java - Spring jpa使用自定义Query时抛出ClassCastException

Question

我试图从查询中仅获取特定的列集，但是，我收到了 ClassCastException。

2019-03-16 00:03:19.321 ERROR 8440 --- [nio-8080-exec-1] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to com.metadata.entity.FileItem] with root cause

java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to com.metadata.entity.FileItem
    at com.metadata.service.FileItemService.findFileWithVersion(FileItemService.java:39) ~[classes/:na]
    at com.metadata.service.FileItemService$$FastClassBySpringCGLIB$$7a805cbb.invoke(<generated>) ~[classes/:na]
    at org.springframework.cglib.proxy.MethodProxy.invoke(MethodProxy.java:218) ~[spring-core-5.1.4.RELEASE.jar:5.1.4.RELEASE]

这是我的结构的基础知识。

文件项

import lombok.AllArgsConstructor;
import lombok.Builder;
import lombok.Data;
import lombok.NoArgsConstructor;
import org.hibernate.annotations.CreationTimestamp;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.IdClass;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
import java.util.Date;
import java.util.UUID;

@Entity
@IdClass(FileItemId.class)
@Table(name = "files")
@AllArgsConstructor
@NoArgsConstructor
@Data
public class FileItem {

    @Id
    private UUID id;

    @Id
    private long version;

    @Column(name = "total_size")
    private long totalSize;

    private String name;
    private byte[] file;

    @CreationTimestamp
    @Temporal(TemporalType.TIMESTAMP)
    @Column(name = "created_at")
    private Date createdAt;
}

FileItemId.java

import lombok.Data;
import java.io.Serializable;
import java.util.UUID;

@Data
public class FileItemId implements Serializable {
    private UUID id;
    private Long version;
}

FileItemRepository

import com.metadata.entity.FileItem;
import com.metadata.entity.FileItemId;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Modifying;
import org.springframework.data.jpa.repository.Query;

import java.util.List;
import java.util.UUID;

public interface FileItemRepository extends JpaRepository<FileItem, UUID> {
    @Query("select f.id, f.version, f.name, f.totalSize, f.createdAt " +
            "from FileItem f " +
            "where f.id = ?1 and f.version = ?2")
    List<FileItem> findByFileIdAndVersion(UUID id, Long version);

}

Answer 1

我已经很长时间没有使用JPA了，但是你的查询似乎返回了一个数组Object[]

那么你的方法的签名应该是:

@Query("select f.id, f.version, f.name, f.totalSize, f.createdAt " +
            "from FileItem f " +
            "where f.id = ?1 and f.version = ?2")
    Object[] findByFileIdAndVersion(UUID id, Long version);

否则，如果您想保留方法签名:

@Query("select f " +
            "from FileItem f " +
            "where f.id = ?1 and f.version = ?2")
    List<FileItem> findByFileIdAndVersion(UUID id, Long version);

另一个解决方案:

@Query("select new com.mycompany.mypackage.MyDTO(f.id, f.version, f.name, f.totalSize, f.createdAt) " +
            "from FileItem f " +
            "where f.id = ?1 and f.version = ?2")
    MyDTO findByFileIdAndVersion(UUID id, Long version);

其中 MyDTO 是您根据需要编写的自定义 Java bean。