Question

获取错误java.sql.SQLIntegrityConstraintViolationException：ORA-01400：

org.springframework.dao.DataIntegrityViolationException：无法执行语句; SQL [不适用];约束[null];嵌套异常是org.hibernate.exception.ConstraintViolationException：无法执行语句]的根本原因 java.sql.SQLIntegrityConstraintViolationException：ORA-01400：无法插入NULL（＆＃34; APP_LOCAL＆＃34;。＆＃34; PI_ADDRESS＆＃34;。＆＃34; EMP_ID＆＃34;）

以下是代码：

@Entity
@Table(name = "PI_EMPLOYEE")
@NamedQuery(name = "Employee.findAll", query = "SELECT a FROM Employee e")
@DynamicUpdate(value=true)
@DynamicInsert(value=true)
public class Employee{

@Id
@SequenceGenerator(name = "PI_EMPLOYEE_SEQ", sequenceName = 
 "PI_EMPLOYEE_SEQ",allocationSize=1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "PI_EMPLOYEE_SEQ")
@Column(name = "EMP_ID")
private Long employeeId;

@OneToMany(fetch = FetchType.LAZY)
@JoinColumn(name = "EMP_ID")
@BatchSize(size=5)
private List<Address> addresses;

}

@Entity
@Table(name = "PI_ADDRESS")
@DynamicUpdate(value=true)
public class Address {

@Id
@SequenceGenerator(name = "PI_ADDRESS_SEQ", sequenceName = "PI_ADDRESS_SEQ",allocationSize=1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "PI_ADDRESS_SEQ")
@Column(name = "ADDRESS_ID")
private Long addressId;

@Column(name = "EMP_ID")
private Long employeeId;

}   



 public class EmployeeRepositoryImpl implements EmployeeRepository {
 @Override
 public Employee saveOrUpdate(Employee employee) {
    Employee employeeToSave = employee;
    if (employee == null) {
        throw new IllegalArgumentException("Employee should not be null");
    }
    try {
        Address address = employee.getAddress();
        employee.setAddress(null);
        employeeToSave = employeeRepository.save(employee);
        if (null != address) {
            setEmployeeID(employeeToSave.getEmployeeId(), address);
            addressRepository.save(address);
            employeeToSave.setAddress(address);
        }
        return employeeToSave;
    } catch (Exception e) {
        LOGGER.error("Exception occurred while saving Employee", e);
        throw new BaseException(e);
    }

}

private void setEmployeeID(Long employeeId, Object object) {
    try {
        ReflectionUtil.setFieldValue(object, "employeeId", employeeId);
    } catch (NoSuchFieldException | SecurityException | IllegalArgumentException | IllegalAccessException e) {
        LOGGER.error("Exception occured while setting employeeId " + e);
        throw new BaseException(e);
    }
  }

}

Answer 1

我通过在父端放置nullable = false并在子端放置insertable = false和updatable = false并在EmployeeRepositoryImpl中删除样板代码以在地址上设置employeeId来解决此问题。

@Entity
@Table(name = "PI_EMPLOYEE")
@NamedQuery(name = "Employee.findAll", query = "SELECT a FROM Employee e")
@DynamicUpdate(value=true)
@DynamicInsert(value=true)
public class Employee{

@Id
@SequenceGenerator(name = "PI_EMPLOYEE_SEQ", sequenceName = 
"PI_EMPLOYEE_SEQ",allocationSize=1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = 
"PI_EMPLOYEE_SEQ")
 @Column(name = "EMP_ID")
 private Long employeeId;

@OneToMany(fetch = FetchType.LAZY)
@JoinColumn(name = "EMP_ID",nullable =false)
@BatchSize(size=5)
private List<Address> addresses;

}
}

@Entity
@Table(name = "PI_ADDRESS")
@DynamicUpdate(value=true)
public class Address {

@Id
@SequenceGenerator(name = "PI_ADDRESS_SEQ", sequenceName = 
"PI_ADDRESS_SEQ",allocationSize=1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = 
"PI_ADDRESS_SEQ")
@Column(name = "ADDRESS_ID",nullable =false,insertable =false,updatable=false)
private Long addressId;

@Column(name = "EMP_ID")
private Long employeeId;

 }

获取错误java.sql.SQLIntegrityConstraintViolationException：ORA-01400：无法将NULL插入（＆＃34; APP_LOCAL＆＃34;。＆＃34; PI_ADDRESS＆＃34;。＆＃34; EMP_ID＆＃34;）

1 个答案: