spring-boot - 有没有办法为 JPA 存储库 : findByEnumContaining(String enum) [enum is subString for possible ENUM values] 添加抽象方法

Question

这是我的 JPA @Repository，在这里我们可以得到 list<Person>与 findByFullNameContaining(String query) - 通过在查询中提供 fullName 的子字符串

@Repository
public interface PersonRepository extends CrudRepository<Person,String> {
    Optional<Person> findByFullName(String fullName);
    List<Person> findByDepartment(Department department);
    List<Person> findByFullNameContaining(String query);
    
}

类似地，我们可以通过提供 ENUM 的子字符串值来为 ENUM 值执行某些操作吗？怎么办？

例如。

public enum Department {
    NORTH,
    SOUTH,
    EAST,
    WEST
}

List<Person> findByDepartmentContaining(String query);

JPA @Entity 人:

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.EnumType;
import javax.persistence.Enumerated;
import javax.persistence.Id;
import javax.persistence.Table;
import javax.persistence.Transient;
import javax.validation.constraints.NotNull;
import javax.validation.constraints.Size;

@Entity
@Table(name = "Person")
public class Person {

    @Id 
    @NotNull(message = "Mobile number is required")
    @Size(min = 10, max = 10, message = "Mobile no. must be 10 digits")
    @Column(name = "person_id", unique = true, length = 10)
    private String mobileNum;

    @Transient
    @NotNull(message = "Password is required")
    @Size(min = 1, message = "Password cannot be empty")
    private String password="****";

    @NotNull(message = "Name cannot be empty")
    @Size(min = 1, max = 255, message = "fullName must be 1-255 characters long")
    @Column(name = "full_name")
    private String fullName;

    @Column(name = "department")
    @Enumerated(EnumType.STRING)
    @NotNull(message = "Department must be specified")
    private Department department = Department.UNKNOWN;

    public Person() {
    }

    public Person(String mobileNum, String fullName, String password, Department department) {
        this.mobileNum = mobileNum;
        this.password = password;
        this.fullName = fullName;
        this.department = department;
    }

    public String getMobileNum() {
        return mobileNum;
    }

    public void setMobileNum(String mobileNum) {
        this.mobileNum = mobileNum;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public String getFullName() {
        return fullName;
    }

    public void setFullName(String fullName) {
        this.fullName = fullName;
    }

    public Department getDepartment() {
        return department;
    }

    public void setDepartment(Department department) {
        this.department = department;
    }

    @Override
    public String toString() {
        return "Person [fullName=" + fullName + ", mobileNum=" + mobileNum + ", password=" + password + ", department=" + department + "]";
    }

}

Answer 1

当声明这样的方法时，这个问题没有具体说明问题是什么。

但是尝试使用 spring boot 2.7 和 PostgreSql 数据库应用程序抛出以下运行时错误:

java.lang.IllegalArgumentException: Parameter value [%some value%] did not match expected type [com.Department (n/a)]
    at org.hibernate.query.spi.QueryParameterBindingValidator.validate(QueryParameterBindingValidator.java:54) ~[hibernate-core-5.6.5.Final.jar:5.6.5.Final]

至少 hibernate 的问题是它期望实体中的 Department 字段始终作为参数传递给它自己的对象类型，即 Department。

我认为没有办法避免这种情况，因为这是 Hibernate 的开箱即用功能。

不过，我认为正确的方法不是定义这种类型的方法，而是定义以下方法。 Department 已经存在于应用程序代码中，因此在需要调用查询时它是已知的。所以我认为以下解决方案将被视为更好的做法:

//Repository method to be used in PersonRepository
List<Person> findByDepartmentIn(List<Department> departments); 

然后可以通过以下方式从服务层调用存储库。

//Service method to be used
public List<Person> findByDepartmentIn(String searchBy) {

    List<Department> departments = Arrays.stream(Department.values()).filter(dep -> dep.getName().contains(searchBy)).collect(Collectors.toList());

    return personRepository.findDepartmentIn(departments);
}