Question

我想计算最近6个月的平均温度。

现在，我有这样的东西：

SELECT AVG(`temp`) FROM `temperature` WHERE YEAR(date) = 2020 AND MONTH(date) = 1 AND `sensor_id` = "00000b858c95"

返回所选月份的平均温度……这是正确的吗？

我的桌子：

CREATE TABLE `temperature` (
  `id` int(11) NOT NULL,
  `sensor_id` varchar(32) NOT NULL,
  `temp` varchar(32) NOT NULL,
  `date` datetime NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;

--
-- Zrzut danych tabeli `temperature`
--

INSERT INTO `temperature` (`id`, `sensor_id`, `temp`, `date`) VALUES
(1, '00000b845b2b', '38.3', '2019-12-05 20:42:06'),
(2, '00000b858c95', '-1.3', '2019-12-05 20:42:06'),
(3, '00000a035951', '24.7', '2019-12-05 20:42:06'),
(4, '00000b845b2b', '38.4', '2019-12-05 20:43:06'),
(5, '00000b858c95', '-1.2', '2019-12-05 20:43:06'),
(6, '00000a035951', '24.7', '2019-12-05 20:43:06'),
(7, '00000b845b2b', '38.4', '2019-12-05 20:44:06'),
(8, '00000b858c95', '-1.2', '2019-12-05 20:44:06'),
(9, '00000a035951', '24.7', '2019-12-05 20:44:06'),
(10, '00000b845b2b', '38.4', '2019-12-05 20:45:05'),
(11, '00000b858c95', '-1.2', '2019-12-05 20:45:05'),
(12, '00000a035951', '24.7', '2019-12-05 20:45:05'),
(13, '00000b845b2b', '38.5', '2019-12-05 20:46:06'),
(14, '00000b858c95', '-1.3', '2019-12-05 20:46:06'),
(15, '00000a035951', '24.7', '2019-12-05 20:46:06'),
(16, '00000b845b2b', '38.6', '2019-12-05 20:47:06'),
(17, '00000b858c95', '-1.3', '2019-12-05 20:47:06'),
(18, '00000a035951', '24.8', '2019-12-05 20:47:06'),
(19, '00000b845b2b', '38.7', '2019-12-05 20:48:06'),
(20, '00000b858c95', '-1.3', '2019-12-05 20:48:06'),
(21, '00000a035951', '24.9', '2019-12-05 20:48:06'),
(22, '00000b845b2b', '39.1', '2019-12-05 21:00:05'),
(23, '00000b858c95', '-1.4', '2019-12-05 21:00:05'),
(24, '00000a035951', '25.9', '2019-12-05 21:00:05'),
(25, '00000b845b2b', '37.9', '2019-12-05 22:00:06'),
(26, '00000b858c95', '-1.4', '2019-12-05 22:00:06'),

....

我希望它在过去6个月中返回6个结果，每个月返回一个。

Answer 1

我想计算最近6个月的平均温度。

您似乎想要简单的日期算术：

SELECT AVG(temp) FROM temperature WHERE date >= current_date - interval 6 month

对于今天2020-01-18，这将选择2019-07-17的记录。

或者，如果您想要当月和过去的6个月（即今天，从2019-07-01开始）：

WHERE date >= date_format(current_date, '%Y-%m-01') - interval 6 month

如果您想在过去6个月中每月记录一次，则需要按月分组，就像这样：

SELECT DATE_FORMAT(date, '%Y-%m-01') date_month, AVG(temp) avg_temp
FROM temperature
WHERE date >= date_format(current_date, '%Y-%m-01') - interval 6 month
GROUP BY DATE_FORMAT(date, '%Y-%m-01')

MYSQL最近6个月的平均温度

1 个答案: