Question

我想获取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这样的表：

表名：社区

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;

表名：user_neighborhood

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：我设法使用以下子查询使查询在第一次出现时看起来正确：

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')

但是，它也会返回（我）已经在的某些社区。对我来说，为什么只有一些是没有意义的。

Answer 1

为什么要在子查询中精确添加user_id！= 1？我认为，如果您知道要获取的用户ID，可以说user_id为10，则在子查询中使用其中user_id = 10，例如：

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')

但是，如果要获取没有用户的所有邻居，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')

希望这会有所帮助！

MySQL查询以获取用户未加入的所有社区

1 个答案: