschrift (
  a_id          SERIAL PRIMARY KEY,Zusatz        VARCHAR(255),Strasse       VARCHAR(30) NOT NULL,Hausnummer    VARCHAR(30) NOT NULL,plz           VARCHAR(5) NOT NULL,Ort           VARCHAR(80) NOT NULL,Bundesland    VARCHAR(20),Land          VARCHAR(20) NOT NULL,create_Date   DATE NOT NULL,modify_Date   DATE
);
INSERT INTO Anschrift VALUES (1,null,'Musterstrasse','13M','12345','Berlin','Deutschland','2001-09-28');
INSERT INTO Anschrift VALUES (2,'bei Müller','Musterweg','1-3','54321','Musterhausen','Muster-Hausen','2002-03-11');

DROP TABLE IF EXISTS ErklaerendePerson;
CREATE TABLE ErklaerendePerson (
  ep_id                 SERIAL PRIMARY KEY,Geschlecht            Geschlecht,Vorname               VARCHAR(30) NOT NULL,Familienname          VARCHAR(30) NOT NULL,Geburtsname           VARCHAR(30) NOT NULL,Titel                 VARCHAR(10),Geburtsdatum          Date NOT NULL,Geburtsort            VARCHAR(30),Anschrift             INTEGER REFERENCES Anschrift(a_id),Email                 VARCHAR(80),Telefon               VARCHAR(20),create_Date           DATE  NOT NULL,modify_Date           DATE
);
INSERT INTO ErklaerendePerson VALUES (1,'M','Max','Mustermann','Dipl.-Inf.','01.01.1901',1,'Max.Mustermann@max.de','0111 12 34 56 789','2001-09-28');

DROP TABLE IF EXISTS BenanntePerson;
CREATE TABLE BenanntePerson (
  bp_id         SERIAL PRIMARY KEY,Geschlecht    Geschlecht,Vorname       VARCHAR(30) NOT NULL,Familienname  VARCHAR(30) NOT NULL,Geburtsdatum  Date NOT NULL,Geburtsort    VARCHAR(30),Anschrift     INTEGER REFERENCES Anschrift(a_id),Telefon       VARCHAR(20),Email         VARCHAR(80),modify_Date   DATE
);
INSERT INTO BenanntePerson VALUES (1,'Maxine','Musterpaar','12.07.1971','0111 444 55 66','Maxine.Musterpaar@xxx.de','2001-09-28');

DROP TABLE IF EXISTS Erklaerung;
CREATE TABLE Erklaerung (
  ose_id            SERIAL PRIMARY KEY,ep_id             INTEGER REFERENCES ErklaerendePerson(ep_id),bp_id             INTEGER REFERENCES BenanntePerson(bp_id),Anmerkung         VARCHAR(120),Erklaerungstyp    Erklaerungstyp NOT NULL,create_Date       DATE NOT NULL,modify_Date       DATE
);
INSERT INTO Erklaerung VALUES (1,'blablabla','VOLLUMFASSEND','2020.02.20');
COMMIT;

实体BenanntePerson

@Entity
@Table(name="BenanntePerson")
@Data
public class BenanntePerson implements Serializable {
    @Id
//    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "bp_id")
    private Integer bpId;

    @Column(name = "Vorname",nullable = true)
    private String vorname;

    @Column(name = "Familienname",nullable = true)
    private String familienname;

    @Temporal(TemporalType.DATE)
    @Column(name = "Geburtsdatum",nullable = false)
    private Date geburtsdatum;

    @Column(name = "Geburtsort",nullable = false)
    private String geburtsort;

    @OneToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "a_id",nullable = false)
    private Anschrift anschrift;

    @Column(name = "Email",nullable = false)
    private String email;

    @Column(name = "Telefonnummer",nullable = false)
    private String telefonnummer;

    @OneToMany(mappedBy = "benanntePerson")
    private Set<Erklaerung> erklaerungen;

    @Column(name = "create_Date",insertable = true,updatable = false,nullable = false)
    private Timestamp createDate;

    @Column(name = "modify_Date",nullable = false)
    private Timestamp modifyDate;
}

关系的 Erklaerung 实体:

@Entity
@Table(name="Erklaerung")
@Data
public class Erklaerung implements Serializable {
    @Id
    //    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "ose_id")
    private Integer oseId;

    @OneToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "ep_id",nullable = false)
    private ErklaerendePerson erklaerendePerson;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "bp_id",nullable = false)
    private BenanntePerson benanntePerson;

    @Enumerated(value = EnumType.STRING)
    @Column(name = "Erklaerungstyp",nullable = false)
    private Erklaerungstyp erklaerungstyp;

    @Column(name = "Anmerkung",nullable = true)
    private String anmerkung;

    @Column(name = "create_Date",nullable = false)
    private Timestamp modifyDate;
}

对应的JPA DAO:

@Repository
public interface BenanntePersonJpaDao extends JpaRepository<BenanntePerson,Integer> {
    @NotNull
    List<BenanntePerson> findByFamilienname(@NotNull String familienname);
    @NotNull
    List<BenanntePerson> findByVorname(@NotNull String vorname);
}

为什么 PSQL 会抱怨??

解决方法

这只是因为 @JoinColumn(name = "a_id",nullable = false) 与数据库表不匹配,因为没有 'a_id' 列。它在表定义中称为“Anschrift”。因此,在表定义中将“Anschrift”重命名为“a_id”可以解决问题。

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