java - SqlExceptionHelper.logExceptions int 类型的错误值 : WRAP

Question

您好，我正在关注《行动》第 5 版书中的 spring，当我想从成分表中获取参数时遇到问题。在以下方法中，我尝试使用成分Repository.findAll() 填充列表:

@GetMapping
public String showDesignForm(Model model){
    List<Ingredient> ingredients = new ArrayList<>();
    ingredientRepository.findAll().forEach(ingredients::add);

    Type[] types = Ingredient.Type.values();
    for (Type type : types){
        model.addAttribute(type.toString().toLowerCase(),
                filterByType(ingredients, type));
    }
    model.addAttribute("taco", new Taco());
    return "welcomePage";
}

但发现以下错误:

org.apache.catalina.core.StandardWrapperValve.invoke Servlet.service() for servlet [dispatcher] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.dao.DataIntegrityViolationException: could not execute query; SQL [select ingredient0_.id as id1_0_, ingredient0_.name as name2_0_, ingredient0_.type as type3_0_ from Ingredient ingredient0_]; nested exception is org.hibernate.exception.DataException: could not execute query] with root cause
    org.postgresql.util.PSQLException: Bad value for type int : WRAP

这就是我的成分实体的外观:

@Data
@RequiredArgsConstructor
@NoArgsConstructor(access=AccessLevel.PRIVATE, force=true)
@Entity
public class Ingredient {
    @Id
    @Column(name = "id", columnDefinition = "varchar(4)")
    private final String id;
    @Column(columnDefinition = "varchar(25)")
    private final String name;
    @Column(columnDefinition = "varchar")
    private final Type type;
    public static enum Type {
        WRAP, PROTEIN, VEGGIES, CHEESE, SAUCE
    }
}

我该如何解决这个问题？提前致谢

Answer 1

使用 JPA 的枚举可以通过 @Enumerated 注释使用不同的获取方法。

默认情况下，这些值似乎是通过序数值获取的，这就是您收到此错误的原因，因为您将值存储为 string 。

尝试以下操作，告诉 JPA 您的枚举值将保留为 String :

@Column(columnDefinition = "varchar")
@Enumerated(EnumType.STRING)
private final Type type;