Spring Data JPA@Query-选择最大值

Question

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Query
import org.springframework.data.repository.query.Param;
import org.springframework.stereotype.Repository;

@Repository
interface IntermediateInvoiceRepository extends JpaRepository<Invoice, String> {

@Query("SELECT max(i.sequence) " +
        "FROM Invoice as i " +
        "WHERE i.fleetId = :fleetId" +
        "   AND i.sequence IS NOT NULL")
Long findMaxSequence(@Param("fleetId") String fleetId);

}

如何在JPA2.0中使用where子句编写MAX查询？

错误是：

2018-09-14T09:27:57,180Z  [main] ERROR o.s.boot.SpringApplication     - Application startup failed
org.springframework.data.mapping.PropertyReferenceException: No property findMaxSequence found for type Invoice!

发票类别（为简洁起见进行了简化）：

@Entity
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
@Table(name = "invoices", indexes = {
        @Index(name = "IDX_FLEET", columnList = "fleetId", unique = false)
        ,
        @Index(name = "IDX_USERSSS", columnList = "userId", unique = false)
        ,
        @Index(name = "IDX_TIME", columnList = "invoiceDate", unique = false)
        ,
        @Index(name = "IDX_SEQUENCE", columnList = "sequence", unique = false)
})
@JsonIgnoreProperties(ignoreUnknown = true)
public class Invoice implements Serializable {

    private static final long serialVersionUID = 1L;

    @GeneratedValue(generator = "uuid")
    @GenericGenerator(name = "uuid", strategy = "uuid2")
    @Column(columnDefinition = "CHAR(36)")
    @Id
    private String id;

    @Column
    private long sequence;

>

可能是findOne在序列列上使用DESC排序的解决方案吗？

@query（“Select i.sequence”+“from Invoice as i”+“，其中i.fleetId=：fleetId”+“Order BY i.sequence DESC”）长getMaxSequence(@param（“fleetId”）String fleetId)；

但我需要以某种方式将resultset限制为1

Answer 1

由于您使用的是JPA存储库，请使用：

org.springframework.data.jpa.repository.Query

注释而不是

org.springframework.data.mongodb.repository.Query

您可以创建查询方法，而不使用@query注释，如：