通过最新日期获取最新结果,其中三列在mysql中具有相同的值

问题描述 投票:0回答:3

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);

fiddle相同

编辑:添加了小提琴

mysql sql database greatest-n-per-group
3个回答
1
投票

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
)

1
投票

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。


0
投票

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
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