MySQL查询以获取用户未加入的所有社区

我想获得用户尚未加入的所有社区（基于不同的ZIP）

我有一个用户表和其他几个类似的表：

select *
from neighborhood
where neighborhood_id NOT IN (select neighborhood_id from user_neighborhood where user_id != 1)
AND neighborhood_postal_code IN ('2000', '2100')

表名：邻里

CREATE TABLE neighborhood(
    `neighborhood_id` INT(11) NOT NULL AUTO_INCREMENT,
    `name` VARCHAR(255) NOT NULL,
    `description` TEXT DEFAULT NULL,
    `neighborhood_postal_code` VARCHAR(255) NOT NULL,
    `region_neighborhood` VARCHAR(255) NOT NULL,
    `created_at` DATETIME DEFAULT CURRENT_TIMESTAMP,
    PRIMARY KEY (`neighborhood_id`),
    INDEX `neighborhood_region_neighborhood_FI_1` (`region_neighborhood`)
) ENGINE = InnoDB;

CREATE TABLE user_neighborhood(
    `user_id` INT(11) NOT NULL,
    `neighborhood_id` INT(11) NOT NULL,
    `activity_circle` INT(1) DEFAULT 0,
    `duo_circle` INT(1) DEFAULT 0,
    FOREIGN KEY (`user_id`) REFERENCES `users` (`user_id`),
    FOREIGN KEY (`neighborhood_id`) REFERENCES `neighborhood` (`neighborhood_id`)
) ENGINE = InnoDB;

表名：user\u邻域

CREATE TABLE neighborhood(
    `neighborhood_id` INT(11) NOT NULL AUTO_INCREMENT,
    `name` VARCHAR(255) NOT NULL,
    `description` TEXT DEFAULT NULL,
    `neighborhood_postal_code` VARCHAR(255) NOT NULL,
    `region_neighborhood` VARCHAR(255) NOT NULL,
    `created_at` DATETIME DEFAULT CURRENT_TIMESTAMP,
    PRIMARY KEY (`neighborhood_id`),
    INDEX `neighborhood_region_neighborhood_FI_1` (`region_neighborhood`)
) ENGINE = InnoDB;

CREATE TABLE user_neighborhood(
    `user_id` INT(11) NOT NULL,
    `neighborhood_id` INT(11) NOT NULL,
    `activity_circle` INT(1) DEFAULT 0,
    `duo_circle` INT(1) DEFAULT 0,
    FOREIGN KEY (`user_id`) REFERENCES `users` (`user_id`),
    FOREIGN KEY (`neighborhood_id`) REFERENCES `neighborhood` (`neighborhood_id`)
) ENGINE = InnoDB;

我尝试了以下查询，但结果不正确：

SELECT n.*
FROM `neighborhood` as n
left join user_neighborhood as un on n.neighborhood_id = un.neighborhood_id
where un.user_id != 1 and n.neighborhood_postal_code IN ('2000', '2100')

更新：我第一次使用如下子查询设法使查询看起来正确：

select *
from neighborhood
where neighborhood_id NOT IN (select neighborhood_id from user_neighborhood where user_id != 1)
AND neighborhood_postal_code IN ('2000', '2100')

---

然而，它也返回（一些）我已经在的社区。我不太明白为什么只有一些

你到底为什么要添加用户id！=子查询中的1？我认为，如果您知道要获取的用户的id（假设user_id为10），则在子查询中使用where user_id=10，如：

select *
from neighborhood
where neighborhood_id NOT IN (select distinct neighborhood_id from user_neighborhood where user_id = 10)
AND neighborhood_postal_code IN ('2000', '2100')

但如果要获取所有没有用户的邻居，则可以使用以下查询：

select *
from neighborhood
where neighborhood_id NOT IN (select distinct neighborhood_id from user_neighborhood)
AND neighborhood_postal_code IN ('2000', '2100')

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希望这有帮助

这是可行的，它完全有道理，它必须在id=而不是！=。。为了这个，我伤了头太久了。谢谢