好的，我还尝试使用以下方法执行此操作
@IdClass：
@Entity@Table(name = "my_entity")@JsonIgnoreProperties(ignoreUnknown = true)@IdClass(Pk.class)public class MyEntity { @Id private Integer type; @Id private String userId; public Pk getId() { return id; } public void setId(Pk id) { this.id = id; }}@IdClass(Pk.class)public class Pk implements Serializable { private static final long serialVersionUID = -3090221844117493661L; private Integer type; private String userId; public Pk() { } public Pk(String userId, Integer type) { this.setUserId(userId); this.setType(type); } public Integer getType() { return type; } public String getUserId() { return userId; } public void setType(Integer type) { this.type = type; } public void setUserId(String userId) { this.userId = userId; } // Auto-generated by Eclipse. @Override public int hashCode() { final int prime = 31; int result = 1; result = prime * result + ((type == null) ? 0 : type.hashCode()); result = prime * result + ((userId == null) ? 0 : userId.hashCode()); return result; } // Auto-generated by Eclipse. @Override public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; Pk other = (Pk) obj; if (type != other.type) return false; if (userId == null) { if (other.userId != null) return false; } else if (!userId.equals(other.userId)) return false; return true; }}这确实给了我一个不同的错误消息a
NullPointerException，这向我暗示Spring Data
JPA无法为构建查询
findAllByUserId(...)。我改为对该查询方法进行了自定义实现：
public interface MyEntityRepository extends JpaRepository<MyEntity, Pk>, MyEntityRepositoryCustom {}public interface MyEntityRepositoryCustom { List<MyEntity> findAllByUserId(String userId);}public class MyEntityRepositoryImpl implements MyEntityRepositoryCustom { @PersistenceContext private EntityManager em; @Override public List<MyEntityRepositoryCustom> findAllByUserId(String userId) { return em .createQuery("select o from MyEntity o where o.userId=:userId", MyEntity.class).setParameter("userId", userId).getResultList(); }}…而且， 它起作用了！
