JPA CriteriaBuilder子查询多选

Question

我在jpa中有关于Subquery类的问题。我需要创建带有两个自定义字段的子查询，但是子查询没有多选方法，而select方法有表达式输入参数（在查询中这是选择）和constact方法不合适。

关于加入子查询结果我也有疑问，有可能吗？怎么样？

我有：

链Enitity

public class Chain {
@Id
@Column(name = "chain_id")
@GeneratedValue(generator = "seq_cha_id", strategy = GenerationType.SEQUENCE)
@SequenceGenerator(name = "seq_cha_id", sequenceName = "SEQ_CHA_ID", allocationSize = 1)
private Long id;
@Column(name = "user_id")
private Long userId;
@Column(name = "operator_id")
private Long operatorId;

@Column(name = "subject")
private String subject;

@OneToMany(fetch = FetchType.LAZY, mappedBy = "chain")
private List<Message> messages;
@Column(name = "status")
private Status status;

public Long getOperatorId() {

    return operatorId;
}

public void setOperatorId(Long operatorId) {
    this.operatorId = operatorId;
}

public Status getStatus() {

    return status;
}

public void setStatus(Status status) {
    this.status = status;
}

public Long getUserId() {

    return userId;
}

public void setUserId(Long userId) {
    this.userId = userId;
}

public Long getId() {

    return id;
}

public void setId(Long id) {
    this.id = id;
}

public String getSubject() {
    return subject;
}

public void setSubject(String theme) {
    this.subject = theme;
}

public List<Message> getMessages() {
    return messages;
}

public void setMessages(List<Message> messages) {
    this.messages = messages;
}
} 

消息Enitity

public class Message {


@Id
@Column(name = "message_id")
@GeneratedValue(generator = "seq_mess_id", strategy = GenerationType.SEQUENCE)
@SequenceGenerator(name = "seq_mess_id", sequenceName = "SEQ_MESS_ID", allocationSize = 1)
private Long id;

@Column(name = "user_id")
private Long userId;

@Column(name = "message", nullable = true, length = 4000)
private String message;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "chain_id")
private Chain chain;

@Column(name = "creation_date")
private Date date;
@Column(name = "status")
private Status status;

public Long getId() {

    return id;
}

public void setId(Long id) {
    this.id = id;
}

public Long getUserId() {

    return userId;
}

public void setUserId(Long userId) {
    this.userId = userId;
}

public String getMessage() {
    return message;
}

public void setMessage(String message) {
    this.message = message;
}

public Chain getChain() {
    return chain;
}

public void setChain(Chain chain) {
    this.chain = chain;
}

public Date getDate() {
    return date;
}

public void setDate(Date date) {
    this.date = date;
}

public Status getStatus() {
    return status;
}

public void setStatus(Status status) {
    this.status = status;
} 

}

用于查询的包装器

public class MessageWrapper {
    private final Long chainId;
    private final Long messageId;

    public MessageWrapper(Long chainId, Long messageId) {
        this.chainId = chainId;
        this.messageId = messageId;
    }
}

我需要创建这个查询（这是查询的一部分，我从谓词得到的另一部分。JPQL不合适）

select ch.*
from hl_chain ch,
  (select mes.CHAIN_ID,
     max(message_id) message_id
   from HL_MESSAGE mes
   group by chain_id) mes
where mes.chain_id = ch.CHAIN_ID
order by message_id;

在子查询中我做

    Subquery<MessageWrapper> subquery = criteriaQuery.subquery(MessageWrapper.class);
    Root<Message> subRoot = subquery.from(Message.class);
    subquery.select(cb.construct(MessageWrapper.class, subRoot.get(Message_.chain), cb.max(subRoot.get(Message_.id))));

但是，子查询没有在params中使用CompoundSelection进行选择，我不能使用CriteriaBuilder构造方法。

Answer 1

您可以从JPA调用本机查询，例如：

Query q = em.createNativeQuery("SELECT p.firstname, p.lastname FROM Person p");
List<Object[]> persons= q.getResultList();

for (Object[] p : persons) {
    System.out.println("Person "
            + p[0]
            + " "
            + p[1]);
}

Answer 2

CriteriaBuilder cb = session.getCriteriaBuilder();
CriteriaQuery<MessageWrapper> q = cb.createQuery(MessageWrapper.class);
Root<Chain> c = q.from(Chain.class);
Join<Chain, Message> m = p.join("messages");
q.groupBy(c.get("id"));
q.select(cb.construct(MessageWrapper.class, c.get("id"), cb.max(m.get("id"))));