Question

要求使用SQL和/或JPQL语句检索按日期（降序）排序的给定帐户的最新记录。它应该为所有用户返回一条记录（最新记录），例如如果“用户a”从“用户b”，“用户c”和“用户d”发送/接收多条消息，它将返回每个用户的最新记录（共3条）。

我对JPQL的尝试：

SELECT DISTINCT dm FROM DirectMessage dm WHERE (dm.messageFrom.id=1 OR dm.messageTo.id=1)  GROUP BY dm.id, dm.messageFrom.id, dm.messageTo.id ORDER BY dm.date DESC

具有架构的具体示例：

如果我希望返回“用户a”的最新记录，则返回的正确记录数将为3（以下这3条记录的每条插入语句）：

1.INSERT INTO direct_message VALUE(2, 2, 1, 'hello back', now());
2.INSERT INTO direct_message VALUE(4, 3, 1, 'hello back', now());
3.INSERT INTO direct_message VALUE(6, 4, 1, 'hello back', now());

脚本

CREATE TABLE account(
    id bigint UNSIGNED NOT NULL auto_increment,
    username varchar(75) NOT NULL,
    name varchar(70) NOT NULL,
    date datetime NOT NULL,
    CONSTRAINT pkey_account_id PRIMARY KEY(id)
);

INSERT INTO account VALUES (1,'user_a','User A', now());
INSERT INTO account VALUES (2,'user_b','User B', now());
INSERT INTO account VALUES (3,'user_c','User C', now());
INSERT INTO account VALUES (4,'user_d','User D', now());



    CREATE TABLE direct_message(
        id bigint UNSIGNED NOT NULL auto_increment,
        message_from_id bigint UNSIGNED NOT NULL,
        message_to_id bigint UNSIGNED NOT NULL,
        message_text varchar(200) NOT NULL,
        date datetime NOT NULL,
        CONSTRAINT pkey_dm_id PRIMARY KEY(id),
        CONSTRAINT fk_direct_message_to FOREIGN KEY (message_to_id) REFERENCES account(id),
        CONSTRAINT fk_message_from FOREIGN KEY (message_from_id) REFERENCES account(id)
    );

    INSERT INTO direct_message VALUE(1, 1, 2, 'hello', now());
    INSERT INTO direct_message VALUE(2, 2, 1, 'hello back', now());
    INSERT INTO direct_message VALUE(3, 1, 3, 'hello', now());
    INSERT INTO direct_message VALUE(4, 3, 1, 'hello back', now());
    INSERT INTO direct_message VALUE(5, 1, 4, 'hello', now());
    INSERT INTO direct_message VALUE(6, 4, 1, 'hello back', now());
    INSERT INTO direct_message VALUE(7, 2, 4, 'hello', now());
    INSERT INTO direct_message VALUE(8, 4, 2, 'hello back', now());

Answer 1

我想你想要

select dm.*
from direct_message dm
where @userid in (dm.message_from_id, dm.message_to_id) and
      dm.date = (select max(dm2.date)
                 from direct_message dm2
                 where (dm2.message_from_id, dm2.message_to_id) in ( (dm.message_from_id, dm.message_to_id), (dm.message_to_id, dm.message_from_id) )
                );

编辑：

有不同的消息具有完全相同的日期/时间戳似乎很奇怪，但是您可以使用id来区分它们：

select dm.*
from direct_message dm
where 1 in (dm.message_from_id, dm.message_to_id) and
      dm.id = (select dm2.id
               from direct_message dm2
               where (dm2.message_from_id, dm2.message_to_id) in ( (dm.message_from_id, dm.message_to_id), (dm.message_to_id, dm.message_from_id) )
               order by dm2.date desc
               limit 1
              );

SQL和/或JPQL语句检索按日期desc排序的单个记录

1 个答案: