Question

我有一个实体产品

@Entity
@Table(name = "Products")
public class ProductEntity implements Serializable{
    @Id
    @GenericGenerator(name = "generator", strategy = "increment")
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Long id;
    @Column(name = "model", nullable = false)
    private String model;
    //setters and getters
}

以下是我尝试按示例查询的方式：

ProductEntity product = new ProductEntity();
product.setId(productId);
Example<ProductEntity> ex = Example.of(product);
Page<ProductEntity> pages = productRepository.findAll(ex, page);

但是我得到PostgreSQL错误： PSQLException：错误：语法错误在或附近＆＃34; L＆＃34;。

此处创建的查询：

Hibernate: select productent0_.id as id1_13_, productent0_.brand_id as brand_id9_13_, productent0_.imgPath as imgPath2_13_, productent0_.model as model3_13_, productent0_.numberOfEvaluation as numberOf4_13_, productent0_.priceBuy as priceBuy5_13_, productent0_.priceSell as priceSel6_13_, productent0_.rating as rating7_13_, productent0_.sumOfEvaluation as sumOfEva8_13_, productent0_.type_id as type_id10_13_ from Products productent0_ where (productent0_.id=1L ) limit ?

systax错误在这里：

where (productent0_.id=1L )

它增加了＆＃39; L＆＃39;长值的字符。那么，我该如何解决这个问题？

更新：有什么问题？仍然无法解决这个问题。

更新：我的存储库：

public interface CrudProductRepository extends JpaRepository<ProductEntity, Long>{
}

JpaRepository的包：

import org.springframework.data.jpa.repository.JpaRepository;

数据JPA。按示例查询。无法使用Long值进行搜索

0 个答案: