我是sql新手,需要您的帮助。我的数据库中有测试结果数据,我想获取最新结果。
----------------------------------------------------------------------------------
| ID | TestDate | App | Service | Environment | Critical | High | Medium | Low |
|----------------------------------------------------------------------------------|
| 1 | 2020-03-01 | app | service-a | sit | 1 | 2 | 3 | 5 |
| 2 | 2020-03-02 | app | service-b | sit | 0 | 3 | 2 | 1 |
| 3 | 2020-03-03 | app | service-a | sit | 0 | 1 | 5 | 3 |
我只想获取服务a的最新结果,然后获取服务b的结果。我尝试过
SELECT MAX(TestDate), App, Service, Environment, Critical, High, Medium, Low from table
GROUP BY TestDate, App, Service, Environment, Critical, High, Medium, Low;
但是它仍然返回两个service-a值。删除GROUP BY
中的某些列将产生非聚合的列错误。
这里有一些示例数据可供您测试。
CREATE TABLE IF NOT EXISTS `test_results` (
`id` INT NOT NULL AUTO_INCREMENT,
`app` VARCHAR(10) NULL,
`environment` VARCHAR(45) NULL,
`service` VARCHAR(256) NULL,
`service_version` VARCHAR(20) NULL,
`test_date` DATETIME NOT NULL,
`critical` VARCHAR(256) NULL,
`high` VARCHAR(256) NULL,
`medium` VARCHAR(256) NULL,
`low` VARCHAR(256) NULL,
PRIMARY KEY (`id`));
INSERT INTO `test_results` (`app`, `environment`, `service`, `service_version`, `test_date`, `critical`, `high`, `medium`, `low`)
VALUES ("app1", "sit", "service-a", "1.0.0", "2020-03-01 01:03:08", 1, 2, 3, 4),
("app1", "sit", "service-a", "1.0.1", "2020-03-03 01:03:08", 5, 7, 3, 1),
("app1", "sit", "service-b", "1.1.2", "2020-03-02 01:03:08", 5, 9, 6, 4),
("app1", "sit", "service-c", "1.0.5", "2020-03-02 01:03:08", 3, 1, 9, 4);
和fiddle相同
编辑:添加了小提琴
您可以过滤而不是汇总;
select t.*
from mytable
where t.testdate = (
select max(t1.testdate) from mytable t1 where t1.service = t.service
)
以下是比使用max()
的子查询更有效的解决方案;
SELECT t.*
FROM `table` t
LEFT JOIN `table` t1 ON t1.service = t.service AND t1.testdate > t.testdate
WHERE t1.id IS NULL;
只有成功执行not的返回行才能成功加入LEFT JOIN
,确保我们只看到testdate
是该service
的最大值的结果。
此解决方案将允许您在相同的日期和时间获得多个测试结果,并且仍然采用最新的测试结果-只要您的ID在每条新记录上都递增。
SELECT
results.*
FROM `test_results` AS results
JOIN `test_results` AS latest_results
ON latest_results.id = (
SELECT id FROM `test_results`
WHERE service = results.service
ORDER BY test_date DESC, id DESC LIMIT 1)
WHERE
latest_results.id = results.id