查找 2 个日期之间的每条记录都有特定值的所有记录

Question

我有一张桌子，上面有一些可用的度假屋（列value ，值1表示可用）。 我怎样才能找到 2 个日期之间可用的所有房屋（列unit_id ）。

桌子

CREATE TABLE `houseavailability` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `date` varchar(100) DEFAULT NULL,
  `value` varchar(100) DEFAULT NULL,
  `unit_id` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `houseavailability_unit_id_IDX` (`unit_id`,`date`) USING BTREE,
  KEY `houseavailability_unit_id_IDX_solo` (`unit_id`) USING BTREE
) ENGINE=InnoDB AUTO_INCREMENT=16648943 DEFAULT CHARSET=latin1;

测试数据

INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814115, '2022-07-23', '1', '1004004');
INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814116, '2022-07-24', '1', '1004004');
INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814117, '2022-07-25', '1', '1004004');
INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814118, '2022-07-26', '1', '1004004');
INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814119, '2022-07-27', '1', '1004004');
INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814120, '2022-07-28', '1', '1004004');
INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814121, '2022-07-29', '1', '1004004');
INSERT INTO houseavailability
(id, `date`, value, unit_id)
VALUES(15814122, '2022-07-30', '0', '1004004');

试图

SELECT houseavailability.*
FROM houseavailability
WHERE houseavailability.date BETWEEN '2022-07-23' AND '2022-07-30'
AND houseavailability.unit_id = 1004004;

http://sqlfiddle.com/#!9/094547/2

Answer 1

例如，查找在整个指定时间段内可用的unit_id 。

SELECT unit_id
FROM houseavailability
WHERE date BETWEEN '2022-07-23' AND '2022-07-30'
GROUP BY unit_id
HAVING sum(value) = datediff('2022-07-30','2022-07-23') + 1;

Answer 2

您可以尝试在HAVING中使用条件聚合true来比较您的日期条件之间是否所有满足此条件的行。

查询 1 ：

SELECT unit_id
FROM houseavailability
WHERE date BETWEEN '2022-07-23' AND '2022-07-30'
GROUP BY unit_id
HAVING COUNT(DISTINCT date) = COUNT(DISTINCT CASE WHEN value = '1' THEN date END)

结果：

DISTINCT总计 function 将只计算一次，如果重复的日子在您的表中有 1 个值，但如果您在遇到这种情况时想要计算多次，您可以从聚合 function 中删除DISTINCT 。

编辑

由于您的unit_id和date列存在UNIQUE约束，因此您无需对聚合 function 使用DISTINCT 。

SELECT unit_id
FROM houseavailability
WHERE date BETWEEN '2022-07-23' AND '2022-07-30'
GROUP BY unit_id
HAVING COUNT(*) = COUNT(CASE WHEN value = '1' THEN date END)