Question

我在Wildfly 12中使用Java EE。我有以下SQL表格方案：

CREATE TABLE AUTHOR ("ID" INTEGER primary key, "FIRSTNAME" VARCHAR(50) not null, "SECONDNAME" VARCHAR(50) not null);
CREATE TABLE BOOK ("ID" INTEGER primary key, "TITLE" VARCHAR(50) not null, "AUTHOR" INTEGER, FOREIGN KEY ("AUTHOR") REFERENCES AUTHOR("ID"));

我想执行以下插入：

INSERT INTO AUTHOR("SECONDNAME","FIRSTNAME") values ('a', 'b');
INSERT INTO BOOK("TITLE","AUTHOR") values ('bookTitle', 1);

我的persistence.xml：

<property name="hibernate.hbm2ddl.auto" value="create-drop" />
<property name="hibernate.show_sql" value="true" />
<property name="hibernate.dialect" value="org.hibernate.dialect.H2Dialect"/>
<property name="javax.persistence.schema-generation.create-source" value="script"/>
<property name="javax.persistence.schema-generation.drop-source" value="script"/>
<property name="javax.persistence.schema-generation.create-script-source" value="META-INF/create.sql"/>
<property name="javax.persistence.schema-generation.drop-script-source" value="META-INF/drop.sql"/>
<property name="javax.persistence.sql-load-script-source" value="META-INF/load.sql"/>

当我显式插入ID时，这是有效的。但是，必须自动生成ID。我的实体类：

@Entity
@XmlRootElement
public class Author {

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

// rest of fields, getters, and setters omitted

}

@Entity
@XmlRootElement
public class Book {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Long id;
// Again, rest of the fields, getters and setters omitted
}

我尝试了所有代策略（AUTO，IDENTITY，SEQUENCE）。正如我所说的，如果我在SQL中显式设置ID，那么SQL可以工作，但在使用entityManager.persist时会遇到同样的问题......

@POST
@Path("/create")
public void createAuthor(Author author) {
    entityManager.persist(author);
}

以上所有内容都给了我：

Caused by: org.h2.jdbc.JdbcSQLException: NULL not allowed for column "ID"; SQL statement: ...

任何帮助？

Answer 1

您只是声明ID列是INTEGER和PRIMARY KEY。您从未设置该列应自动递增，因此错误消息是正确的。将您的DDL更改为：

CREATE TABLE AUTHOR (
    "ID" INTEGER primary key auto_increment,
    "FIRSTNAME" VARCHAR(50) not null,
    "SECONDNAME" VARCHAR(50) not null
);

CREATE TABLE BOOK (
    "ID" INTEGER primary key auto_increment,
    "TITLE" VARCHAR(50) not null,
    "AUTHOR" INTEGER, FOREIGN KEY ("AUTHOR") REFERENCES AUTHOR("ID")
);

列“ID”H2 Hibernate

1 个答案: