Php SQL仅在两个用户之间检索结果

我需要检索两个用户之间的对话，但我不知道如何。。。我的结果有不必要的对话，我当前的sql:

select *
from user_talk t1
inner join user_talk t2 on t2.id = t1.id
where
    t1.talk_id = t2.talk_id and
    (
        t1.user_id = 1 or
        t2.user_id = 4
    );

此sql显示用户1和4之间的结果，但也显示用户1和用户2之间的结果，以及用户2和65之间的结果，但我只需要用户1和4

如何编写正确的sql

编辑：用户对话

CREATE TABLE IF NOT EXISTS `reverse`.`user_talk` (
  `id` BIGINT UNSIGNED NOT NULL AUTO_INCREMENT,
  `talk_id` INT(10) UNSIGNED NOT NULL,
  `user_id` INT(10) UNSIGNED NOT NULL,
  `type` VARCHAR(10) NOT NULL DEFAULT 'INVITED',
  `status` VARCHAR(10) NOT NULL DEFAULT 'ACTIVE',
  `created_at` TIMESTAMP NULL,
  `updated_at` TIMESTAMP NULL,
  PRIMARY KEY (`id`),
  INDEX `fk_usertalk_1_idx` (`user_id` ASC),
  INDEX `fk_usertalk_2_idx` (`talk_id` ASC),
  INDEX `user_in_talk_idx` (`talk_id` ASC, `user_id` ASC),
  CONSTRAINT `fk_usertalk_1`
    FOREIGN KEY (`user_id`)
    REFERENCES `reverse`.`user` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_usertalk_2`
    FOREIGN KEY (`talk_id`)
    REFERENCES `reverse`.`talk` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB

谈话

CREATE TABLE IF NOT EXISTS `reverse`.`talk` (
  `id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
  `created_at` TIMESTAMP NULL,
  `updated_at` TIMESTAMP NULL,
  PRIMARY KEY (`id`))
ENGINE = InnoDB

消息

CREATE TABLE IF NOT EXISTS `reverse`.`message` (
  `id` BIGINT UNSIGNED NOT NULL AUTO_INCREMENT,
  `talk_id` INT(10) UNSIGNED NOT NULL,
  `sender_id` INT(10) UNSIGNED NOT NULL,
  `receiver_id` INT(10) UNSIGNED NOT NULL,
  `body` TEXT NOT NULL,
  `status` VARCHAR(10) NOT NULL DEFAULT 'UNREAD',
  `created_at` TIMESTAMP NULL,
  `updated_at` TIMESTAMP NULL,
  PRIMARY KEY (`id`),
  INDEX `fk_message_1_idx` (`sender_id` ASC),
  INDEX `fk_message_2_idx` (`receiver_id` ASC),
  INDEX `fk_message_3_idx` (`talk_id` ASC),
  CONSTRAINT `fk_message_1`
    FOREIGN KEY (`sender_id`)
    REFERENCES `reverse`.`user` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_message_2`
    FOREIGN KEY (`receiver_id`)
    REFERENCES `reverse`.`user` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_message_3`
    FOREIGN KEY (`talk_id`)
    REFERENCES `reverse`.`talk` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB

（请试试这个

select *
  from user_talk t1
       inner join user_talk t2 on t1.id = t2.id
 where (t1.user_id = 1 or t1.user_id = 4)
   and (t2.user_id = 1 or t2.user_id = 4);

更新：只需对下面显示的数据运行此SQL

并且得到了这个reult，只适用于

user\u id

1和4

---

选择用户对话，从用户1到用户4以及从用户4到用户1：

SELECT *
FROM user_talk t1
INNER JOIN user_talk t2 ON t1.talk_id = t2.talk_id
WHERE 
    (t1.user_id = 1 AND t2.user_id = 4)
    OR
    (t1.user_id = 4 AND t2.user_id = 1)

我解释：

• 您需要从2个用户中选择user_talk，因此我从2个表中选择
• 你需要选择对话，所以我想对话id应该是相同的
• 从用户1到4或从用户4到1需要它们

---

描述您的模式。您是否也只需要对话或用户数据？@FLefèvre查看表sql。目前只有用户对话数据。谢谢！请描述您的

对话

表。但我认为我的答案应该有效。是的，您的答案很好。谢谢！sql结果与我的答案中的sql相同。