Spring JPA 存储库无法解析属性实体

Question

我正在尝试对 table 的两列进行简单搜索。

我的客户实体 class 片段：

@Entity
@Table(name = "CUSTOMER")
public class Customer {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "CUSTOMER_ID", nullable = false)
    private int customer_id;

    @Column(name = "FIRST_NAME", nullable = false)
    private String first_name;

    @Column(name = "LAST_NAME", nullable = false)
    private String last_name;

    @Column(name = "PHONE_NUMBER", nullable = false)
    private String phone_number;

    @Column(name = "EMAIL")
    private String email;

我的 customerRepository class 片段：

@Repository
public interface CustomerRepository extends JpaRepository<Customer, Integer> {

    List<Customer> findAllByFirst_nameAndLast_name(String firstName, String lastname);

}

在编译和运行 springBoot 时我得到这个错误：

nested exception is java.lang.IllegalArgumentException: Failed to create query for method public abstract java.util.List com.owl.owlserver.repositories.CustomerRepository.findAllByFirst_nameAndLast_name(java.lang.String,java.lang.String)! No property first found for type Customer!

所以它无法检测到客户对象中的字段，我应该以某种方式将客户 class 导入存储库吗？

Answer 1

我建议按照标准将属性保存在驼峰式外壳中

Because we treat the underscore character as a reserved character, we strongly advise following standard Java naming conventions (that is, not using underscores in property names but using camel case instead).

    @Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "CUSTOMER_ID", nullable = false)
private int customerId;

@Column(name = "FIRST_NAME", nullable = false)
private String firstName;

@Column(name = "LAST_NAME", nullable = false)
private String lastName;

@Column(name = "PHONE_NUMBER", nullable = false)
private String phoneNumber;

@Column(name = "EMAIL")
private String email;

然后按照Property Expressions

查询

List<Customer> findAllByFirstNameAndLastName(String firstName, String lastname);

Answer 2

问题出在你的snake_case命名规范上，建议使用驼峰命名规范。

正如 Spring 文档所说：

Because we treat the underscore character as a reserved character, we strongly advise following standard Java naming conventions (that is, not using underscores in property names but using camel case instead).

您可以看到相关的 JIRA 问题 here -- 属性名称不支持下划线。

Answer 3

SpringJPA 的重点是使用 Repository 中定义的方法来简化查询构建。约定是使用驼峰式命名实体中的字段，在编写查询方法时也是如此。只需将实体字段的命名从 snake case 更改为 camel case 即可。

Spring JPA 存储库无法解析属性实体

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Spring JPA 存储库无法解析 属性 实体

Spring JPA repository cannot resolve property of entity

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Spring JPA 存储库无法解析属性实体